Thermodynamics Fundamentals and Engineering Applications Reynolds, William C., Colonna TEST BANK

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PREFACE ............................................................................................................................ 4 SAMPLE EXAM 1 .............................................................................................................. 5 Exercise 1 (true/false questions) ....................................................................................... 5 Exercise 2 ......................................................................................................................... 5 Exercise 3 ......................................................................................................................... 6 Exercise 4 ......................................................................................................................... 6 Exercise 5 ......................................................................................................................... 6 Exercise 6 ......................................................................................................................... 7 SAMPLE EXAM 2 .............................................................................................................. 8 Exercise 1 (true/false questions, with justification).......................................................... 8 Exercise 2 ......................................................................................................................... 8 Exercise 3 ......................................................................................................................... 9 Exercise 4 ......................................................................................................................... 9 Exercise 5 ....................................................................................................................... 10 Exercise 6 ....................................................................................................................... 11 SAMPLE EXAM 3 ............................................................................................................ 12 Exercise 1 (true/false questions) ..................................................................................... 12 Exercise 2 ....................................................................................................................... 13 Exercise 3 ....................................................................................................................... 13 Exercise 4 ....................................................................................................................... 14 Exercise 5 ....................................................................................................................... 14 Exercise 6 ....................................................................................................................... 16 SAMPLE EXAM 4 ............................................................................................................ 16 Exercise 1 (true/false questions) ..................................................................................... 16 Exercise 2 ....................................................................................................................... 17 Exercise 3 ....................................................................................................................... 17 Exercise 4 ....................................................................................................................... 18 Exercise 5 ....................................................................................................................... 18 Exercise 6 ....................................................................................................................... 18 SAMPLE EXAM 5 ............................................................................................................ 19 Exercise 1 (true/false questions) ..................................................................................... 19 Exercise 2 ....................................................................................................................... 20 Exercise 3 ....................................................................................................................... 20 Exercise 4 ....................................................................................................................... 21 Exercise 5 ....................................................................................................................... 21 Downloaded from Scholarfriends.com

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Exercise 6 ....................................................................................................................... 21 Exercise 7 ....................................................................................................................... 21 SAMPLE EXAM 6 ............................................................................................................ 23 Exercise 1 (true/false questions) ..................................................................................... 23 Exercise 2 ....................................................................................................................... 23 Exercise 3 ....................................................................................................................... 24 Exercise 4 ....................................................................................................................... 24 Exercise 5 ....................................................................................................................... 24 Exercise 6 ....................................................................................................................... 25 Exercise 7 ....................................................................................................................... 25

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PREFACE This is a collection of exemplary exams aimed at testing the first five chapters of the book, therefore material typically covered in an undergraduate course. Each exercise is followed by its worked-out solution, in red. These exams vary in difficulty and should take three to four hours to solve, depending on the level of the student.

I would much appreciate being notified of any error that may be detected by instructors and their students, as well as of any suggestion for improvements.

P. Colonna

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SAMPLE EXAM 1 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) For a polytropic ideal gas, the specific internal energy difference between two states over an isobaric process (𝑃 = const. ) is given by Δ𝑢 = 𝑐𝑃 (𝑇2 − 𝑇1 ).

2) A fluid can be condensed by isobaric cooling starting from a state defined by a pressure that is greater than the critical pressure.

3) The specific enthalpy at the outlet of a compressor is higher than that at its inlet.

4) The internal energy of an expanding gas contained in a non-adiabatic cylinder-piston always decreases.

5) The rotor of an adiabatic turbine transfers entropy to the working fluid.

6) The 𝑐𝑃 of a fluid depends only on temperature.

7) The accumulation term in an accounting equation is positive if the final value of the accounted variable is greater that the initial value.

8) A gas turbine engine can be more efficient than a Carnot engine if these engines do not operate between the same minimum and maximum temperatures.

9) A fluid kept at constant specific volume within a perfectly tight piston-cylinder system cannot transfer energy as work.

10) The internal energy of a macroscopic system is the sum of all types of microscopic energy modes and of macroscopic potential and kinetic energy.

11) Boiling a fluid in a closed rigid container increases its entropy.

12) The isentropic efficiency of a real compressor can reach 100% if the compressor is designed very well.

EXERCISE 2 In the US, the efficiency of a power plant is often expressed in terms of “heat rate”. The heat rate is the amount of energy as heat input to a power plant to generate one kilowatt hour (kWh) of electricity, that is heat rate ≡

Thermal Energy in BTU Electrical Energy out in kW h Downloaded from Scholarfriends.com

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Determine the thermal efficiency of a power plant defined as 𝜂≡

𝑊̇output , 𝑄̇input

if its heat rate is 11000 BTU/kWh. Use the unit equivalent 1 BTU = 1055 J and show all the steps to obtain it.

EXERCISE 3 A certain type of actuator consists of a closed piston-cylinder system in which the force on the piston due to the pressurized gas inside the cylinder is controlled by heating or cooling.

The piston starts from state 1 and moves to another state. The gas obeys the ideal gas law.

During the actuation, the heater is controlled such that the pressure varies with the position of the piston according to the prescribed function 𝑃(𝑥) = 𝑃1 𝑉1 (𝜋𝑟𝑥)−1 Derive the expression for the energy transfer as heat 𝑄(𝑥) for the process in terms of piston position 𝑥 , initial pressure 𝑃1 , initial position 𝑥1 , and internal radius of the cylinder 𝑟.

EXERCISE 4 The volume of a piston-cylinder vessel containing 100 kg of CO2 is kept fixed at 1 m3, while the container is cooled from 𝑇 = 354 K until it reaches saturated vapor conditions.

After that, the fluid is cooled at constant pressure until it reaches the saturated liquid condition. Using the provided 𝑃 − ℎ thermodynamic chart for carbon dioxide, estimate the energy transfer as heat and indicate the process by drawing a line on the 𝑃 − ℎ chart.

EXERCISE 5 A refrigeration system works according to the inverse Rankine cycle (or so-called heat pump cycle) using CO2 as the working fluid. The fluid undergoes the following processes:    

pressurization in a compressor; cooling by ambient air in a heat exchanger until it reaches saturated liquid conditions, expansion to a liquid-vapor state at low pressure and temperature through a throttling valve until it reaches a given pressure, and, from this liquid-vapor state, fluid receives energy as heat from the environment to be cooled until it vaporizes completely thus reaching the compressor inlet state.

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The diagram of the system is

The system operates at steady state and the pressure drop in the heat exchangers can be neglected.

The working fluid enters the compressor in saturated vapor conditions at a pressure of 15 bar and leaves the compressor at a pressure of 50 bar and temperature of 340 K. Use the tables and the 𝑃 − ℎ chart provided in the appendix to evaluate the needed thermodynamic properties.

Hint: there is no need to perform interpolation for the property value of state 1, as the values of the two closest states in the table are almost identical. a) For each component (Compressor, Condenser, Valve, Evaporator) make a separate sketch to support the derivation of its energy balance; b) write the energy balance for each component; c) calculate the coefficient of performance of the system defined as COP ≡

𝑄̇evaporator 𝑊̇compressor

d) Determine the mass flow rate of CO2 that allows to obtain a cooling power of 20 kW.

EXERCISE 6 Gas enters a device operating at steady state at a mass flow rate of 𝑀̇ and at temperature 𝑇1 and pressure 𝑃1 , and exits the device at 𝑇2 and 𝑃2 . Both the pressure and the temperature at the inlet are higher than at the outlet. The device converts the energy of the gas into shaft power 𝑊̇ and is cooled by heat transfer at the rate 𝑄0̇ to the environment at 𝑇0 . The Downloaded from Scholarfriends.com

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properties of the gas at the conditions inside the device can be calculated assuming that it is an ideal gas with constant specific heat. a) Make a sketch of the system to support the derivation of the energy balance; b) make a sketch of the system to support the derivation of the entropy balance; c) derive an expression for the maximum shaft power 𝑊̇ that can be obtained from the 𝑃 𝑇 system in terms of 𝑀̇ , 𝑇0 , 𝑇1 , pressure ratio 𝛽𝑃 = 𝑃1 , temperature ratio 𝛽𝑇 = 𝑇1 and gas 2

2

constants.

SAMPLE EXAM 2 EXERCISE 1 (TRUE/FALSE QUESTIONS, WITH JUSTIFICATION) Provide a theoretical explanation of why the answer to the question is true or false, with mathematical relations if needed.

1) Temperature is an intensive property 2) A gas is compressed in a cylinder. Energy is transferred as heat from the gas to the surroundings of the cylinder. The internal energy accumulation of the gas is always lower than zero.

3) The state of a simple compressible substance is fixed if at least three independent thermodynamic variables are specified.

4) The kinetic energy of the gases expelled by the nozzle of a rocket is the same in a reference frame moving with the rocket and in the one fixed to the earth, considered as inertial.

5) The temperature in the inside of a fridge decreases, thus the entropy of the substances within it also decreases. The global production of entropy associated with this process is negative.

EXERCISE 2 When considering the motion of a fluid on a molecular level, the state of the fluid can no longer be described by the Navier-Stokes equations. By integrating the phase space of possible states of the system, both over velocity and location, one can find mean properties.

One of these properties is the shear stress 𝜏. The solution of the Couette-Poiseuille flow problem with a heated wall provides the expression for shear stress as 𝜏=

𝜌𝒱 √𝜋𝛽𝑤

where 𝜌 is the number density (number of moles per volume), 𝒱 is the bulk velocity of the fluid and 𝛽𝑤 some thermodynamic parameter. a) Find the unit of the parameter 𝛽𝑤 . Express it only in SI units of primary quantities. For m example, Newton should be expressed kg ∙ s2.

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b) Find the value of parameter 𝛽𝑤 using the data 𝒱 = 1.86

miles , h

𝜌 = 1.157

mol , ft 3

𝜏 = 20 Pa, 1 km = 0.621 miles, 1 m = 3.28 ft.

EXERCISE 3 An ideal gas is heated in a piston-cylinder system. The friction between the piston and the cylinder can be neglected. The piston is initially locked. The energy is transferred as heat in such a way that the temperature of the gas remains constant. The system has a circular cross-section with radius 𝑟 and the surrounding environment is at sea-level atmospheric conditions. The pressure in the cylinder is much higher than that of the environment. At a certain time instant the lock is removed and the piston accelerates.

Obtain an expression for the velocity of the piston head as a function of the piston head mass 𝑀p , the gas constant 𝑅, the temperature of the gas, the radius of the piston 𝑟, the mass of the gas in the cylinder, the atmospheric pressure, the change in piston head position Δ𝑥 and the initial position of the piston head 𝑥1 .

EXERCISE 4 To boil pasta for dinner, we introduce 1.5 litres of saturated liquid water in a special pot with a massless lid that can move freely (no friction) and is perfectly tight. This way saturated conditions can be maintained. The environment is at sea level.

In order to transfer energy as heat to it, we place the pot on top of a stove which has the following power settings A: 1300 W B: 1400 W C: 1500 W At the same time, energy as heat is transferred from the pot to the kitchen (surroundings) at a rate of 100 W. The cooking time is 8 minutes. To ensure that all the pasta is submerged in the water, it is necessary that the volume of water is always at least 1.2 litres. The effect of the pasta in the water can be neglected in this simple analysis. a) What is the power setting providing the highest rate of energy transfer heat that can be used?

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b) Starting from the saturated liquid state, how much time would it take for the water to completely evaporate if the stove is kept at the power setting calculated as the answer to question a)?

EXERCISE 5 A combined cycle power plant is a system formed by two thermal engines, which is commonly used for electricity generation. The operation of the two thermal engines is based on two different thermodynamic cycles. The working principle of the combined cycle is that the high temperature cycle discharges thermal energy as heat to the low temperature cycle.

Consider a concentrated solar power combined cycle system whose simplified system diagram is reported in the figure below. The high-temperature cycle is called Brayton cycle, while the low temperature cycle is called Rankine cycle.

The working fluid of the high-temperature cycle is air and that of the low temperature cycle is water.

Specifications:         

Environment: 𝑃1 = 1 bar, 𝑇1 = 25 ℃; Air turbine inlet: 𝑃3 = 17 bar, 𝑇3 = 1233 ℃; Air turbine isentropic efficiency: 𝜂34s = 0.85; Air compressor efficiency: 𝜂12s = 0.80; Energy transfer as heat to the air Brayton cycle: 𝑄̇in = 𝑄̇23 = 96.44 MW Air temperature at the stack: 𝑇5 = 120 ℃; Steam turbine inlet (superheated): 𝑃6 = 20 bar, 𝑇6 = 477 ℃; Steam turbine isentropic efficiency: 𝜂67s = 0.82; Pressure in the steam condenser: 𝑃7 = 𝑃8 = 0.2 bar;

The power consumption of the feed-water pump can be neglected: state 8 = state 9.

The thermodynamic state of the pump inlet/condenser outlet is saturated liquid water.

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a) In addition to the usual sketches of the system which are part of the 8-step method for energy analysis, draw a qualitative sketch of the processes forming the two cycles in the two 𝑇𝑠 charts below, one of air, and one of water.

b) Calculate the thermal efficiency 𝜂 ≡ 𝑊̇net ⁄𝑄̇in of the combined cycle. Calculate also the thermal efficiency of the two cycles individually.

Assume the following:     

Air can be considered a polytropic ideal gas with constant isobaric specific heat, kJ 𝑐𝑃 = 1.0 kg∙K. The heat exchangers are adiabatic (only energy transfer as heat between fluids and not with the surroundings).

No state change of the working fluids within the ducts connecting the components.

Pressure losses in all heat exchangers can be neglected.

The power consumption of the feed-water pump can be neglected: state 8 = state 9.

The pump inlet/condenser outlet states are saturated liquid water.

c) Compute the second law efficiency based on the maximum and minimum cycle temperature of the combined cycle and those of the two cycles and discuss these results.

EXERCISE 6 A tank is separated into two compartments by an adiabatic and frictionless piston. One compartment contains 0.2 m3 of air and the other compartment contains 0.1 kg of oxygen, both initially at 27 ℃ and 90 kPa. The compartment of the tank containing oxygen is insulated. The compartment of the tank containing air is slowly supplied with energy as heat through a wall whose temperature is controlled and kept constant at 300 ℃. During Downloaded from Scholarfriends.com

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the process the two compartments are in mechanical equilibrium. The energy transfer as heat stops when the pressure of the air rises to 150 kPa. The molar mass of oxygen is 32 g/mol and its specific heat at constant pressure is 940 J/(kg ∙ K). Assume that the ideal gas assumption holds for both gases. a) Determine the temperature of the air and of the oxygen at the end of the heating process. b) Calculate the energy transferred as heat into the system and the entropy production during the process. c) Suppose you could alter the temperature of the wall where the heat is transferred. What would be the optimal temperature to minimize entropy production?

SAMPLE EXAM 3 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) While heating a saturated liquid at constant volume, its pressure and temperature remain constant.

2) A thermal energy reservoir (TER) can supply or absorb a finite amount of energy as heat without undergoing any change in temperature.

3) An isolated system consists of two volumes made of different materials with different initial temperatures, and in contact with one another. After some time, the system has reached an equilibrium state. Both materials have the same final temperature and the same final specific internal energy.

4) A thermodynamic engine can function without discharging energy as heat.

5) The coefficient of performance of a system can be obtained by dividing the useful energy output by the total energy input.

6) Any intensive property of matter depends on the considered mass.

7) The specifications sheet of a car engine shows that its net efficiency is 45% at an ambient temperature of 23 ℃. Your friend Jeremy tells you that the peak temperature of the gases in the cylinders is 230 ℃. Is this true or false?

8) The internal energy of a gas is related to the energy associated with the microscopically disorganized motion of its molecules.

9) Energy is transferred as heat to a saturated fluid mixture that is stored in a rigid container. The quality of the initial state is 0.2, and it is 0.8 in the final state. The pressure Downloaded from Scholarfriends.com

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and temperature of the fluid do not change during this process.

10) The entropy change of any fluid can be evaluated analytically if only the specific heats at the initial and final states are given.

11) An isolated control volume containing an ideal gas undergoes an isothermal process.

No internal energy gets accumulated within the control volume.

12) An adiabatic mixer merges two streams of a certain fluid entering into the mixer at the different temperatures. The entropy of the outlet stream is equal to the sum of the entropy at the inlets.

EXERCISE 2 Non-dimensional quantities play an important role in energy systems modeling and design.

An example of a dimensionless quantity used for pumps is the pump specific speed 𝑁𝑠 , and its definition is 𝑁𝑠 =

𝑁√𝑄 (𝑔𝐻)3/4

Where    

𝑁 is the rotational speed of the pump, in rad/s, 𝑄 is the volume flow rate, in m3 / s, 𝑔 is the gravitational acceleration, in m/s2 , 𝐻 is the head of the pump, in m.

a) Prove that the specific speed is a dimensionless quantity. b) Calculate the specific speed of the liquid oxygen (LOX) pump used in the SpaceX Raptor Engine, given the following information:    

𝑔 = 9.81 m/s 2 , 𝑄 = 346 000 gal/hr, 𝑁 = 14 000 rpm, 𝐻 = 7 250 ft.

Use the following unit conversions:  1 ft = 0.3048 m  1 gal = 0.00379 m3 EXERCISE 3 An ideal gas is heated in a piston-cylinder system where the piston is able to move freely.

The system has a circular cross-section with radius 𝑟 and the surrounding environment is at sea-level atmospheric conditions. The mass of the piston can be considered negligible.

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energy transferred as heat, the pressure, the isochoric specific heat 𝑐𝑣 , the gas constant 𝑅, and the radius 𝑟.

EXERCISE 4 A constant-volume vessel of 20 liters holds 2 kg of oxygen in thermodynamic equilibrium at an initial temperature of 102 K. The vessel is heated until there is only saturated vapor oxygen. a) Determine the temperature and pressure once the oxygen has reached its final equilibrium state. b) Determine the amount of energy transferred as heat for this process c) Sketch the process on the provided 𝑃 − ℎ diagram for oxygen EXERCISE 5 Consider the following simple Rankine cycle system operating with water as the working fluid, Pressure losses in pipes, ducts and heat exchangers are negligible. The turbine and pump isentropic efficiencies are reported in the table. Pressure losses are negligible everywhere in the system.

Turbine isentropic efficiency

0.6

Pump isentropic efficiency

0.8

The following tables list the information needed to perform the computation of the cycle thermodynamic states at the inter-component interfaces for two different cases.

Case A ID

state

𝑻 / ℃ 𝑷 / 𝐛𝐚𝐫

𝒙

1

Pump inlet

49.3

0

2

Pump outlet

3

Turbine inlet

4

Turbine outlet

4 150

𝒔 / 𝐤𝐉 /(𝐤𝐠 ∙ 𝐊)

𝒉 / 𝐤𝐉/𝐤𝐠

0.6966

207.412

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ID

state

𝑻 / ℃ 𝑷 / 𝐛𝐚𝐫

𝒙

1

Pump inlet

151.85

0

2

Pump outlet

3

Turbine inlet

4

Turbine outlet

𝒔 / 𝐤𝐉 /(𝐤𝐠 ∙ 𝐊)

𝒉 / 𝐤𝐉/𝐤𝐠

6.348

2547.35

140 376.85

140

NOTE: NOT ALL THE CELLS IN THE TABLES NEED TO BE FILLED IN TO SOLVE THE PROBLEM. The two thermodynamic cycles in the temperature-entropy diagram of water are

Hint: turbine thermodynamic states The turbine isentropic efficiency is ∆ℎ ∆ℎ𝑠 Where ∆ℎ is the turbine specific work, and ∆ℎ𝑠 is the turbine isentropic specific work. 𝜂t,𝑠 =

Hint: second law efficiency The second law efficiency is 𝜂II ≡

𝜂th 𝑇̅ 1 − ̅C 𝑇H

Where 𝜂th is the cycle thermal efficiency. 𝑇̅C is the so-called equivalent thermodynamic temperature for thermal energy rejection (condenser). 𝑇̅H is the equivalent thermodynamic temperature for thermal energy transfer to the steam (boiler).

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They are computed as 𝑇̅C =

ℎ4 − ℎ1 ℎ3 − ℎ2 and 𝑇̅H = . 𝑠4 − 𝑠1 𝑠3 − 𝑠2

In order to grasp the need for introducing these temperatures, look at the shape of the cycles in the temperature-entropy diagrams. Do they look similar to a Carnot cycle? a) Calculate the cycle thermal efficiency for both case A and case B. Discuss the difference between these values. b) Calculate the cycle second law efficiency for both case A and case B. EXERCISE 6

A metal wall is exposed to temperature 𝑇1 at one side and to temperature 𝑇2 at the other side. Demonstrate which of the two temperatures must be higher for the energy transfer as heat to occur in the direction indicated in the figure (i.e., from left to right). Assume that the process is steady-state, that the wall is infinitely long, and that no other energy transfer occurs.

SAMPLE EXAM 4 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) A reversible process is isentropic.

2) A person in a raising elevator lifting a box over a distance of half meter will do more work than a person on the ground lifting the same box over the same distance.

3) The terms of the equation for the steady-state energy balance of an open system are the same as those for a closed system.

4) The efficiency of a wind turbine cannot be larger than the Carnot cycle efficiency.

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5) The critical point can be evaluated using the ideal gas model.

6) An explosion can be regarded as a sudden entropy production.

7) When mixing adiabatically hot and cold gases to a new equilibrium state, the entropy increases.

8) In a polytropic process the pressure can change independently of the volume.

9) A system formed by a simple compressible substance enclosed in a constant volume can transfer energy as work and heat 10) Removing energy as heat from a closed system reduces its entropy.

11) At ambient room conditions, methane is liquid.

12) Energy transfer as heat under a finite temperature difference is reversible.

13) Cooling a gas in closed volume at constant pressure increases its volume.

14) An ideal valve transfers more energy as work than a non-ideal valve.

EXERCISE 2 (Same as Exercise 1.1 of the book) An 80 kg student wishes to walk on snow without sinking and decides to make a pair of snow shoes. The student determined experimentally that the maximum pressure the snow can withstand is 0.6 kPa. What is the minimum surface required by one snow shoe?

Would the student be able to sell snow shoes to his fellow classmates? Explain your answer.

EXERCISE 3 The temperature of 10 kg of air inside a piston-cylinder is increased from 25 ℃ to 80 ℃ using an electrical heater, while the pressure inside the cylinder is maintained constant.

During this process 50 kJ of energy is transferred as heat from the system to the environment.

Find the required amount of energy that must be added to the system by the heater, for cylinder pressures of 1, 2 and 5 bar.

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EXERCISE 4 A constant volume vessel of 100 liters holds 1 kg of water at an initial pressure of 30 bar and initial temperature 𝑇1 . The vessel is then cooled to 350 K. a) Determine the initial temperature of the system 𝑇1 and the pressure after cooling. b) Determine the energy transferred as heat during the process. Make a sketch of the process on the provided 𝑃 − ℎ chart.

EXERCISE 5 A small solar engine is used for power conversion. Water enters the pump as saturated liquid at 50 ℃ and is pumped up to 2 bar by a small centrifugal pump (process 1-2). Solar energy is gathered using solar collectors, connected to the boiler. The boiler evaporates the water at 2 bar (process 2-3), and saturated vapor at this pressure enters a small turbine (process 3-4). The steam leaves the turbine at 50 ℃ and with a liquid fraction of 6 %, and is subsequently condensed (process 4-1). The mass flow rate is 140 kg/hr and the pump is driven by a 0.4 kW motor.

a) Calculate the net mechanical power output of this solar engine. b) Compute the energy conversion efficiency taken as the net shaft work output divided by the energy transfer as heat to the fluid in the boiler. c) Estimate the area of the solar collectors, assuming the collectors can absorb 800 W/m2 .

EXERCISE 6 Consider the two closed adiabatic boxes illustrated in the following sketch.

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W

W

The first box accommodates a frictionless shaft transferring mechanical work to a spinning propeller. The second box accommodates a frictionless shaft transferring mechanical work to lift a weight.

Assume that the same amount of energy as work is provided to the two shafts for a fixed amount of time starting at 𝑡s and ending at 𝑡e . a) Describe in conceptual terms (in words) the changes occurring in the boxes during the transfer of energy as work to the shafts (𝑡s < 𝑡 < 𝑡e ). b) Describe in conceptual terms (words) the changes occurring in the boxes once the transfer of energy as work to the shafts stops (t > te).

SAMPLE EXAM 5 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) Volume is an intensive property.

2) A simple Brayton cycle engine has a higher efficiency than the Carnot engine.

3) When compressing a gas adiabatically its internal energy increases.

4) Color is a thermodynamic property.

5) A closed system allows mass exchange with the surroundings.

6) The production of entropy in the universe is zero.

7) The production of energy in the universe is zero.

8) Temperature has the same dimensional units as energy.

9) The temperature is proportional to volume in an isobaric process of an ideal gas.

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10) The value of the internal energy of an ideal gas is always larger than the value of its enthalpy at a given temperature.

11) Extracting energy as heat from a closed system reduces its entropy.

12) Energy transfer as heat is a reversible process.

13) A refrigerator is a heat pump.

14) In a polytropic process the pressure can change independently of the volume.

EXERCISE 2 (Similar to exercise 1.2 of the book) A wind farm of 30 identical windmills is converting the kinetic energy of the wind into 50 MW of electrical power. The velocity of the wind blowing through the rotor disc of each windmill is 𝑉 = 7 m/s. The air density is 𝜌 = 1.2 kg/m3 . If the efficiency of each windmill is 0.63, what is the windmills rotor radius 𝑅?

EXERCISE 3 (Similar to exercise 4.13 of the book) The throttling calorimeter is a device for determining the thermodynamic state of a liquid-vapour mixture by means of temperature and pressure measurements. The procedure is to deviate a small amount of the mixture from the main flow, pass it through a valve, and take the measurements as shown in the figure below.

Assume that the fluid flowing through the calorimeter is water.

a) Make a sketch of the process between state 1 and 2 on a 𝑇 − 𝑠 diagram, indicating the state points and the line corresponding to the throttling process. b) Compute the quality of the mixture at state 1, given the pressure and temperature measurements displayed in the figure below.

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EXERCISE 4 13 kg of ammonia at its critical state is placed in a constant volume container, which is then cooled until the temperature is 310 K. a) How much energy was transferred as heat from the ammonia in this process? b) How much of the tank volume is occupied by liquid at the end of the process? c) Sketch the process on the provided 𝑃 − ℎ diagram for ammonia.

EXERCISE 5 Consider the intake stroke of an internal combustion engine under the following idealizations.     

Negligible kinetic and potential energy everywhere Uniform state inside the cylinder at any instant One-dimensional flow Ideal gas Non-adiabatic cylinder walls

a) Make a sketch of the system, including i) its boundary, ii) all the quantities involved in the mass and energy balance, and arrows indicating the positive direction for the energy transfer terms. b) Write the mass balance on the control volume. c) Write the energy balance on the control volume in terms of temperature 𝑇, mass 𝑀, volume 𝑉, energy transfer rate as heat 𝑄̇ and gas constants 𝑐𝑃 and 𝑐𝑣 .

EXERCISE 6 Ten grams of computer chips with a specific heat 𝑐𝑃 = 𝑐𝑣 = 0.3 kJ/(kg ∙ K) are initially at 20 ℃ . These chips are placed in an adiabatic container with 5 grams of saturated liquid R134a at −40 ℃ in order to cool them down. The fluid remains saturated over the process.

Assume that during the cooling process the pressure remains constant. a) Calculate the entropy change of the chips; b) Calculate the entropy change of the R134a; c) Compute the entropy production due to this process; d) state whether this process is possible or not.

EXERCISE 7 The majority of current aircraft engines employ gas turbines based on the open Brayton thermodynamic cycle. Hence, being able to analyze the performance of this type of thermal engines is important for the aspiring aerospace engineer, and it is easier to start from the

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analysis of the ideal Brayton thermodynamic cycle. The sketch of the process flow diagram and of the temperature-entropy diagram for the ideal Brayton cycle are

The corresponding processes are as follows: 1-2: Isentropic compression 2-3: Isobaric transfer of energy as heat 3-4: Isentropic expansion 4-1: Isobaric transfer of energy as heat Assume that air is the working fluid and that it obeys the ideal gas law with constant specific heats (polytropic ideal gas).

Derive the efficiency of an engine operating according to the ideal Brayton cycle, as a 𝑐𝑝 𝑃 function of the cycle pressure ratio Π ≡ 𝑃12 and the ratio of the specific heats 𝛾 ≡ 𝑐 . 𝑣

Hint: for a polytropic ideal gas 𝛾−1

𝑇1 𝑃2 𝛾 =( ) 𝑇2 𝑃1

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SAMPLE EXAM 6 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) A modern petrol engine cannot be more efficient than a Carnot engine.

2) The value of the internal energy of an ideal gas is always lower than the value of the enthalpy at a given temperature.

5) Temperature is an intensive property.

4) The specific heat of air does not vary with the temperature.

5) In an adiabatic process energy as heat is transferred from high to low temperature.

6) A diesel engine with high compression ratio can be more efficient than a Carnot engine.

7) An explosion can be regarded as a sudden entropy production.

8) By definition 𝑐𝑝 is always smaller than 𝑐𝑣 .

9) The production of entropy in the universe is zero.

10) In a polytropic process the pressure can change independently of the volume.

11) Removing energy as heat from a closed system reduces its entropy.

12) A system formed by a simple compressible substance enclosed in a constant volume cannot transfer energy as work.

13) In a gas turbine engine the fluid enthalpy at the outlet of the turbine is larger than at the inlet.

EXERCISE 2 What is the kinetic energy in MJ of an aircraft weighing 150 000 lbm (pound mass) and flying at 880 km/h?

For the conversion of units, use the method of multiplying by suitable units, e.g., 1 kg 1= 2.205 lbm TESTBANK

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EXERCISE 3 Consider the isochoric (constant-volume) compression of a gas from pressure 𝑃1 to 𝑃2 in a piston-cylinder system. The increase of pressure is achieved by controlled energy transfer as heat 𝑄 to the gas, while the piston is kept fixed. a) Make a sketch of the considered system with its control volume, the appropriate nomenclature for all quantities involved in the energy balance, and arrows indicating the positive direction of energy transfers. b) Treating the gas as polytropic and ideal, derive the expression for the energy input as heat 𝑄 as a function of the pressure ratio 𝑃2 /𝑃1 .

EXERCISE 4 5 kg of CO2 at its critical state is placed in a perfectly rigid container, which is then cooled until the temperature is 260 K. a) What is the volume occupied by liquid at the end of the process? b) How much energy is transferred as heat during the process?

EXERCISE 5 Consider the simple steam Rankine cycle system operating at steady state, whose process flow diagram is depicted in the figure with the corresponding processes in the T-s diagram.

Water is compressed in a pump (1-2, single point in the 𝑇 − 𝑠 diagram as the temperature and entropy increase are negligible), receives energy as heat, which first increases its temperature and then evaporates it and superheats it (2-3), steam is expanded in a turbine (3-4) and then condensed (4-1). Assume that the flow is one-dimensional at all intercomponent thermodynamic states, that the kinetic energy and potential energy of the fluid at the state points is negligible. Assume in addition that the pump and the turbine are adiabatic.

The following assumptions are valid:  

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The work of the pump is negligible (compressing liquid water does not require much work compared to the energy converted into mechanical work by the turbine) The enthalpy of the water at the outlet of the pump is 138 kJ/kg.

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  

The evaporation pressure is 80 bar, the condensation temperature is 30 oC The steam at the inlet of the turbine at 400 oC The isentropic efficiency of the turbine is 0.85

For each of the 4 components, 1) Make a sketch of the pertinent control volume, with the appropriate nomenclature for all quantities involved in the energy balance and arrows indicating the positive direction of energy transfers and derive the steady-state energy balance per unit mass flow of working fluid; 2) Calculate the efficiency of the system.

EXERCISE 6 Consider a turbine operating at steady state. The working fluid enters the control volume in state 1 and leaves it in state 2. The boundary of the control volume that separates the turbine from a large environment is at temperature 𝑇0 . The turbine reversibly rejects energy as heat to the environment. Assume that the kinetic and potential energy of the entering and leaving flow can be neglected. a) Derive the expression for the mechanical power output as a function of thermodynamic properties of the inlet and outlet states, of the temperature of the environment and of the rate of entropy production within the process. b) Derive the expression for the maximum mechanical power of the turbine.

EXERCISE 7 Consider a two-part isolated system whereby the total volume of the system and of the parts are constant.

Demonstrate that if the two parts are approaching thermal equilibrium, the energy transfer as heat between the two parts must be from the part at higher temperature to the part at lower temperature. The relation between the thermodynamic temperature and the entropy is 𝜕𝑢 𝑇=( ) 𝜕𝑠 𝑣

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TEST BANK accompanying the book THERMODYNAMICS - FUNDAMENTALS AND ENGINEERING APPLICATIONS P. Colonna Delft University of Technology, The Netherlands

October 16TH, 2019

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PREFACE ............................................................................................................................ 4 SAMPLE EXAM 1 .............................................................................................................. 5 Exercise 1 (true/false questions) ....................................................................................... 5 Exercise 2 ......................................................................................................................... 6 Exercise 3 ......................................................................................................................... 6 Exercise 4 ......................................................................................................................... 7 Exercise 5 ....................................................................................................................... 10 Exercise 6 ....................................................................................................................... 13 SAMPLE EXAM 2 ............................................................................................................ 16 Exercise 1 (true/false questions, with justification)........................................................ 16 Exercise 2 ....................................................................................................................... 17 Exercise 3 ....................................................................................................................... 17 Exercise 4 ....................................................................................................................... 19 Exercise 5 ....................................................................................................................... 22 Exercise 6 ....................................................................................................................... 28 SAMPLE EXAM 3 ............................................................................................................ 32 Exercise 1 (true/false questions) ..................................................................................... 32 Exercise 2 ....................................................................................................................... 33 Exercise 3 ....................................................................................................................... 33 Exercise 4 ....................................................................................................................... 35 Exercise 5 ....................................................................................................................... 38 Exercise 6 ....................................................................................................................... 45 SAMPLE EXAM 4 ............................................................................................................ 48 Exercise 1 (true/false questions) ..................................................................................... 48 Exercise 2 ....................................................................................................................... 48 Exercise 3 ....................................................................................................................... 50 Exercise 4 ....................................................................................................................... 52 Exercise 5 ....................................................................................................................... 54 Exercise 6 ....................................................................................................................... 57 SAMPLE EXAM 5 ............................................................................................................ 59 Exercise 1 (true/false questions) ..................................................................................... 59 Exercise 2 ....................................................................................................................... 59 Exercise 3 ....................................................................................................................... 61 Exercise 4 ....................................................................................................................... 64 Exercise 5 ....................................................................................................................... 66

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Exercise 6 ....................................................................................................................... 67 Exercise 7 ....................................................................................................................... 70 SAMPLE EXAM 6 ............................................................................................................ 72 Exercise 1 (true/false questions) ..................................................................................... 72 Exercise 2 ....................................................................................................................... 72 Exercise 3 ....................................................................................................................... 73 Exercise 4 ....................................................................................................................... 73 Exercise 5 ....................................................................................................................... 75 Exercise 6 ....................................................................................................................... 77 Exercise 7 ....................................................................................................................... 78

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PREFACE This is a collection of exemplary exams aimed at testing the first five chapters of the book, therefore material typically covered in an undergraduate course. Each exercise is followed by its worked-out solution, in red. These exams vary in difficulty and should take three to four hours to solve, depending on the level of the student.

I would much appreciate being notified of any error that may be detected by instructors and their students, as well as of any suggestion for improvements.

P. Colonna

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SAMPLE EXAM 1 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) For a polytropic ideal gas, the specific internal energy difference between two states over an isobaric process (𝑃 = const. ) is given by Δ𝑢 = 𝑐𝑃 (𝑇2 − 𝑇1 ).

FALSE 𝑅Δ𝑇 Δ𝑢 = Δℎ − 𝑃Δ𝑣 = 𝑐𝑃 Δ𝑇 − 𝑃 ( ) = (𝑐𝑃 − R)Δ𝑇 = 𝑐𝑣 Δ𝑇 𝑃 2) A fluid can be condensed by isobaric cooling starting from a state defined by a pressure that is greater than the critical pressure.

FALSE 3) The specific enthalpy at the outlet of a compressor is higher than that at its inlet.

TRUE 4) The internal energy of an expanding gas contained in a non-adiabatic cylinder-piston always decreases.

FALSE Δ𝑢 = ±𝑄 − 𝑊 5) The rotor of an adiabatic turbine transfers entropy to the working fluid.

FALSE 6) The 𝑐𝑃 of a fluid depends only on temperature.

FALSE (only for an ideal gas) 7) The accumulation term in an accounting equation is positive if the final value of the accounted variable is greater that the initial value.

TRUE (Accumulation = Final) Initial + Production) 8) A gas turbine engine can be more efficient than a Carnot engine if these engines do not operate between the same minimum and maximum temperatures.

TRUE 9) A fluid kept at constant specific volume within a perfectly tight piston-cylinder system cannot transfer energy as work.

TRUE 10) The internal energy of a macroscopic system is the sum of all types of microscopic energy modes and of macroscopic potential and kinetic energy.

FALSE (Only microscopic types of energy) 11) Boiling a fluid in a closed rigid container increases its entropy.

TRUE 12) The isentropic efficiency of a real compressor can reach 100% if the compressor is designed very well.

FALSE

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EXERCISE 2 In the US, the efficiency of a power plant is often expressed in terms of “heat rate”. The heat rate is the amount of energy as heat input to a power plant to generate one kilowatt hour (kWh) of electricity, that is heat rate ≡

Thermal Energy in BTU Electrical Energy out in kW h

Determine the thermal efficiency of a power plant defined as 𝜂≡

𝑊̇output , 𝑄̇input

if its heat rate is 11000 BTU/kWh. Use the unit equivalent 1 BTU = 1055 J and show all the steps to obtain it. 𝑊̇output 𝑊output 1 kWh 𝜂≡ = = = 𝑄input 11000 BTU 𝑄̇input

J 1000 s 3600 s 1055 J 11000 BTU ∙ 1 BTU

= 31 %

EXERCISE 3 A certain type of actuator consists of a closed piston-cylinder system in which the force on the piston due to the pressurized gas inside the cylinder is controlled by heating or cooling.

The piston starts from state 1 and moves to another state. The gas obeys the ideal gas law.

During the actuation, the heater is controlled such that the pressure varies with the position of the piston according to the prescribed function 𝑃(𝑥) = 𝑃1 𝑉1 (𝜋𝑟𝑥)−1 Derive the expression for the energy transfer as heat 𝑄(𝑥) for the process in terms of piston position 𝑥 , initial pressure 𝑃1 , initial position 𝑥1 , and internal radius of the cylinder 𝑟.

1) Draw a diagram identifying the system to be analyzed by enclosing it within dashed lines.

Q U W x 2) Show the forces, acting on the system, that you will use in the evaluation of energy transfer as work.

The only force results from the pressure of the gas on the cylinder 3) Identify the reference frame for evaluations of macroscopic kinetic and potential energy and work.

See sketch.

4) List all simplifying idealizations 6

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  

Kinetic and potential energy are negligible compared to other energy transfers; The properties of the gas can be modeled with the ideal gas law; Friction can be neglected.

5) Show all significant energy transfers on your sketch to define the symbols, using arrows to indicate the directions chosen for positive energy transfer, and show the significant changes in energy storage See sketch.

6) State whether the energy balance is over some definite time interval or on a rate basis.

The energy balance is over the time interval from state 1 to a generic state in which the position of the piston is 𝑥.

7) Write the energy balance; there should be a one-to-one correspondence between the terms in your equation and those on your diagram. ∆𝑈 = 𝑄 − 𝑊 8) Bring in process, state, or other information as necessary to complete the analysis. 𝑃(𝑥) = 𝑃1 𝑉1 (𝜋𝑟𝑥)−1 The work of the gas on the piston can be expressed as 2

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𝑊(𝑥) = ∫ 𝑃𝑑𝑉 = 𝑃1 𝑉1 ∫ (𝜋𝑟𝑥)−1 𝑑𝑉 1

𝑥

1

2

𝑟 𝑑𝑥 = 𝑃1 𝜋𝑟 3 𝑥1 [ln(𝑥) − ln(𝑥1 )] 𝑥 1

= 𝑃1 𝑉1 ∫ (𝜋𝑟𝑥)−1 (𝜋𝑟 2 )𝑑𝑥 = 𝑃1 𝑉1 ∫ 𝑥1

= 𝑃1 𝜋𝑟 3 𝑥1 ln

𝑥 𝑥1

The internal energy is 𝑐𝑣 𝑐𝑣 𝑐𝑣 ∆𝑈 = 𝑀 𝑐𝑣 ∆𝑇 = (𝑃2 𝑉2 − 𝑃1 𝑉1 ) = [𝑃1 𝑉1 (𝜋𝑟𝑥)−1 𝑉2 − 𝑃1 𝑉1 ] = 𝑃1 𝑉1 (𝑟 − 1) 𝑅 𝑅 𝑅 𝑐𝑣 2 (𝑟 = 𝑃1 𝜋𝑟 𝑥1 − 1) 𝑅 The expression for the energy transfer as heat that causes the prescribed pressure variation as a function of the piston position is therefore 𝑥 𝑐𝑣 + 𝑃 𝜋𝑟 2 𝑥1 (𝑟 − 1) 𝑥1 𝑅 1 𝑥 𝑐𝑣 𝑥 𝑐𝑣 = 𝑃1 𝜋𝑥1 𝑟 2 [𝑟 ln + (𝑟 − 1)] = 𝑃1 𝜋𝑥1 𝑟 2 [𝑟 ln + (𝑟 − 1)] 𝑥1 𝑅 𝑥1 𝑅

𝑄(𝑥) = 𝑊 + ∆𝑈 = 𝑃1 𝜋𝑟 3 𝑥1 ln

EXERCISE 4 The volume of a piston-cylinder vessel containing 100 kg of CO2 is kept fixed at 1 m3, while the container is cooled from 𝑇 = 354 K until it reaches saturated vapor conditions.

After that, the fluid is cooled at constant pressure until it reaches the saturated liquid

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condition. Using the provided 𝑃 − ℎ thermodynamic chart for carbon dioxide, estimate the energy transfer as heat and indicate the process by drawing a line on the 𝑃 − ℎ chart.

Isochoric process: 1-2

V =const.

Q12 U12 The energy balance gives 𝑄12 = ∆𝑈12 STATE 1 The fluid is superheated and

1 m3 m3 𝑣1 = 𝑣2 = = 0.01 , 100 kg kg 𝑇1 = 354 K. From 𝑃 − ℎ chart of CO2 we read the pressure and specific enthalpy and they are 𝑃1 = 5.5 MPa = 5.5 × 106 Pa ℎ1 = 515

kJ kg

therefore, 𝑢1 = (ℎ − 𝑃𝑣)1 = 515

kJ 106 m3 kJ − 5.5 × 3 Pa × 10−2 = 460 kg 10 kg kg

STATE 2 The fluid is in the saturated vapor state at 𝑣1 = 𝑣2 = 0.01 m3 ⁄ kg.

From the 𝑃 − ℎ chart we obtain 𝑃2 = 3.5 MPa, ℎ2 = 430 𝑢2 = (ℎ − 𝑃𝑣)2 = 430

kJ , kg

kJ 106 m3 kJ − 3.5 × 3 Pa × 10−2 = 395 . kg 10 kg kg

From the energy balance, the specific energy transfer as heat for the isochoric cooling is therefore 𝑞12 = 𝑢1 − 𝑢2 = (460 − 395)

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kJ kJ = 65 . kg kg

isobaric process: 2-3

Q23

W23 P = const. U23 The energy balance gives 𝑄23 = 𝑊23 + ∆𝑈23 .

Given that 𝑃 = const., we have that Δ𝑈 = 𝑄 − 𝑃ΔV. With 𝐻 ≡ 𝑈 + 𝑃𝑉, we have that, for the isobaric condensation, ∆𝐻 = ∆𝑈 + 𝑃Δ𝑉 = 𝑄 + 𝑃Δ𝑉 − 𝑃Δ𝑉 = 𝑄.

This in this case gives 𝑄23 = 𝑀(ℎ2 − ℎ3 ).

STATE 3 From the 𝑃 − ℎ chart for CO2 we can evaluate the specific enthalpy difference between the saturated vapor and liquid states at 𝑃2 = 𝑃3 ℎ2 − ℎ3 = ℎLV = (430 − 200)

kJ kJ = 230 . kg kg

The total energy transfer as heat from state 1 to state 3 is therefore 𝑄 = 𝑄12 +𝑄23 = 𝑀[(𝑢1 − 𝑢2 ) + (ℎ2 − ℎ3 )] = 100 kg × (65 + 230)

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kJ = 29 500 kJ. kg

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The graphical representation of the cooling process on the 𝑃 − ℎ diagram is as follows

EXERCISE 5 A refrigeration system works according to the inverse Rankine cycle (or so-called heat pump cycle) using CO2 as the working fluid. The fluid undergoes the following processes:    

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pressurization in a compressor; cooling by ambient air in a heat exchanger until it reaches saturated liquid conditions, expansion to a liquid-vapor state at low pressure and temperature through a throttling valve until it reaches a given pressure, and, from this liquid-vapor state, fluid receives energy as heat from the environment to be cooled until it vaporizes completely thus reaching the compressor inlet state.

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The diagram of the system is

The system operates at steady state and the pressure drop in the heat exchangers can be neglected.

The working fluid enters the compressor in saturated vapor conditions at a pressure of 15 bar and leaves the compressor at a pressure of 50 bar and temperature of 340 K. Use the tables and the 𝑃 − ℎ chart provided in the appendix to evaluate the needed thermodynamic properties.

Hint: there is no need to perform interpolation for the property value of state 1, as the values of the two closest states in the table are almost identical. a) For each component (Compressor, Condenser, Valve, Evaporator) make a separate sketch to support the derivation of its energy balance;

𝑊̇comp

Compressor 1-2 𝑀̇ℎ1

𝑀̇ℎ2

Condenser 2-3 𝑄̇cond 𝑀̇ℎ2

𝑀̇ℎ3

𝑀̇ℎ3

𝑀̇ℎ4

Valve 3-4

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Evaporator 4-1 𝑄̇evap 𝑀̇ℎ4

𝑀̇ℎ1

b) write the energy balance for each component; 𝑊̇comp = 𝑀̇(ℎ2 − ℎ1 ) 𝑄̇evap = 𝑀̇(ℎ1 − ℎ4 ) 𝑄̇cond = 𝑀̇(ℎ2 − ℎ3 ) 𝑀̇ℎ3 = 𝑀̇ℎ4 c) calculate the coefficient of performance of the system defined as 𝑄̇evaporator COP ≡ 𝑊̇compressor COP ≡

𝑄̇evaporator ℎ1 − ℎ4 = 𝑊̇compressor ℎ2 − ℎ1

State 1 is in the saturated vapor phase, hence it can be obtained by interpolating among the values kJ ℎ(𝑃 = 14.69 bar, 𝑞 = 1) = 436.89 kg and kJ ℎ(𝑃 = 15.69 bar, 𝑞 = 1) = 437 . kg The enthalpy variation is so small that taking the arithmetic average, or any of the two values is good enough. However, the results of the linear interpolation is ℎ1 = ℎ(𝑃 = 15 bar, 𝑞 = 1) = 436.92

kJ . kg

State 2 is in the superheated vapor state and from the 𝑃 − ℎ chart we have that ℎ2 = ℎ(𝑇 = 340 K, 𝑃 = 5 MPa) = 500

kJ . kg

State 4 is in the vapor-liquid state and its enthalpy can be found by interpolating the enthalpy for values ℎ3 = ℎ4 = ℎ(𝑃 𝑠at = 5 MPa) thus between ℎ′ (𝑃 𝑠at = 4.829 MPa) = 233.70 and

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kJ kg

ℎ′′ (𝑃 𝑠at = 5.192 MPa) = 242.55

kJ , kg

resulting in ℎ3 = ℎ4 = ℎ(𝑃 = 5 MPa, 𝑞 = 0) = 237.9

kJ kg

The coefficient of performance is therefore kJ kJ 𝑄̇evaporator ℎ1 − ℎ4 (437 − 238) kg 199 kg COP ≡ = = = = 3.0 kJ ℎ2 − ℎ1 (500 − 437) kJ 𝑊̇compressor 66 kg kg d) Determine the mass flow rate of CO2 that allows to obtain a cooling power of 20 kW. The cooling power is the wanted effect (see sketch): energy from the environment is transferred to the working fluid, which, as a consequence, evaporates. The result is that the environment is kept cool.

The mass flow rate of CO2 that allows to obtain a cooling power of 20 kW is computed as 𝑀̇ =

𝑄̇evap 20 kW kg = = 0.1 . kJ (ℎ1 − ℎ4 ) s 199.1 kg

EXERCISE 6 Gas enters a device operating at steady state at a mass flow rate of 𝑀̇ and at temperature 𝑇1 and pressure 𝑃1 , and exits the device at 𝑇2 and 𝑃2 . Both the pressure and the temperature at the inlet are higher than at the outlet. The device converts the energy of the gas into shaft TESTBANK

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power 𝑊̇ and is cooled by heat transfer at the rate 𝑄0̇ to the environment at 𝑇0 . The properties of the gas at the conditions inside the device can be calculated assuming that it is an ideal gas with constant specific heat. a) Make a sketch of the system to support the derivation of the energy balance; 𝑊̇

𝑀̇ℎ1

𝑀̇ℎ2

𝑄0̇ b) make a sketch of the system to support the derivation of the entropy balance;

𝑀̇𝑠1

𝑀̇𝑠2

℘̇𝑠

𝑇0

𝑄̇ / 𝑇0

c) derive an expression for the maximum shaft power 𝑊̇ that can be obtained from the 𝑃 𝑇 system in terms of 𝑀̇ , 𝑇0 , 𝑇1 , pressure ratio 𝛽𝑃 = 𝑃1 , temperature ratio 𝛽𝑇 = 𝑇1 and gas 2

2

constants.

The balance equations are written on a rate basis. The energy balance for the CV is 𝑀̇ℎ1 − 𝑊̇ = 𝑀̇ℎ2 + 𝑄̇ .

The entropy balance for the CV is 𝑀̇𝑠1 + ℘̇𝑠 = 𝑀̇ 𝑠2 +

𝑄̇ . 𝑇0

Combining the two equations results in 𝑊̇ = 𝑀̇ℎ1 − 𝑀̇ℎ2 − 𝑄̇ = 𝑀̇(ℎ1 − ℎ2 ) − 𝑇0 𝑀̇(𝑠1 − 𝑠2 ) − ℘̇𝑠 𝑇0 .

According to the second law of thermodynamics, the maximum amount of power is obtained for ℘̇𝑠 = 0, hence 𝑊̇max = 𝑀̇(ℎ1 − ℎ2 ) − 𝑇0 𝑀̇(𝑠1 − 𝑠2 ).

The fluid obeys the ideal gas law with constant specific heat therefore

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𝑊̇max = 𝑀̇[𝑐𝑃 (𝑇1 − 𝑇2 ) − 𝑇0 (𝑠1 − 𝑠2 )] 𝑇1 𝑣1 = 𝑀̇ [𝑐𝑃 (𝑇1 − 𝑇2 ) − 𝑇0 [𝑐𝑣 ln ( ) + 𝑅 ln ( )]] 𝑇2 𝑣2 ̇ 𝑇1 𝑃1 = 𝑀 [𝑐𝑃 (𝑇1 − 𝑇2 ) − 𝑇0 [𝑐𝑃 ln ( ) − 𝑅 ln ( )]] 𝑇2 𝑃2 𝑊̇max = 𝑀 [𝑐𝑃 𝑇1 (1 −

1 ) − 𝑇0 (𝑐𝑃 ln 𝛽𝑇 − 𝑅 ln 𝛽𝑃 )] 𝛽𝑇

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SAMPLE EXAM 2 EXERCISE 1 (TRUE/FALSE QUESTIONS, WITH JUSTIFICATION) Provide a theoretical explanation of why the answer to the question is true or false, with mathematical relations if needed.

1) Temperature is an intensive property TRUE Intensive properties are independent of the size of the control volume or mass, and this is the case for temperature. Temperature is related to the average kinetic energy of the molecules of the substance forming the control volume or mass. For this reason, it does not depend on its size, thus it cannot be an extensive property.

2) A gas is compressed in a cylinder. Energy is transferred as heat from the gas to the surroundings of the cylinder. The internal energy accumulation of the gas is always lower than zero.

FALSE An energy balance for the control volume encompassing the system shows that the sign of the energy accumulation term depends on the magnitude of the work and of the energy-transfer-as-heat terms, therefore the statement cannot be unconditionally true.

This example clarifies the conceptual difference between internal energy and energy transfer as heat. Internal energy can increase even if there is energy transfer as heat from the gas.

3) The state of a simple compressible substance is fixed if at least three independent thermodynamic variables are specified.

FALSE For a simple compressible substance, the only energy transfer modes are work and energy transfer as heat. According to the state principle, the number of independent thermodynamic variables that needs to be specified is therefore two.

4) The kinetic energy of the gases expelled by the nozzle of a rocket is the same in a reference frame moving with the rocket and in the one fixed to the earth, considered as inertial.

FALSE Suppose the velocity of the rocket in the earth frame is Vr and the velocity of the exhaust gases in the rocket frame is Veg. The kinetic energy of the exhaust gases is proportional to Vr2 in the rocket frame, whereas it is proportional to ( Veg  Vr )2 in the earth frame. Thus, they are different.

5) The temperature in the inside of a fridge decreases, thus the entropy of the substances within it also decreases. The global production of entropy associated with this process is negative.

FALSE The global production of entropy is the summation of the entropy associated with the temperature decrease inside the fridge (negative) and that resulting from the processes that make it possible. The second law of thermodynamics prescribes that the global production of entropy is positive, therefore the entropy production of the processes causing the decrease of entropy inside the fridge must be larger than that associated with the decrease of the fridge internal temperature.

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EXERCISE 2 When considering the motion of a fluid on a molecular level, the state of the fluid can no longer be described by the Navier-Stokes equations. By integrating the phase space of possible states of the system, both over velocity and location, one can find mean properties.

One of these properties is the shear stress 𝜏. The solution of the Couette-Poiseuille flow problem with a heated wall provides the expression for shear stress as 𝜏=

𝜌𝒱 √𝜋𝛽𝑤

where 𝜌 is the number density (number of moles per volume), 𝒱 is the bulk velocity of the fluid and 𝛽𝑤 some thermodynamic parameter. a) Find the unit of the parameter 𝛽𝑤 . Express it only in SI units of primary quantities. For m example, Newton should be expressed kg ∙ s2. mol m mol m ∙ s mol ∙ s s 3 3 ∙ s 𝛽𝑤 = ⇒ [𝛽𝑤 ] = m = m m = = N m ∙ kg m ∙ kg √𝜋 ∙ 𝜏 kg ∙ 2 s m2 m2 𝜌𝒱

b) Find the value of parameter 𝛽𝑤 using the data 𝒱 = 1.86

miles , h

𝜌 = 1.157

mol , ft 3

𝜏 = 20 Pa, 1 km = 0.621 miles, 1 m = 3.28 ft. mol m −3 miles m s −1 ∙ (0.3048 ) 1.86 1610 (3600 ) 𝜌𝒱 ft h miles h m3 𝛽𝑤 = = N √𝜋 ∙ 𝜏 √𝜋 ∙ 20 m2 s = 0.957 m ∙ kg 1.157

EXERCISE 3 An ideal gas is heated in a piston-cylinder system. The friction between the piston and the cylinder can be neglected. The piston is initially locked. The energy is transferred as heat in such a way that the temperature of the gas remains constant. The system has a circular cross-section with radius 𝑟 and the surrounding environment is at sea-level atmospheric conditions. The pressure in the cylinder is much higher than that of the environment. At a certain time instant the lock is removed and the piston accelerates.

Obtain an expression for the velocity of the piston head as a function of the piston head mass 𝑀p , the gas constant 𝑅, the temperature of the gas, the radius of the piston 𝑟, the mass

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of the gas in the cylinder, the atmospheric pressure, the change in piston head position Δ𝑥 and the initial position of the piston head 𝑥1 .

Application of the 8-step method for energy analysis.

1) Sketches of the relevant systems.

The sketch of the forces acting on the piston is

2) The analysis is performed on control masses.

3) The reference frame is inertial as shown on the sketch.

4) List of the idealizations  the friction between the piston and the cylinder is negligible;  there is no accumulation of potential energy within the gas or the piston;  the temperature control is perfect, thus able to keep the temperature of the gas strictly constant;  the atmospheric pressure remains constant during the process;  the gas inside the piston is ideal.

5) Indicate the transfer and accumulation terms appearing in the balance equations are indicated on the sketch, the positive direction of the transfer terms is also indicated.

6) The analysis is performed on a time interval basis.

7) The energy balances are 𝑄 = 𝑊 + Δ𝑈, 𝑊 = Δ𝐸k + 𝑊 ′ .

8) The other equations needed to solve the problem are as follows. 𝑃𝑉 = 𝑀g 𝑅𝑇.

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Both work terms in the piston energy balance can be expressed as functions of state variables as 𝑉2

𝑊 = ∫ 𝑃𝑑𝑉 = 𝑀g 𝑅𝑇 ln 𝑉1

𝑉2 , 𝑉1

𝑉2

𝑊 = ∫ 𝑃atm 𝑑𝑉 = 𝑃atm (𝑉2 −𝑉1). 𝑉1

hence Δ𝐸k = 𝑀𝑔 𝑅𝑇 ln

𝑉2 − 𝑃atm (𝑉2 −𝑉1 ). 𝑉1

Then, introducing the definition of the kinetic energy, we obtain 1 𝑉2 𝑀p 𝒱p2 = 𝑀g 𝑅𝑇 ln − 𝑃atm (𝑉2 −𝑉1 ).

2 𝑉1 Solving for the piston velocity gives 2 𝑉2 𝒱p = √ [𝑀g 𝑅𝑇 ln − 𝑃atm (𝑉2 −𝑉1 )] . 𝑀p 𝑉1 Noting that 𝑉(𝑥) = 𝜋𝑟 2 𝑥, the expression for the piston velocity as a function of the specified variables is 2 Δ𝑥 𝒱p = √ [𝑀𝑔 𝑅𝑇 ln (1 + ) − 𝑃atm 𝜋𝑟 2 Δ𝑥] . 𝑀p 𝑥1 EXERCISE 4 To boil pasta for dinner, we introduce 1.5 litres of saturated liquid water in a special pot with a massless lid that can move freely (no friction) and is perfectly tight. This way saturated conditions can be maintained. The environment is at sea level.

In order to transfer energy as heat to it, we place the pot on top of a stove which has the following power settings A: 1300 W B: 1400 W C: 1500 W At the same time, energy as heat is transferred from the pot to the kitchen (surroundings) at a rate of 100 W. The cooking time is 8 minutes. To ensure that all the pasta is submerged in the water, it is necessary that the volume of water is always at least 1.2 litres. The effect of the pasta in the water can be neglected in this simple analysis.

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a) What is the power setting providing the highest rate of energy transfer heat that can be used?

Application of the 8-step method for energy analysis.

1) Sketch of the system with boundaries

2) The analysis is performed considering a control mass.

3) The reference frame is inertial 4) Idealizations:  the pressure inside the pot is the same as that outside during the whole process (1 atm): the process is isobaric;  the mass of the lid is negligible (no weight);  there is no friction between the lid and the inner surface of the pot;  changes in potential, kinetic etc. energy are negligible;  the moving sealing of the vessel is perfect (closed volume); 5) The transfer and accumulation terms appearing in the balance equations are indicated on the sketch, the positive direction of the transfer terms is also indicated.

6) The energy balance is written on a time interval basis 7) The energy balance is 𝑄out + Δ𝑈 = 𝑄in − 𝑊.

In case of isobaric heating, Δ𝑈 = Δ𝐻, therefore 𝑄out + Δ𝐻 = 𝑄in − 𝑊.

Given that the mass of the lid can be neglected and that no friction occurs between the edge of the lid and the inner surface of the pot, the energy balance is 𝑄out + Δ𝐻 = 𝑄in 8) Other information and equations needed to solve the problem are as follows.

Since the cooking occurs at sea-level, 𝑃 = 𝑃atm = 101325 Pa, and the pressure of the water inside the pot remains constant during the process.

At the initial state 1, water is in the saturated liquid state, therefore from Appendix A.3

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ℎ1 = ℎL = 419

kJ kg

𝑣1 = 𝑣 L = 0.00104

m3 kg

The total mass of water is 𝑀H2 O,TOT =

𝑉H2 O,1 0.0015 m3 = = 1.44 kg. m3 𝑣1 0.00104 kg

The mass of liquid water at the end of the cooking time, state 2, must occupy a volume of 1.2. The state of the water is saturated. The mass of liquid water in state 2 is therefore 𝑉HL2 O,2 0.0012 m3 L 𝑀H2 O,2 = = = 1.15 kg m3 𝑣L 0.00104 kg The quality of the vapor-liquid fluid in state 2 (end of the cooking) is obtained from its definition and is therefore 𝑥2 ≡

𝑀HL 2 O,2 𝑀V 𝑀 − 𝑀L = = 1− = 0.2, 𝑀 𝑀 𝑀H2 O,TOT

The specific enthalpy of the water when it reaches state 2, can be calculated with the quality, once the value of the specific enthalpy in the saturate vapor state is obtained from Appendix A.3, and this is ℎV = 2675

kJ , kg

therefore ℎ2 = (1 − 𝑥2 )ℎL + 𝑥2 ℎV = 0.8 × 419

kJ kJ kJ + 0.2 × 2675 = 870 . kg kg kg

The energy balance over the given period of time allows to calculate the power that would lead to obtain exactly 1.2 litre of liquid water in equilibrium with vapor. Since the energy transfer rate as heat to the environment is constant and 𝑄̇out = 100 W, we have that 𝑄in = 𝑄out + Δ𝐻 = 0.1

kJ kJ × 8 × 60 s + (870 − 419) × 1.44 kg = 697 kJ s kg

Given that the energy transfer as heat occurs at a constant rate, we have that 𝑄in = 𝑄̇in Δ𝑡 → 𝑄̇in =

𝑄in 697 kJ = = 1.45 kW Δ𝑡 8 × 60 s

With power setting corresponding to the maximum rate of energy transfer (C: 1.5 kW) the level of liquid water at the end of the cooking would be lower than the prescribed 1.2 litre.

The power setting that can be use is therefore B, therefore 1.40 kW.

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b) Starting from the saturated liquid state, how much time would it take for the water to completely evaporate if the stove is kept at the power setting calculated as the answer to question a)?

If the water is completely evaporated, at the end of this process its thermodynamic state (state 3) is that of saturated vapor ℎ3 = ℎV = 2675

kJ . kg

Considering that the rate of energy transfer is constant, the energy balance over the time interval is 𝑄̇in Δ𝑡 = Δ𝐻 + 𝑄̇out Δ𝑡, which allows to compute the time that it takes to completely evaporate the water as

Δ𝑡 =

Δ𝐻 𝑄̇in − 𝑄̇out

kJ × 1.44 kg kg = 2499 s = 41.7 min (1.4 − 0.1) kW

(2675 − 419) =

EXERCISE 5 A combined cycle power plant is a system formed by two thermal engines, which is commonly used for electricity generation. The operation of the two thermal engines is based on two different thermodynamic cycles. The working principle of the combined cycle is that the high temperature cycle discharges thermal energy as heat to the low temperature cycle.

Consider a concentrated solar power combined cycle system whose simplified system diagram is reported in the figure below. The high-temperature cycle is called Brayton cycle, while the low temperature cycle is called Rankine cycle.

The working fluid of the high-temperature cycle is air and that of the low temperature cycle is water.

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Specifications:         

Environment: 𝑃1 = 1 bar, 𝑇1 = 25 ℃; Air turbine inlet: 𝑃3 = 17 bar, 𝑇3 = 1233 ℃; Air turbine isentropic efficiency: 𝜂34s = 0.85; Air compressor efficiency: 𝜂12s = 0.80; Energy transfer as heat to the air Brayton cycle: 𝑄̇in = 𝑄̇23 = 96.44 MW Air temperature at the stack: 𝑇5 = 120 ℃; Steam turbine inlet (superheated): 𝑃6 = 20 bar, 𝑇6 = 477 ℃; Steam turbine isentropic efficiency: 𝜂67s = 0.82; Pressure in the steam condenser: 𝑃7 = 𝑃8 = 0.2 bar;

The power consumption of the feed-water pump can be neglected: state 8 = state 9.

The thermodynamic state of the pump inlet/condenser outlet is saturated liquid water. a) In addition to the usual sketches of the system which are part of the 8-step method for energy analysis, draw a qualitative sketch of the processes forming the two cycles in the two 𝑇𝑠 charts below, one of air, and one of water.

b) Calculate the thermal efficiency 𝜂 ≡ 𝑊̇net ⁄𝑄̇in of the combined cycle. Calculate also the thermal efficiency of the two cycles individually.

Assume the following:    

Air can be considered a polytropic ideal gas with constant isobaric specific heat, kJ 𝑐𝑃 = 1.0 kg∙K. The heat exchangers are adiabatic (only energy transfer as heat between fluids and not with the surroundings).

No state change of the working fluids within the ducts connecting the components.

Pressure losses in all heat exchangers can be neglected.

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

The power consumption of the feed-water pump can be neglected: state 8 = state 9.

The pump inlet/condenser outlet states are saturated liquid water.

Application of the 8-step method for energy and entropy analysis.

1) Sketch of the system with boundaries and sketches of the components 𝑄̇23,in

𝑊̇34,t

𝑚̇1 ℎ1 𝑊̇67,t 𝑚̇5 ℎ5

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𝑄̇78,out

2) The analysis is performed considering a control volume for each component and also one for the entire system for the overall energy balance check.

3) The reference frame is inertial.

4) Idealizations:  the system is at steady state;  no state change of the working fluids in the ducts connecting the components;  no change of working fluid compositions;  the flow can be considered one-dimensional at the connections between components;  negligible kinetic and potential energy at the intercomponent states;  the ideal gas assumption holds for air;  there are no pressure losses in any of the heat exchangers;  the pump, compressor and turbines are adiabatic;  the heat exchangers are adiabatic;  the work performed by the pump of the steam bottoming cycle plant is negligible;  the efficiency of the electrical generators is 100 %.

5) Direction of positive energy transfers as indicated on the sketches 6) Energy and entropy balances on a rate basis.

7) The energy balance equations for each of the components are     

Gas Turbine Compressor (1-2): Gas Turbine Heater (2-3): Gas Turbine Expander (3-4): Heat Recovery Steam Generator - air (4-5): Heat Recovery Steam Generator - water (9-6): 𝑀̇water ℎ6  Steam turbine (6-7):  Steam Condenser (7-8):  Feedwater Pump (8-9):

𝑀̇air ℎ1 + 𝑊̇12 = 𝑀̇air ℎ2 𝑀̇air ℎ2 + 𝑄̇23 = 𝑀̇air ℎ3 𝑀̇air ℎ3 = 𝑊̇34 + 𝑀̇air ℎ4 𝑀̇air ℎ4 = 𝑀̇air ℎ5 + 𝑄̇HRSG 𝑀̇water ℎ9 + 𝑄̇HRSG = 𝑀̇water ℎ6 = 𝑊̇67 + 𝑀̇water ℎ7 𝑀̇water ℎ7 = 𝑄̇78 + 𝑀̇water ℎ8 𝑀̇water ℎ8 + 𝑊̇89 = 𝑀̇water ℎ9

8) Bring in other information as necessary, identify and solve for the unknowns.

Starting from ideal gas relations, the isentropic temperature at the gas turbine compressor outlet can be determined as 𝛾−1

𝑃2 𝛾 𝑇2𝑠 = 𝑇1 ( ) = 669.5 K. 𝑃1 The compressor outlet temperature is calculated from definition of isentropic efficiency, under the assumption of polytropic ideal gas, i.e., 𝜂12s ≡

ℎ2𝑠 − ℎ1 𝑐𝑃 (𝑇2𝑠 − 𝑇1 ) = , ℎ2 − ℎ1 𝑐𝑃 (𝑇2 − 𝑇1 )

and is 𝑇2 = 762.3 K. TESTBANK

25

The mass-flow rate of air through the gas turbine can be calculated from the energy balance valid for the heater, namely 𝑀̇air =

𝑄̇23 𝑄̇23 = = 130 kg/s ℎ3 − ℎ2 𝑐𝑃 (𝑇3 − 𝑇2 )

Given that the Brayton cycle is an open cycle, the inlet pressure and the outlet pressure are equal to that of the environment. Furthermore, pressure losses within the heat exchangers can be neglected, hence 𝑃1 = 𝑃4 = 𝑃5 . The isentropic temperature at the turbine outlet can also be determined by means of the ideal gas relation valid for a reversible adiabatic expansion 𝛾−1

𝑃4 𝛾 𝑇4𝑠 = 𝑇3 ( ) = 670.4 K. 𝑃3 Similarly to the calculation of the outlet temperature of the compressor, the outlet temperature of the gas turbine expander results from 𝜂34s ≡

ℎ3 − ℎ4 𝑐𝑃 (𝑇3 − 𝑇4 ) = ⟹ 𝑇4 = 795.7 K. ℎ3 − ℎ4𝑠 𝑐𝑃 (𝑇3 − 𝑇4𝑠 )

The energy transferred as heat from the hot exhaust of the gas turbine evaporates the pressurized water in the HRSG. The thermal power needed to bring the water from state 9 to state 6 can be calculated with the energy balance on the hot side of the HRSG (see sketch) as 𝑄̇HRSG = 𝑀̇air (ℎ4 − ℎ5 ) = 𝑀̇air 𝑐𝑃 (𝑇4 − 𝑇5 ) = 52.3 MW The enthalpy at state 8 and 9 are assumed to be the same as the power consumption of the feedwater pump can be neglected. Given that state 8 is assumed to be saturated liquid, its enthalpy can be obtained from Appendix A.3 as ℎ8 = ℎsat (𝑃8 ) = 73.1

kJ = ℎ9 . kg

The enthalpy of the superheated steam at state 6 (steam turbine inlet), as well as the entropy can be found from Appendix A.3 as ℎ6 = ℎ(𝑃6 = 20 bar, 𝑇6 = 477 ℃) = 3417 = 7.367

kJ , kg

𝑠6 = 𝑠(𝑃6 = 20 bar, 𝑇6 = 477 ℃)

kJ . kg ∙ K

The mass-flow rate of water of the steam power plant can be determined from the energy balance valid for the water side of the HRSG as 𝑀̇water =

𝑄̇HRSG = 15.6 kg/s ℎ6 − ℎ9

The value of the steam turbine outlet enthalpy corresponding to an isentropic expansion can be estimated, albeit with a large degree of approximation from the 𝑇𝑠 chart of 26

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Appendix A.3, knowing that 𝑃7𝑠 = 𝑃7 = 0.02 MPa and that 𝑠7𝑠 = 𝑠6 ≈ 7.4 kJ/kg ∙ K , which gives ℎ7𝑠 = 2450

kJ kg

This value is then used, together with the value and definition of isentropic turbine efficiency to determine the enthalpy at the outlet of the steam turbine, i.e., 𝜂67s ≡

ℎ6 − ℎ7 kJ ⟹ ℎ7 = 2625 . ℎ6 − ℎ7𝑠 kg

The values of all the power transfers occurring in the system, which are needed to compute the thermal efficiencies, can now be calculated as products of the appropriate mass flow rate and enthalpy difference between the inlet and the outlet of the component where the power transfer takes place. We therefore have  Gas Turbine Compressor (1-2):  Gas Turbine Expander (3-4):  Steam Turbine (6-7): 12.4 MW

𝑊̇12 = 𝑀̇air 𝑐𝑃 (𝑇2 − 𝑇1 ) = 60.3 MW 𝑊̇34 = 𝑀̇air 𝑐𝑃 (𝑇3 − 𝑇4 ) = 92.4 MW 𝑊̇67 = 𝑀̇water (ℎ6 − ℎ7 ) =

The net power output of the gas turbine, of the steam turbine and of the combined cycle power plants are therefore  Gas Turbine Power Plant:  Steam Turbine Power Plant:  Combined Cycle Power Plant:

𝑊̇GT = 𝑊̇34 − 𝑊̇12 = 32 MW 𝑊̇ST = 𝑊̇67 − 𝑊̇89 = 12.4 MW 𝑊̇CC = 𝑊̇GT + 𝑊̇ST = 44.4 MW

The thermal efficiency of the combined cycle power plant and of the gas turbine and steam turbine power plants are therefore  Combined Cycle Power Plant: 𝜂CC =

𝑊̇CC = 0.46 𝑄̇23

𝜂GT =

𝑊̇GT = 0.33 𝑄̇23

 Gas Turbine Power Plant:

 Steam Turbine Power Plant: 𝜂ST =

𝑊̇ST = 0.24 𝑄̇HRSG

c) Compute the second law efficiency based on the maximum and minimum cycle temperature of the combined cycle and those of the two cycles and discuss these results.

The temperature at states 7, 8, 9 can be obtained from the saturated steam tables of Appendix A.3, using the pressure as entry variable, yielding 𝑇7 = 𝑇8 = 𝑇9 = 290.65 K. The Carnot efficiency is determined from its definition

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𝜂Carnot ≡

𝑇max − 𝑇min . 𝑇max

We have therefore that  Combined Cycle Power Plant:  Gas Turbine Power Plant:  Steam Turbine Power Plant:

𝜂CC,carnot = 0.807 𝜂GT,Carnot = 0.802 𝜂ST,Carnot = 0.612

These values provide the maximum thermodynamic efficiency of any engine operating between the given maximum and minimum temperatures. The II-law efficiency is therefore the correct merit parameter that can be used to benchmark the actual power plants. Note therefore that the maximum efficiency for any energy conversion device cannot be 100 %.

The second-law efficiency of each power plant is calculated from the provided definition, and these are  Combined Cycle Power Plant: 𝜂CC,II =  Gas Turbine Power Plant: 𝜂GT,II =  Steam Turbine Power Plant: 𝜂ST,II =

𝜂CC 𝜂CC,carnot 𝜂GT 𝜂GT,Carnot 𝜂ST 𝜂ST,Carnot

= 0.57 = 57 %

= 0.41 = 41 %

= 0.39 = 39 %

The second law efficiency allows to rank the power plants in terms of what it is possible to achieve by comparison with an ideal Carnot cycle power plant operating with the same maximum and minimum temperatures. The combined cycle power plant is therefore the best, reaching almost 60 % of the Carnot efficiency, while the two power plants taken separately perform in a similar manner (around 40 % of the maximum achievable performance), even if the temperature levels at which they operate are greatly different.

EXERCISE 6 A tank is separated into two compartments by an adiabatic and frictionless piston. One compartment contains 0.2 m3 of air and the other compartment contains 0.1 kg of oxygen, both initially at 27 ℃ and 90 kPa. The compartment of the tank containing oxygen is insulated. The compartment of the tank containing air is slowly supplied with energy as heat through a wall whose temperature is controlled and kept constant at 300 ℃. During the process the two compartments are in mechanical equilibrium. The energy transfer as heat stops when the pressure of the air rises to 150 kPa. The molar mass of oxygen is 32 g/mol and its specific heat at constant pressure is 940 J/(kg ∙ K). Assume that the ideal gas assumption holds for both gases. a) Determine the temperature of the air and of the oxygen at the end of the heating process.

Application of the 8-step method for energy and entropy analysis.

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1) Sketch of the system with boundaries and arrows indicating the positive direction of the quantities appearing as terms of the energy balance equations.

2) The analysis is performed considering a control mass (closed volume).

3) The reference frame is inertial 4) Idealizations:  the tank is rigid and closed;  changes in potential, kinetic etc. energy are negligible;  the container as well as the piston are adiabatic with exception of the heat transfer through one boundary as indicated on the sketch;  The energy as heat is supplied slowly such that the pressure in the compartments is always uniform.  Specific heats are assumed as independent of temperature.

5) Direction of positive energy transfers as indicated on the sketch.

6) Energy and entropy balances on a time interval basis.

7) The energy balance equations are  system: 𝑄 = ∆𝑈air + ∆𝑈O2 ,  air compartment: 𝑄 = ∆𝑈air + 𝑊,  oxygen compartment: 𝑊 = ∆𝑈O2 .

The entropy balance equations are  system:

𝑄 ⁄𝑇 + ℘𝑠 = ∆𝑆air + ∆𝑆O2 , w

𝑄 ⁄𝑇 + ℘𝑠 = ∆𝑆air , w  oxygen compartment: 0 = ∆𝑆O2 .  air compartment:

The entropy production within the oxygen compartment is zero because the process is isentropic, therefore entropy balance for the whole system is the same as the entropy balance for the air compartment.

8) Bring in process, state, or other information as necessary to complete the analysis.

First the properties of the gasses need to be determined using Appendix A.1. The value of the isochoric specific heat can be calculated by subtracting the value of the gas constant from the value of the isobaric specific heat. The resulting table is

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air

oxygen

𝑅 / J /(kg ∙ K)

287

260

𝑐𝑃 / J /(kg ∙ K)

1004

940

𝑐𝑣 / J /(kg ∙ K)

717

680

𝛾

1.4

1.38

If we define as state 2 the state of the system at the end of the process, the temperature of oxygen at state 2 can be calculated with the temperature relationship for isentropic processes valid for ideal gasses, resulting in γ−1

PO ,2 γ TO2 ,2 = TO2 ,1 ( 2 ) = 345.5 K. PO2 ,1 Using the ideal gas law and the calculated temperature, the specific volume of oxygen at state 1 and state 2 can be computed as 𝑣O2 ,1 = ( 𝑣O2 ,2 = (

𝑀𝑅𝑇1 ) = 0.0866 m3 , 𝑃1 O2

𝑀𝑅𝑇2 ) = 0.0598 m3 . 𝑃2 O2

The mass of the air can be determined with the ideal gas law as 𝑀air =

𝑣air,1 𝑃1 = 0.209 kg. 𝑅𝑇1

The volume of the air once it reaches the final state is given by 𝑣air,2 = 𝑣air,1 + 𝑣O2 ,1 − 𝑣O2 ,2 = 0.2268 m3 .

The temperature of the air in the final state can be determined by means of the ideal gas equation of state as 𝑣2 𝑃2 𝑇air,2 = ( ) = 567.0 K. 𝑅𝑀 𝑎𝑖𝑟 b) Calculate the energy transferred as heat into the system and the entropy production during the process.

The energy transferred as heat can be calculated with the system energy balance as 𝑄 = ∆𝑈air + ∆𝑈O2 = [𝑀𝑐𝑣 (𝑇2 − 𝑇1 )]air + [𝑀𝑐𝑣 (𝑇2 − 𝑇1 )]O2 = 43.27 kJ. The entropy production results from the system entropy balance over the time interval and is ℘𝑠 = 𝑀air Δ𝑆air −

30

𝑄 𝑇2 𝑃2 𝑄 = [𝑀 (𝑐𝑝 𝑙𝑛 ( ) − 𝑅 ln ( ))] − = 27.98 J/K. 𝑇𝑤 𝑇1 𝑃1 air 𝑇w

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c) Suppose you could alter the temperature of the wall where the heat is transferred. What would be the optimal temperature to minimize entropy production?

The lowest entropy production occurs if the temperature of the wall is infinitesimally higher than the temperature of the air throughout the process, in other words if the heat transfer were reversible. The temperature difference cannot be zero, otherwise no energy can be transferred as heat.

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SAMPLE EXAM 3 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) While heating a saturated liquid at constant volume, its pressure and temperature remain constant.

FALSE 2) A thermal energy reservoir (TER) can supply or absorb a finite amount of energy as heat without undergoing any change in temperature.

TRUE 3) An isolated system consists of two volumes made of different materials with different initial temperatures, and in contact with one another. After some time, the system has reached an equilibrium state. Both materials have the same final temperature and the same final specific internal energy.

FALSE 4) A thermodynamic engine can function without discharging energy as heat.

FALSE 5) The coefficient of performance of a system can be obtained by dividing the useful energy output by the total energy input.

FALSE 6) Any intensive property of matter depends on the considered mass.

FALSE 7) The specifications sheet of a car engine shows that its net efficiency is 45% at an ambient temperature of 23 ℃. Your friend Jeremy tells you that the peak temperature of the gases in the cylinders is 230 ℃. Is this true or false?

FALSE 8) The internal energy of a gas is related to the energy associated with the microscopically disorganized motion of its molecules.

TRUE 9) Energy is transferred as heat to a saturated fluid mixture that is stored in a rigid container. The quality of the initial state is 0.2, and it is 0.8 in the final state. The pressure and temperature of the fluid do not change during this process.

TRUE 10) The entropy change of any fluid can be evaluated analytically if only the specific heats at the initial and final states are given.

FALSE 11) An isolated control volume containing an ideal gas undergoes an isothermal process.

No internal energy gets accumulated within the control volume.

TRUE 12) An adiabatic mixer merges two streams of a certain fluid entering into the mixer at the different temperatures. The entropy of the outlet stream is equal to the sum of the

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entropy at the inlets.

FALSE EXERCISE 2 Non-dimensional quantities play an important role in energy systems modeling and design.

An example of a dimensionless quantity used for pumps is the pump specific speed 𝑁𝑠 , and its definition is 𝑁𝑠 =

𝑁√𝑄 (𝑔𝐻)3/4

Where    

𝑁 is the rotational speed of the pump, in rad/s, 𝑄 is the volume flow rate, in m3 / s, 𝑔 is the gravitational acceleration, in m/s2 , 𝐻 is the head of the pump, in m.

a) Prove that the specific speed is a dimensionless quantity.

1

𝑁𝑠 =

𝑁√𝑄

3 ⇒ (𝑔𝐻)4

[𝑁𝑠 ] =

1 m3 2 [s] [ s ] 3

3 m4 [ 2 ] [m]4 s

3

m2 [s] = = [−] 3

m2 [s]

b) Calculate the specific speed of the liquid oxygen (LOX) pump used in the SpaceX Raptor Engine, given the following information:    

𝑔 = 9.81 m/s 2 , 𝑄 = 346 000 gal/hr, 𝑁 = 14 000 rpm, 𝐻 = 7 250 ft.

Use the following unit conversions:  1 ft = 0.3048 m  1 gal = 0.00379 m3

𝑁𝑠 =

𝑁√𝑄

gal 0.00379 h×m3 1 2π rad×min √346 000× 14 000× min × 60 × × 3600 × s h gal×s

3 = (𝑔𝐻)4

≈ 0.495

3

m m 4 (9.81× 2 ×7 520×ft × 0.3048× ) ft s

EXERCISE 3 An ideal gas is heated in a piston-cylinder system where the piston is able to move freely.

The system has a circular cross-section with radius 𝑟 and the surrounding environment is at sea-level atmospheric conditions. The mass of the piston can be considered negligible.

Obtain an expression for the change in piston head position as a function of the amount of

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energy transferred as heat, the pressure, the isochoric specific heat 𝑐𝑣 , the gas constant 𝑅, and the radius 𝑟.

Application of the 8-step method for energy analysis.

1) Sketch of the relevant system

2) The analysis is performed on control masses.

3) The reference frame is inertial as shown on the sketch.

4) The idealizations are:  the mass of the piston is negligible;  the friction between the piston and the cylinder is negligible;  there is no accumulation of potential energy within the gas or the piston;  the system is closed (perfectly sealed);  the atmospheric pressure remains constant during the process;  the gas inside the piston is ideal.

5) The transfer and accumulation terms appearing in the balance equations are indicated on the sketch. The positive direction of the transfer terms is also indicated.

6) The analysis is performed on a time interval basis.

7) The energy balance for a small amount of energy transfer as heat and work is which can be integrated over a period of time resulting in

8) Bring in process, state, or other information as necessary to complete the analysis.

For an ideal gas, the complete thermodynamic model is given by 𝑃𝑣 = 𝑅𝑇,

Δ𝑢 = 𝑐𝑣 ∫ 𝑇𝑑𝑇.

which can be written in terms of extensive properties as

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𝑃𝑉 = 𝑀𝑅𝑇,

Δ𝑈 = 𝑀𝑐𝑣 Δ𝑇.

Combining these two equations yields 𝑃2 𝑉2 𝑃1 𝑉1 Δ𝑈 = 𝑀𝑐𝑣 ( − ). 𝑅𝑀 𝑅𝑀 The piston head can move freely, thus the pressure exerted by the piston on the gas is the atmospheric pressure at all times (isobaric process), consequently Δ𝑈 =

𝑃𝑎𝑡𝑚 𝑐𝑣 (𝑉2 − 𝑉1 ). 𝑅

The volume of a cylinder is given by 𝑉 = π𝑟 2 𝑥.

The accumulation of internal energy within the cylinder can therefore be written as Δ𝑈 =

𝑃atm 𝑐𝑣 π𝑟 2 (𝑥2 − 𝑥1 ). 𝑅

The work done by the gas on the piston can be expressed as 𝑉2

𝑥2

𝑉1

𝑥1

𝑊 = ∫ 𝑃𝑑𝑉 = 𝑃atm π𝑟 2 ∫ 𝑑𝑥 = 𝑃atm π𝑟 2 (𝑥2 − 𝑥1 ).

Substituting the expressions for the change in internal energy and for the work done by the gas on the piston into the energy balance over the time interval t 2 − t1 yields 𝑃atm 𝑐𝑣 π𝑟 2 𝑐𝑣 (𝑥2 − 𝑥1 ) + 𝑃atm π𝑟 2 (𝑥2 − 𝑥1 ) = ( + 1) 𝑃atm π𝑟 2 (𝑥2 − 𝑥1 ). 𝑄= 𝑅 𝑅 Rearranging terms allows to obtain the explicit relation between the piston position, and the amount of energy transfer as heat as 𝑄 Δ𝑥 = (𝑥2 − 𝑥1 ) = 𝑐 ( 𝑅𝑣 + 1) 𝑃atm π𝑟 2 EXERCISE 4 A constant-volume vessel of 20 liters holds 2 kg of oxygen in thermodynamic equilibrium at an initial temperature of 102 K. The vessel is heated until there is only saturated vapor oxygen. a) Determine the temperature and pressure once the oxygen has reached its final equilibrium state.

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Application of the 8-steps method for energy analysis.

1) Sketch of the system with boundaries.

2) The analysis is performed considering a control mass.

3) The reference frame is inertial.

4) Idealizations:  Constant volume  The initial and final thermodynamic states of the oxygen are in equilibrium.  Closed system, but not isolated because there is energy transfer as heat  Changes in potential, kinetic etc. energy are negligible 5) Direction of positive energy transfers and energy accumulation as indicated on the sketch.

6) Energy balance over the time interval 𝑡2 − 𝑡1 .

7) The energy balance for the control mass is 𝑄 = 𝛥𝑈 = 𝑀(𝑢2 − 𝑢1 ), or, equivalently, for an isochoric process 𝑄 = 𝑀(ℎ2 − 𝑃2 𝑣 − ℎ1 + 𝑃1 𝑣).

8) Other information and equations needed to solve the problem are as follows.

State 1 and state 2 can be determined once the specific volume of oxygen during the heating process is calculated, thus 𝑣=

𝑉 20 l 0.02 m3 m3 = = = 0.01 . 𝑀 2 kg 2 kg kg

STATE 1 The state of a simple compressible substance is fixed if two appropriate independent variables are fixed, in this case 𝑇1 = 102 K,

𝑣1 = 0.01

m3 . kg

State 1 is saturated, with quality between 0 and 1. First the value for the saturated liquid and vapor states at 𝑇1 = 102 K must be determined by interpolation using data for saturated oxygen in Appendix A.8. The values used for the interpolation and the resulting values are 𝑇/K

𝑣 L /(𝑚3 /kg) 𝑣 V /(𝑚3 /kg)

𝑃/MPa

ℎL /(kJ/kg)

ℎV /(kJ/kg)

100

0.000917

0.09592

0.254

−116.45

86.16

105

0.00094

0.06612

0.3785

−107.64

88.85

𝟏𝟎𝟐

𝟎. 𝟎𝟎𝟎𝟗𝟐𝟔

𝟎. 𝟎𝟖𝟒𝟎

𝟎. 𝟑𝟎𝟑𝟖

−𝟏𝟏𝟐. 𝟗𝟐𝟔

𝟖𝟕. 𝟐𝟑𝟔

Given the specific volume for saturated liquid and vapor at the 𝑇_1 the quality of state 1 can be determined from its definition as

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3 m3 −4 m 0.01 − 9.26 ⋅ 10 𝐿 𝑣 − 𝑣1 kg kg 𝑥1 = 𝑉 = 0.109 3 𝐿 = 𝑣1 − 𝑣1 0.0840 m − 9.26 ⋅ 10−4 m3

kg

kg

The specific enthalpy can then be calculated as ℎ1 = (1 − 𝑥1 )ℎ1𝐿 + 𝑥ℎ1𝑉 = (1 − 0.109) × 112.926 = −91.063

kJ kJ + 0.109 × 87.236 kg kg

kJ kg

STATE 2 In this case, the other variable to determine the thermodynamic state in addition to 𝑣2 = 𝑣1 = 0.01 m3 ⁄ kg is specified by the condition that the fluid is in the saturated vapor state, thus 𝑥 = 1, where 𝑥 is the vapor quality.

The temperature 𝑇2 can then be estimated as 𝑇2 = 𝑇2 (𝑣2 = 𝑣1 , 𝑥2 = 1) be determined by interpolation using data for saturated oxygen in Appendix A.8. The values used for the interpolation and the resulting values are 𝑇/K

𝑣 V /(𝑚3 /kg)

𝑃/MPa

ℎV /(kJ/kg)

135

0.0112

2.225

91.74

40

0.00856

2.788

88.47

𝟏𝟑𝟕. 𝟑

𝟎. 𝟎𝟎𝟏𝟎

𝟐. 𝟒𝟖𝟏

𝟗𝟎. 𝟐𝟓

The temperature and pressure at the end of the heating process are therefore 𝑇2 = 137.3K, and 𝑃2 = 2.48 MPa. b) Determine the amount of energy transferred as heat for this process The energy transferred during the process as heat is calculated from the energy balance using the thermodynamic properties of state 1 and 2 as 𝑄 = (ℎ2 − 𝑃2 𝑣) − (ℎ1 − 𝑃1 𝑣) kJ m3 kJ = 2 kg × (90.25 − 2.48 ⋅ 103 kPa × 0.01 − (−91.063) + 0.3038 kg kg kg 3 m × 103 kPa × 0.01 ) = 319.08 kJ kg c) Sketch the process on the provided 𝑃 − ℎ diagram for oxygen The line representing the process must be along the isochoric line 𝑣 = 0.01 m3 ⁄kg.

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EXERCISE 5 Consider the following simple Rankine cycle system operating with water as the working fluid, Pressure losses in pipes, ducts and heat exchangers are negligible. The turbine and pump isentropic efficiencies are reported in the table. Pressure losses are negligible everywhere in the system.

Turbine isentropic efficiency

0.6

Pump isentropic efficiency

0.8

The following tables list the information needed to perform the computation of the cycle thermodynamic states at the inter-component interfaces for two different cases.

Case A

38

ID

state

𝑻 / ℃ 𝑷 / 𝐛𝐚𝐫

𝒙

1

Pump inlet

49.3

0

2

Pump outlet

3

Turbine inlet

4

Turbine outlet

4 150

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𝒔 / 𝐤𝐉 /(𝐤𝐠 ∙ 𝐊)

𝒉 / 𝐤𝐉/𝐤𝐠

0.6966

207.412

Case B ID

state

𝑻 / ℃ 𝑷 / 𝐛𝐚𝐫

𝒙

1

Pump inlet

151.85

0

2

Pump outlet

3

Turbine inlet

4

Turbine outlet

𝒔 / 𝐤𝐉 /(𝐤𝐠 ∙ 𝐊)

𝒉 / 𝐤𝐉/𝐤𝐠

6.348

2547.35

140 376.85

140

NOTE: NOT ALL THE CELLS IN THE TABLES NEED TO BE FILLED IN TO SOLVE THE PROBLEM. The two thermodynamic cycles in the temperature-entropy diagram of water are

Hint: turbine thermodynamic states The turbine isentropic efficiency is ∆ℎ ∆ℎ𝑠 Where ∆ℎ is the turbine specific work, and ∆ℎ𝑠 is the turbine isentropic specific work. 𝜂t,𝑠 =

Hint: second law efficiency The second law efficiency is 𝜂II ≡

𝜂th 𝑇̅ 1 − ̅C 𝑇H

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Where 𝜂th is the cycle thermal efficiency. 𝑇̅C is the so-called equivalent thermodynamic temperature for thermal energy rejection (condenser). 𝑇̅H is the equivalent thermodynamic temperature for thermal energy transfer to the steam (boiler).

They are computed as 𝑇̅C =

ℎ4 − ℎ1 ℎ3 − ℎ2 and 𝑇̅H = . 𝑠4 − 𝑠1 𝑠3 − 𝑠2

In order to grasp the need for introducing these temperatures, look at the shape of the cycles in the temperature-entropy diagrams. Do they look similar to a Carnot cycle? a) Calculate the cycle thermal efficiency for both case A and case B. Discuss the difference between these values.

In order to calculated the thermal efficiency related to the two considered cycles, defined as η≡

𝑊̇Net , 𝑄̇Boiler

the relevant energy transfers must be determined by applying the 8-step method for energy analysis to both the system and each of its components.

1) Sketch of the system with boundaries and sketches of the components

2) The analysis is performed considering a control volume for each component and also one for the entire system for the overall energy balance check.

3) The reference frame is inertial.

4) Idealizations:  the mass of working fluid does not change (no leakage to the environment);  the system is at steady state;  changes in potential, kinetic etc. energy between the inlet and outlet of each component are negligible;  1-dimensional flow at the inter-component interfaces;  no pressure drop in ducts, pipes, heat exchangers; 5) Direction of positive energy transfers and energy accumulation as indicated on the sketch.

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6) Energy and mass balances on a rate basis.

7-8) Energy balances and other relations needed to complete the analysis.

The energy balance on a rate basis for the system is 𝑊̇pump + 𝑄̇boiler =

𝑑𝑈 + 𝑊̇turbine + 𝑄̇condenser . 𝑑𝑡

The efficiency can be calculated as η≡

𝑀̇(𝑤turbine − 𝑤pump ) 𝑤turbine − 𝑤pump 𝑊̇net = = . 𝑞boiler 𝑀̇ ⋅ 𝑞boiler 𝑄̇boiler

The needed values of the energy transfers per unit mass flow are computed with component energy balances, and by using of the definition of isentropic efficiency for the pump and the turbine. 𝑤pump + ℎ1 = ℎ2 ,

Pump:

ℎ2 = ℎ2,𝑠 ηpump,𝑠 + ℎ1 (1 − ηpump,𝑠 ).

𝑞boiler + ℎ2 = ℎ3.

Boiler: Turbine:

ℎ3 = 𝑤turbine + ℎ4 ,

Condenser:

ℎ4 = 𝑞condenser + ℎ1 .

ℎ4 = ℎ4,𝑠 ⋅ η𝑠,turbine + ℎ3 (1 − η𝑠,turbine ).

Using these energy balances, the specified state properties and the thermodynamic tables in Appendix A.3, all the quantities needed to compute the thermal efficiency can be calculated. The detailed sequence of the calculations is provided as appendix to the solution of this problem.

Case A ID

state

𝑻/𝑲

𝑷 / 𝐌𝐏𝐚

𝒙

𝒉 / 𝐤𝐉/𝐤𝐠 𝒔 / 𝐤𝐉

/(𝐤𝐠 ∙ 𝐊)

1

Pump inlet

322.45

𝟎. 𝟏𝟐

0

𝟐𝟎𝟔. 𝟗

𝟎. 𝟔𝟗𝟔𝟑

2

Pump outlet

4

207.412

0.6966

3

Turbine inlet

4

𝟐𝟕𝟓𝟐. 𝟐

𝟔. 𝟗𝟐𝟖

4

Turbine outlet

𝟐𝟒𝟑𝟏. 𝟏

𝟕. 𝟓𝟓𝟗𝟐

423.15

𝟎. 𝟏𝟐

𝟎. 𝟗𝟑𝟑

The energy transfers per unit mass and efficiency of the Rankine cycle A are therefore 𝑤pump = 0.512 kJ/kg 𝑞boiler = 2545 kJ/kg 𝑤turbine = 321.1 kJ/kg 𝑞condenser = 2224 kJ/kg η = 0.126 As a check, note also that the energy balance for the entire system is verified.

Case B TESTBANK

41

ID

state

𝑻/𝑲

𝑷 / 𝐌𝐏𝐚

𝒙

𝒉 / 𝐤𝐉/𝐤𝐠 𝒔 / 𝐤𝐉

/(𝐤𝐠 ∙ 𝐊)

1

Pump inlet

425

𝟎. 𝟓

0

𝟔𝟒𝟎. 𝟐

𝟏. 𝟖𝟔𝟐

2

Pump outlet

14.0

𝟔𝟓𝟔. 𝟕

𝟏. 𝟖𝟔𝟖

3

Turbine inlet

14.0

𝟐𝟖𝟗𝟗. 𝟒

𝟓. 𝟕𝟗𝟔

4

Turbine outlet

𝟐𝟓𝟒𝟕. 𝟑

𝟔. 𝟑𝟒𝟖

650

The energy transfers per unit mass and efficiency of the Rankine cycle B are therefore 𝑤pump = 15.47 kJ/kg 𝑞boiler = 2244 kJ/kg 𝑤turbine = 352.0 kJ/kg 𝑞condenser = 1907 kJ/kg η = 0.150 The thermal efficiency of cycle B is higher because the specific work that is obtained from the turbine is higher, while the specific energy input to the boiler is lower. The specific work of the pump of cycle B is much higher than that of cycle A because the water must be pumped to a much higher pressure, however this term has little influence on the net specific work output of the system because it is an order of magnitude smaller than that of the turbine. This shows the advantage of the Rankine cycle. The working fluid is pressurized in the liquid state. In gas cycles, working fluid pressurization requires high efficiency in order to achieve high overall conversion efficiency. b) Calculate the cycle second law efficiency for both case A and case B. In order to compute the second law efficiency for the two cycles, the equivalent thermodynamic temperature of the boiler and of the condenser must be determined using its definition. For the condenser, the equivalent thermodynamic temperature is 𝑇̅cold ≡

ℎ4 − ℎ1 , 𝑠4 − 𝑠1

therefore, the values for cycle A and B are 𝑇̅cold,A =

2431.1 − 206.9 = 322.57 K, 7.592 − 0.6963

𝑇̅cold,B =

2547.3 − 640.2 = 424.98 K. 6.348 − 1.861

For the boiler, the equivalent thermodynamic temperature is 𝑇̅hot ≡

ℎ3 − ℎ2 , 𝑠3 − 𝑠2

therefore, the values for cycle A and B are 𝑇̅hot,A =

2752.2 − 207.412 2899.4 − 656.7 = 404.38K𝑇̅hot,B = = 571.16K 6.928 − 0.697 5.796 − 1.868

The second-law efficiency of the two cycles can then be found using its definition

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η𝐼𝐼 ≡

ηnet , 𝑇̅cold 1− ̅ 𝑇ℎ𝑜𝑡

resulting in ηII,A =

0.126 = 0.600, 322.57 1 − 404.38

ηII,B =

0.150 = 0.586 424.98 1− 571.16

Cycle A makes a better use of the energy input even if its thermal efficiency is lower because its thermal efficiency is higher if compared to the Carnot efficiency between the pertinent equivalent thermodynamic temperature, which provides the correct ideal term of comparison. The better thermodynamic quality of cycle A can also be qualitatively appreciated because the shape of the cycle is more similar to that of a Carnot cycle, which is rectangular.

Detailed solution Case A State 1 is saturated liquid, therefore 𝑇1 = 273.15 + 49.3 = 322.45 K and, using data from Appendix A.3 for interpolation 𝑃1 = 𝑃1 (𝑇1 , 𝑥 = 0) = 0.01055 +

322.45 − 320 × (0.01353 − 0.01055) = 0.012 MPa. 325 − 320

Again using the table for saturated properties, we have that ℎ1 = ℎ1 (𝑃4 , 𝑥 = 0) = 206.9 kJ/kg 𝑠1 = 𝑠1 (𝑃4 , 𝑥 = 0) = 0.6963 kJ/kg State 3 is superheated vapor and the temperature and pressure are specified 𝑇3 = 273.15 + 150 = 423.15 K, And the enthalpy and pressure can be obtained from the table in Appendix A.3 by interpolation, resulting in ℎ3 = 2738.1 + 𝑠3 = 6.8954 +

423.15 − 416.8 × (2811.9 − 2738.1) = 2752.2 kJ/kg, 450 − 416.8

423.15 − 416.8 × (7.0659 − 6.8954) = 6.928 kJ/(kg ∙ K).

450 − 416.8

With the entropy at the turbine inlet and the pressure at the turbine outlet,which is equal to the pressure at the pump inlet, the enthalpy corresponding to the isentropic turbine outlet can be determined as follows 𝑥4,𝑠 = (6.928 − 0.6963)/(8.0850 − 0.6963) = 0.843

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ℎ4,𝑠 = 206.9 + 0.843 × (2590.3 − 206.9) = 2216.1 kJ/kg.

Using the turbine isentropic efficiency, the turbine specific work can be calculated, after which the actual enthalpy of the Turbine outlet (4) can be calculated. 𝑤turbine = 0.6 × (2752.215 − 2216.086) = 321.1 kJ/kg.

The enthalpy at the turbine outlet can therefore be calculated and is ℎ4 = 2752.2 − 321.08 = 2431.1 kJ/kg.

The quality at the turbine outlet is therefore 𝑥4 = (2431.138 − 206.9)/(2590.3 − 206.9) = 0.933, thus the entropy is 𝑠4 = 0.6963 + 0.933 × (8.0850 − 0.6963) = 7.592 kJ/(kg ∙ K).

Finally, the specific energy transfers and cycle efficiency are 𝑤pump = 207.412 − 206.9 = 0.512 kJ/kg 𝑞boiler = 2752.215 − 207.412 = 2545 kJ/kg 𝑤turbine = 2752.2 − 2431.1 = 321.1 kJ/kg 𝑞condenser = 2431.138 − 206.9 = 2224.2 kJ/kg The thermal efficiency and the second law efficiency can then be calculated as η = (321.08 − 0.512)/2544.8 = 0.126, 𝑇̅cold = (2431.138 − 206.9)/(7.592 − 0.6963) = 322.57 K, 𝑇̅hot = (2752.215 − 207.412)/(6.928 − 0.697) = 408.38 K, ηII,A = 0.126/(1 − 322.56/408.38) = 0.600 Case B In analogy with Case 1, the properties of state 1 results from 𝑇1 = 273.15 + 151.85 = 425 K, 𝑃1 = 𝑃1 (𝑇1 , 𝑥 = 0) = 0.5002 MPa. ℎ1 = ℎ1 (𝑃4 , 𝑥 = 0) = 640.2 kJ/kg 𝑠1 = 𝑠1 (𝑃4 , 𝑥 = 0) = 1.8606 kJ/kg With the entropy at the pump inlet and the pressure at the pump outlet, the enthalpy corresponding to the isentropic pump outlet can be estimated as ℎ2,𝑠 = 640.2 + (1.8606 − 1.84586)/(2.09191 − 1.84586) × (754.07 − 646.11) = 652.58 kJ/(kg ∙ K).

Using the pump isentropic efficiency, the pump specific work can be calculated and with its value the enthalpy at the pump outlet can also be calculated as follows

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𝑤pump = (652.58 − 640.2)/0.8 = 15.47 kJ/kg ℎ2 = 640.2 + 15.472 = 655.67 kJ/kg With the pressure and enthalpy at the pump outlet, the entropy can be calculated as 𝑠2 = 1.84586 + (655.67 − 646.11)/(754.07 − 646.11) × 2.09191 = 1.86858 kJ/(kg ∙ K).

The enthalpy and entropy at the turbine inlet can be obtained from the table for superheated vapor in Appendix A using the specified values for temperature and pressure as follows 𝑇3 = 273.15 + 376.85 = 650 K ℎ3 = 3022.5 + (14 − 10)/(15 − 10) × (2868.6 − 3022.5) = 2899.4 kJ/kg 𝑠3 = 6.1007 + (14 − 10)/(15 − 10) × (5.7198 − 6.1007) = 5.796 kJ/(kg ∙ K).

With the calculated specific enthalpies, all the remaining specific energy transfers can be calculated as 𝑞boiler = 2899.4 − 655.67 = 2243.7 kJ/kg 𝑤turbine = 2899.4 − 2547.346 = 352.03 kJ/kg 𝑞condenser = 2547.3 − 640.2 = 1907.2 kJ/kg The thermal efficiency and the second law efficiency can then be calculated as η = (352.03 − 15.47)/ 2243.7 = 0.150 , 𝑇̅cold = (2547.3 − 640.2)/(6.348 − 1.8606) = 424.98 K, 𝑇̅hot = (2899.4 − 655.67)/(5.796 − 1.868) = 571.16 K, ηII,A = 0.150/(1 − 424.98/571.16) = 0.586.

EXERCISE 6

A metal wall is exposed to temperature 𝑇1 at one side and to temperature 𝑇2 at the other side. Demonstrate which of the two temperatures must be higher for the energy transfer as heat to occur in the direction indicated in the figure (i.e., from left to right). Assume that

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the process is steady-state, that the wall is infinitely long, and that no other energy transfer occurs.

Application of the 8-step method for energy and entropy analysis.

1) Sketch

2) The analysis is performed considering a control mass.

3) The reference frame is inertial.

4) Idealizations:  steady state;  changes in potential, kinetic etc. energy are negligible.

5) Direction of positive energy transfers as indicated on the sketch.

6) Energy and entropy balances on a rate basis 7) The energy balance equation on a rate basis is 𝑄1̇ =

𝑑𝑈 + 𝑄2̇ . 𝑑𝑡

The steady-state energy balance therefore reads 𝑄1̇ = 𝑄2̇ = 𝑄.̇ The entropy balance is ̇ + ℘̇ 𝑠 = 𝑆in

𝑑𝑆 ̇ , + 𝑆out 𝑑𝑡

which, at steady-state, becomes 𝑄̇1 𝑄̇2 + ℘̇ 𝑠 = . 𝑇1 𝑇2 The rate of entropy production can be obtained by combining the energy and entropy balance equations as ℘̇ 𝑠 =

𝑄̇ 𝑄̇ 1 1 − = 𝑄̇ ( − ) 𝑇2 𝑇1 𝑇2 𝑇1

The second law of thermodynamics requires that 46

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℘̇ 𝑠 ≥ 0, therefore 1 1 𝑄̇ ( − ) ≥ 0 𝑇2 𝑇1 Given that 𝑄̇ has been as defined positive, we have that 1 1 − ≥ 0, 𝑇2 𝑇1 hence it must be that 𝑇1 ≥ 𝑇2 .

The equality holds in the limiting and ideal case of reversible energy transfer as heat: the entropy production associated with energy transfer as heat tends to zero if the energy transfer occurs with a temperature difference that tends to zero.

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SAMPLE EXAM 4 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) A reversible process is isentropic.

TRUE 2) A person in a raising elevator lifting a box over a distance of half meter will do more work than a person on the ground lifting the same box over the same distance.

FALSE 3) The terms of the equation for the steady-state energy balance of an open system are the same as those for a closed system.

FALSE 4) The efficiency of a wind turbine cannot be larger than the Carnot cycle efficiency.

FALSE 5) The critical point can be evaluated using the ideal gas model.

FALSE 6) An explosion can be regarded as a sudden entropy production.

TRUE 7) When mixing adiabatically hot and cold gases to a new equilibrium state, the entropy increases.

TRUE 8) In a polytropic process the pressure can change independently of the volume.

FALSE 9) A system formed by a simple compressible substance enclosed in a constant volume can transfer energy as work and heat FALSE 10) Removing energy as heat from a closed system reduces its entropy.

TRUE 11) At ambient room conditions, methane is liquid.

FALSE 12) Energy transfer as heat under a finite temperature difference is reversible.

FALSE 13) Cooling a gas in closed volume at constant pressure increases its volume.

FALSE 14) An ideal valve transfers more energy as work than a non-ideal valve.

FALSE EXERCISE 2 (Same as Exercise 1.1 of the book) An 80 kg student wishes to walk on snow without sinking and decides to make a pair of snow shoes. The student determined experimentally

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that the maximum pressure the snow can withstand is 0.6 kPa. What is the minimum surface required by one snow shoe?

Would the student be able to sell snow shoes to his fellow classmates? Explain your answer.

1) Make an idealized sketch of the system and identify the boundary.

2) Indicate the inertial reference frame on the sketch.

The snow shoe is taken as the control mass. The reference frame is positioned on the ground and stationary, thus it is inertial (see sketch).

3) Idealizations   

The maximum pressure that the snow can withstand, as measured by the student, is correct and representative of relevant snow conditions.

The pressure distribution over the shoe surface is uniform in magnitude and direction.

The snow behaves as a rigid body until the maximum pressure is reached.

4) Specify the time basis for the analysis.

The momentum balance for the system will be written on a rate basis.

5) Indicate on the sketch all non-zero terms that will appear in the balance equation (arrows indicate the positive direction).

There is no accumulation nor production.

6) Write the balance equation.

In this case, the relevant conservation equation is the momentum balance in the y direction 𝐹 ⏟snow − 𝐹weight = momentum inputs

⏟ 0

.

accumulation of momentum

The system is analyzed when the entirety of the student's weight acts on one single snowshoe, as the student wishes to walk and not simply stand. The momentum balance equation, familiar to students who followed courses in mechanics, is written here using one of the forms of production accounting.

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7) Introduce other modeling information as necessary to bring the analysis to the point where you have the same number of equations as unknowns and can therefore (in principle) solve for the unknowns.

In order to solve the momentum balance for the unknown A_\text{shoe}, some other modeling information is needed. The relation between the pressure and the force exerted by the snow on the shoe, which comes from the definition of pressure, gives 𝐹snow = 𝑃 ⋅ 𝐴shoe .

Moreover, the weight acting upon the snowshoe due to the gravitational force is 𝐹weight = 𝑤 = 𝑀𝑔 The momentum balance in the 𝑦 direction can thus be written as 𝑃𝐴shoe − 𝑀𝑔 = 0, which can be solved for the unknown shoe area, and it gives 𝐴shoe =

𝑀𝑔 . 𝑃

If 𝑀 = 80 kg, 𝑔 = 9.81 m/𝑠 2 and 𝑃 = 0.6 kPa = 0.6 kN / 𝑚2 , we have that 𝐴shoe =

80 × 9.81 m m2 m2 kg 2 = 1.308 N = 1.308 m2 0.6 s 1000 N 𝑁

Given the large surface needed for these snowshoes, it is unlikely the student will be able to sell them to his classmates . We know however that show shoes exist and work, therefore a more complex investigation about the problem is needed.

EXERCISE 3 The temperature of 10 kg of air inside a piston-cylinder is increased from 25 ℃ to 80 ℃ using an electrical heater, while the pressure inside the cylinder is maintained constant.

During this process 50 kJ of energy is transferred as heat from the system to the environment.

Find the required amount of energy that must be added to the system by the heater, for cylinder pressures of 1, 2 and 5 bar.

Assume the air inside the cylinder as an ideal gas with 𝑐𝑃 = 1.006 kJ⁄(kg ∙ K) Application of the 8-step method for energy analysis.

1) Draw a diagram identifying the control volume.

2) Show the forces if the CV moves.

The forces acting on the control mass are not of interest in this case.

3) Indicate reference frame

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The reference frame is inertial. The coordinate x is used to track the position of the right boundary of the system, and thus also of the piston.

4) List all simplifying idealizations.  the air inside the piston can be considered an ideal gas;  the system is closed (perfect seal of the piston);  the pressure inside the cylinder can be assumed constant;  kinetic and potential energy are negligible if compared to the other energy transfers.

5) indicate all the directions of the energy transfers on the sketch See sketch 6) Time basis of the energy balance.

Over the period of time in which the heating occurs 7) Energy balance. 𝑄 = 𝑊 + ∆𝑈 8) Bring in process, state, or other information as necessary to complete the analysis.

The internal energy change for an ideal gas is ∆𝑈 = ∆𝑢 = 𝑀𝑐𝑣 (𝑇2 − 𝑇1 ), while the work done by the gas on the piston is 𝑉2

𝑊 = ∫ 𝑃𝑑𝑉 = 𝑃(𝑉2 − 𝑉1 ), 𝑉1

considering that the pressure is constant. Given that the air can be considered an ideal gas, we have that 𝑊 = 𝑀𝑅(𝑇2 − 𝑇1 ).

The energy balance can therefore be written as 𝑄 = 𝑀𝑅(𝑇2 − 𝑇1 ) + 𝑀𝑐𝑣 (𝑇2 − 𝑇1 ) In addition to the energy transfer as heat due to the electric heater, the system also transfers energy as heat to the environment therefore 𝑄 = 𝑄in − 𝑄out .

The energy transfer as heat to the system is therefore 𝑄in = 𝑀𝑅(𝑇2 − 𝑇1 ) + 𝑀𝑐𝑣 (𝑇2 − 𝑇1 ) + 𝑄out = 𝑀(𝑅 + 𝑐𝑣 )(𝑇2 − 𝑇1 ) + 𝑄out = 𝑀𝑐𝑃 (𝑇2 − 𝑇1 ) + 𝑄out .

The pressure therefore has no influence on the amount of energy as heat that must be transferred to the system.

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𝑄in = 10 kg × 1.006

kJ kWh (80 − 20)K + 50 kJ = 653.6 kJ × 1 = 0.18 kWh (kg ∙ K) 3600 kJ

EXERCISE 4 A constant volume vessel of 100 liters holds 1 kg of water at an initial pressure of 30 bar and initial temperature 𝑇1 . The vessel is then cooled to 350 K. a) Determine the initial temperature of the system 𝑇1 and the pressure after cooling.

Application of the 8-step method for energy analysis.

1) Draw a diagram identifying the control volume.

2) Show the forces if the CV moves.

3) Indicate reference frame The reference frame is inertial.

4) List all simplifying idealizations.  the system is closed (perfect seal of the piston);  the volume is constant;  kinetic and potential energy are negligible if compared to the other energy transfers.

5) indicate all the directions of the energy transfers on the sketch See sketch 6) Time basis of the energy balance.

Over the period of time in which the cooling occurs 7) Energy balance.

0 = 𝑄 + ∆𝑈 8) Bring in process, state, or other information as necessary to complete the analysis. the specific volume of the water in the initial state 1 is 𝑣1 =

𝑉 100 l dm3 m3 m3 = × × = 0.1 𝑀 1 kg 1l 1000 dm3 kg

As water is a simple compressible fluid, and two thermodynamic variables are specified, namely 𝑃1 = 30 bar and 𝑣1 = 0.1 m3 ⁄kg state 1 is fixed. The corresponding temperature can be estimated with values from the tables in Appendix A.3. Inspection of the 𝑃 − ℎ chart reveals that the state is superheated. As the values of pressure and temperature of state 1 are not listed in the table, the temperature must be obtained by interpolation of the appropriate values with 𝑃1 = 3 MPa , therefore with temperatures 650 K and 700 K , hence 3

𝑣1 = 0.09534

52

m + kg

(0.1040 − 0.09534) (700 − 650)K

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m3 kg

(𝑇1 − 650 K) = 0.1

m3 . kg

Solving this equation for 𝑇1 gives 𝑇1 = 676.9 K. The final state 2 is fixed by the conditions 𝑣2 = 𝑣1 = 0.1 m3 ⁄ kg and 𝑇2 = 350 K. On the 𝑃 − ℎ chart, intersecting the isochore with the isotherm, shows that state 2 is saturated and close to the liquid saturation curve. The pressure is therefore 𝑃2 = 𝑃sat (350 K) = 0.04168 MPa. b) Determine the energy transferred as heat during the process. Make a sketch of the process on the provided 𝑃 − ℎ chart.

From the energy balance 𝑄 = −∆𝑈 = −𝑀(𝑢2 − 𝑢1 ) The specific internal energy of state 1 is obtained by interpolation from the values listed table in Appendix A as kJ kJ (2980.5 − 2892.5) kg kJ (676.9 − 650)K = 2939.8 . 𝑢1 = 2892.5 + (700 − 650)K kg kg The quality of the liquid vapor mixture is given by (0.1 − 0.001027) m3 ⁄ kg 𝑣2 − 𝑣2L 𝑥= V = = 0.0258, 𝑣2 + 𝑣2L (3.842 − 0.001027) m3 ⁄ kg hence the specific internal energy of state 2 can be computed and is 𝑢2 = 𝑥𝑢2V + (1 − 𝑥)𝑢2L = 377.28

kJ . kg

The energy transferred as heat to the water is therefore 𝑄 = −𝑀(𝑢2 − 𝑢1 ) = −1 kg (377.28 − 2939.8)

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kJ kJ = 2562.56 . kg kg

53

The process in the 𝑃 − ℎ chart is

EXERCISE 5 A small solar engine is used for power conversion. Water enters the pump as saturated liquid at 50 ℃ and is pumped up to 2 bar by a small centrifugal pump (process 1-2). Solar energy is gathered using solar collectors, connected to the boiler. The boiler evaporates the water at 2 bar (process 2-3), and saturated vapor at this pressure enters a small turbine (process 3-4). The steam leaves the turbine at 50 ℃ and with a liquid fraction of 6 %, and is subsequently condensed (process 4-1). The mass flow rate is 140 kg/hr and the pump is driven by a 0.4 kW motor.

a) Calculate the net mechanical power output of this solar engine.

The net power output of the system is given by the power of the pump subtracted from that of the turbine. The power of the turbine can be obtained by applying the 8-step method for energy analysis to the turbine 1) Draw a diagram identifying the control volume.

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2) Show the forces if the CV moves.

The CV does not move 3) Indicate reference frame The reference frame is inertial.

4) List all simplifying idealizations.  the mass of working fluid does not change (no leakage to the environment);  the system is at steady state;  changes in potential, kinetic etc. energy between the inlet and outlet of each component are negligible;  1-dimensional flow at the inter-component interfaces;  no pressure drop in ducts, pipes, heat exchangers; 6) Time basis of the energy balance.

The energy balance is on a rate basis.

7-8) Energy balances and other relations needed to complete the analysis.

The energy is 𝑀̇ℎ3 = 𝑊̇t + 𝑀̇ℎ4 State 3 is saturated and the pressure is 2 bar, hence, from the table in Appendix A ℎ3 = ℎ(𝑥 = 1, 𝑃3 = 0.2 MPa) = 2706.2

kJ . kg

The enthalpy of state 4 can be calculated because it is a saturated state at 𝑇4 = 323.15 K and its quality is specified. The table of Appendix A provide values of saturated properties for temperatures 320 K and 325 K, therefore an interpolation is necessary in order to obtaine the values of saturated enthalpy at the specified temperature, resulting in the table 𝑻/𝑲

𝒉𝐋 / 𝐤𝐉/𝐤𝐠

𝒉𝐕 / 𝐤𝐉/𝐤𝐠

320

196.2

2585.7

325

217.1

2594.6

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323.15

209.4

2591.3

The enthalpy of state 4 is therefore ℎ4 = 𝑥ℎ4V + (1 − 𝑥4 )ℎ3L = 0.94 × 2591.3

kJ kJ kJ + (1 − 0.94) × 209.4 = 2448.4 . kg kg kg

The power output of the tubine is therefore 𝑊̇t = 𝑀̇(ℎ3 − ℎ4 ) =

140 kg 1 hr kJ × × (2706.2 − 2448.4) = 10 kW. hr 3600 s kg

The net mechanical power of the engine is 𝑊̇net = 𝑊̇t − 𝑊̇p = 10 kW − 0.4 kW = 9.6 kW. b) Compute the energy conversion efficiency taken as the net shaft work output divided by the energy transfer as heat to the fluid in the boiler.

In this case the energy analysis method must be applied to the system and to the condenser in order to obtain the energy transfer as heat to the fluid. The steps are analogous to those for the turbine. The energy balance for the system is 𝑊̇p + 𝑄̇b = 𝑊̇t + 𝑄̇c The sketch for the condenser is

therefore the energy balance reads 𝑀̇ℎ4 = 𝑄̇c + 𝑀̇ℎ1 .

The specific enthalpy at the outlet of the condenser is given by ℎ1 = ℎ(𝑇 = 323.15, 𝑥 = 0) = 209.4 thus 𝑄̇c = 𝑀̇(ℎ4 − ℎ1 ) = 87.1 kW, and 𝑄̇b = 𝑊̇t − 𝑊̇p + 𝑄̇c = 96.7 kW. The thermal efficiency of the engine is therefore

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kJ , kg

𝜂≡

𝑊̇net 9.6 kW = = 0.1 ̇ 96.7 kW 𝑄b

The efficiency is very low, but this engine is simple and no special material is needed because of the very low temperature involved, thus the cost is also possibly low. c) Estimate the area of the solar collectors, assuming the collectors can absorb 800 W/m2 .

The total area needed to obtain 𝑄̇b = 96.7 kW is given by 𝐴=

96.7 × 1000 W = 121 m2 800 W/m2

The amount of surface needed for the solar collectors is large considering the power output of the engine, and this investment cost makes the adoption of this technology less attractive then photovoltaic solar panels.

EXERCISE 6 Consider the two closed adiabatic boxes illustrated in the following sketch.

W

W

The first box accommodates a frictionless shaft transferring mechanical work to a spinning propeller. The second box accommodates a frictionless shaft transferring mechanical work to lift a weight.

Assume that the same amount of energy as work is provided to the two shafts for a fixed amount of time starting at 𝑡s and ending at 𝑡e . a) Describe in conceptual terms (in words) the changes occurring in the boxes during the transfer of energy as work to the shafts (𝑡s < 𝑡 < 𝑡e ).

In the box with the propeller, prior to the transfer of energy as work (𝑡 < 𝑡s ), the propeller and the gas in the box are at rest. Once the shaft starts rotating, work is transferred as kinetic energy to the gas molecules in the box due to the friction between the propeller and the gas, causing an increases of the internal energy of the gas and of its temperature. This increases the random molecular motion and thus the entropy of the gas. The volume is constant so the pressure of the gas also increases. This is an irreversible process.

In the box with the weight, prior to the transfer of energy as work (𝑡 < 𝑡s ), the weight and the gas in the box are at rest. Once energy as work is transferred to the shaft, the weight begins to rise. The shaft work is is converted into potential energy as the weight increases its altitude. No molecular disorganization occurs in the gas, therefore the temperature and

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the entropy of the gas remain constant. Pressure also remains constant. If all the friction occurring tin the system can be neglected this reversible. b) Describe in conceptual terms (words) the changes occurring in the boxes once the transfer of energy as work to the shafts stops (t > te).

The box with the propeller is adiabatic, thus no energy transfer as heat to the environment is possible. The internal energy, temperature and entropy of the gas remain constant and no other change can occur. At 𝑡 > 𝑡e ) the weight starts falling down transferring its potential energy to the shaft as work. The total amount of work that was transferred was accumulated as potential energy in the weight is transferred back to the shaft and the the shaft spins in the opposite direction until the weight reaches its initial position. The temperature, pressure, internal energy and entropy of the gas do not change.

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SAMPLE EXAM 5 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) Volume is an intensive property.

FALSE 2) A simple Brayton cycle engine has a higher efficiency than the Carnot engine.

FALSE 3) When compressing a gas adiabatically its internal energy increases.

TRUE 4) Color is a thermodynamic property.

FALSE 5) A closed system allows mass exchange with the surroundings.

FALSE 6) The production of entropy in the universe is zero.

FALSE 7) The production of energy in the universe is zero.

TRUE 8) Temperature has the same dimensional units as energy.

FALSE 9) The temperature is proportional to volume in an isobaric process of an ideal gas.

TRUE 10) The value of the internal energy of an ideal gas is always larger than the value of its enthalpy at a given temperature.

FALSE 11) Extracting energy as heat from a closed system reduces its entropy.

TRUE 12) Energy transfer as heat is a reversible process.

FALSE 13) A refrigerator is a heat pump.

TRUE 14) In a polytropic process the pressure can change independently of the volume.

FALSE EXERCISE 2 (Similar to exercise 1.2 of the book) A wind farm of 30 identical windmills is converting the kinetic energy of the wind into 50 MW of electrical power. The velocity of the wind blowing through the rotor disc of each windmill is 𝑉 = 7 m/s. The air density is 𝜌 = 1.2 kg/m3 . If the efficiency of each windmill is 0.63, what is the windmills rotor radius 𝑅?

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Application of the 7-steps analysis methodology 1) Definition of the system and its and boundary with a sketch 2) Indication of the reference frame see sketch 3) Idealizations  the wind velocity V is constant and uniform in the z direction;  the frontal surface of the blades and that of the hub are negligible;  the density of the air is constant;  the efficiency of the wind turbine is constant.

4) Indication of the time basis for the balance equations.

The interval is an infinitesimal amount of time 𝑑𝑡.

5) Indication of all the terms appearing in the balance equations.

See sketch.

6) Balance equations. 𝑑𝑊 = 𝑑𝐸k .

7) Other relations needed to complete the analysis The kinetic energy that accumulated within the CV over the time 𝑑𝑡 is 𝑑𝐸k =

1 1 𝑀𝑉 3 = 𝜌𝐴𝑉 2 𝑑𝑧.

2 2

The area of the rotor disk is given by 𝐴 = π𝑅 2 , therefore the energy accumulation term can be written as 𝑑𝐸k =

1 2 π 2 𝜌𝑉 ( 𝐷 ) 𝑑𝑧.

2 4

The rate of kinetic energy flow through the windmill disk is therefore 𝑑𝐸k 1 2 π 2 𝑑𝑧 𝜌π𝐷2 𝑉 3 = 𝜌𝑉 ( 𝐷 ) = . 𝑑𝑡 2 4 𝑑𝑡 8 The amount of this energy converted into electrical power is 𝜌π𝐷2 𝑉 3 𝑊̇𝑒 = η𝑊̇ = η ( ) 8 The power capacity of the wind farm is 30 times the power of one single windmill, hence 𝑊̇wind farm 𝜌π𝐷2 𝑉 3 𝜌π𝑅 2 𝑉 3 = η( ) = η( ).

30 8 2

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Solving this equation for 𝑅 gives 𝑅=√

𝑊̇wind farm = 15πηρ𝑉 3 √

50 MW 15 × π × 0.63 × 1.2

kg m 3 × (7 s) m3

= 63.97 m.

EXERCISE 3 (Similar to exercise 4.13 of the book) The throttling calorimeter is a device for determining the thermodynamic state of a liquid-vapour mixture by means of temperature and pressure measurements. The procedure is to deviate a small amount of the mixture from the main flow, pass it through a valve, and take the measurements as shown in the figure below.

Assume that the fluid flowing through the calorimeter is water.

a) Make a sketch of the process between state 1 and 2 on a 𝑇 − 𝑠 diagram, indicating the state points and the line corresponding to the throttling process.

The process of the fluid flowing through a valve is isenthalpic, therefore, after checking on the 𝑇 − 𝑠 diagram provided in Appendix A, the sketch of the process is

b) Compute the quality of the mixture at state 1, given the pressure and temperature measurements displayed in the figure below.

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Application of the 8-steps method for energy analysis.

1) Make an idealized sketch of the system and identify the boundary, indicating if it moves in time.

2) If the boundary moves, indicate the forces acting on the CV (Work).

The CV boundary is fixed with respect to the inertial reference frame.

3) If the CV moves, indicate the reference frame.

The control volume is still.

4) List the idealizations.  steady state;  kinetic and potential energy of all streams are negligible if compared to other energy transfers in the system;  the calorimeter is adiabatic;  the flow is one-dimensional at sections 1, and 2.

5) Indicate on the sketch all non-zero terms that will appear in the balance equation (arrows indicate the direction of positive transfer).

See sketch.

6) Specify the time basis for the analysis.

The analysis is on a rate basis.

7) Write the balance equation.

The energy balance for the throttling valve is 𝑀̇ℎ1 = 𝑀̇ℎ2 8) Add other equations that can be derived from the available information in order to solve for the unknowns of the problem.

In order to determine state 1, which is in vapor-liquid equilibrium, knowing 𝑇1 and 𝑃1 is not sufficient as for a saturated state we have that 𝑃 = 𝑃sat (𝑇). Another independent thermodynamic variable must be determined, for example its vapor quality.

The specific enthalpy of state 1 is

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ℎ1 = 𝑥1 ℎ1𝑉 + (1 − 𝑥1 )ℎ1𝐿 .

Therefore, if we combine it with the equation of the energy balance, we obtain an expression for the vapor quality, that is 𝑥1 =

ℎ2 − ℎ1𝐿 . ℎ1𝑉 − ℎ1𝐿

This equation shows that 𝑥1 can be determined, hence state 1, if ℎ2 can be estimated from the measurement of 𝑃2 and 𝑇2 . Only if state 2 is in a superheated vapor thermodynamic state, ℎ2 depends on both pressure and temperature.

Thus, the throttling (pressure drop caused by the valve) must be enough to vaporize all the liquid in the stream, that is 𝑃2 must be lower than the saturation temperature at 2, namely 𝑃2 < 𝑃sat (𝑇2 ). Inspection of the 𝑇 − 𝑠 or 𝑃 − ℎ diagrams shows that this limits the use of the throttling calorimeter for the estimation of the vapor quality to states with quality close to 1.

For example, if the fluid is water and we use the thermodynamic model implemented in IF97 (cf. the 𝑇 − 𝑠 diagram in Appendix A.3), the interval of specific enthalpies for which the upstream state is in the two-phase region and the downstream is superheated is limited to the range 2600 kJ / kg ≳ ℎ ≳ 2800 kJ / kg, which corresponds to 𝑥 ≳ 0.85.

In case the readings for water are 𝑃1 = 10 bar,

𝑇1 = 179.9 ℃,

𝑃2 = 1.2 bar,

𝑇2 = 120 ℃,

we can immediately verify that state 1 is saturated from the table for saturated water in Appendix A.3 because 𝑃1 = 𝑃sat (𝑇1 ).

We can then calculate ℎ2 using values from the table for saturated steam in Appendix A.3.

For superheated steam we have ℎ𝑎 = ℎ𝑎 (𝑇𝑎 ) = 393.4 K , 𝑃𝑎 = 0.2 MPa) = 2706.2 kJ / kg.

Given that 𝑇2 = 393.1𝐾 ≈ 𝑇𝑎 , in order to perform a linear interpolation, we need ℎb = ℎb (𝑇a = 𝑇b = 393.4 K, 𝑃b < 𝑃2 ). ℎ𝑏 can also be found by linear interpolation, using the values ℎc = ℎ 𝑠𝑎𝑡 (𝑇c = 373.1 K) = 2675.5 kJ / kg, ℎ𝑑 = ℎ(𝑇d = 400 K , 𝑃d = 0.101325 MPa) = 2730.3 kJ / kg, and we get ℎb = ℎb (𝑇b = 393.4 K, 𝑃b = 0.101325 MPa) = ℎc +

ℎd − ℎc (𝑇 − 𝑇c ) 𝑇d − 𝑇c b

= 2716.9 kJ / kg.

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63

We can finally calculated h_2 by linear interpolation as ℎ2 = ℎ2 (𝑇2 = 393.1 K, 𝑃2 = 0.12 MPa) = ℎb +

ℎa − ℎb (𝑃 − 𝑃b ) = 2714.8 kJ / kg. 𝑃a − 𝑃b 2

The vapor quality of state 1 that can be obtained from the reading of the throttle calorimeter is therefore 𝑥1 =

ℎ2 − ℎ1𝐿 (2714.8 − 762.7)kJ / kg = = 0.97. ℎ1𝑉 − ℎ1𝐿 (2777.1 − 762.7)kJ / kg

EXERCISE 4 13 kg of ammonia at its critical state is placed in a constant volume container, which is then cooled until the temperature is 310 K. a) How much energy was transferred as heat from the ammonia in this process?

Application of the 8-step method for energy analysis.

1) Draw a diagram identifying the control volume.

2) Show the forces if the CV moves.

3) Indicate reference frame The reference frame is inertial.

4) List all simplifying idealizations.  the system is closed (perfect seal of the piston);  the volume is constant;  kinetic and potential energy are negligible if compared to the other energy transfers.

5) indicate all the directions of the energy transfers on the sketch See sketch 6) Time basis of the energy balance.

Over the period of time in which the cooling occurs 7) Energy balance.

0 = 𝑄 + ∆𝑈 8) Bring in process, state, or other information as necessary to complete the analysis.

The energy balance can be rewritten as 𝑄 = Δ𝑈 = 𝑀(𝑢1 − 𝑢2 ), STATE 1 Using the tables of Appendix A.7, we obtain 𝑣1 = 𝑣c = 0.004444

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m3 , kg

and kJ m3 3 (11.333 𝑢1 = ℎ𝑐 − 𝑃𝑐 𝑣c = 1262.38 − × 10 kPa) × (0.004444 ) kg kg kJ = 1212.02 . kg STATE 2 Given that the container is closed and rigid 𝑣2 = 𝑣1 , and the temperature is 𝑇2 = 310 K. Inspection of the 𝑇 − 𝑠 or 𝑃 − ℎ diagrams in Appendix A.7 shows that state 2 is in the vapor-liquid region, hence its quality can be computed as m3 m3 0.004444 − 0.001708 L 𝑣c − 𝑣2 kg kg 𝑥2 = V = = 0.032 3 L m m3 𝑣2 − 𝑣2 0.09230 − 0.001708 kg

kg

The specific internal energy can therefore be calculated as 𝑢2 = (1 − 𝑥2 )𝑢2L + 𝑥2 𝑢2V .

With 𝑢 = ℎ − 𝑃𝑣, by interpolating the values corresponding to 𝑇 = 308 K and 𝑇 = 312 K we have that 𝑢2L = 515.36

kJ m3 kJ − (1.4 × 103 kPa) × (0.001708 ) = 513 , kg kg kg

and 𝑢2V = 1631.95 − (1.4 × 103 kPa) × (0.09230

m3 kJ ) = 1502.7 , kg kg

thus 𝑢2 = (1 − 𝑥2 )𝑢2L + 𝑥2 𝑢2V = (1 − 0.032)

kJ kJ kJ + 0.032 × = 544.64 kg kg kg

Thus the energy transferred as heat of the system is: 𝑄 = 𝑀(𝑢1 − 𝑢2 ) = 13 kg × (1212.02

kJ kJ − 544.64 ) = 8.67 𝑀𝐽. kg kg

b) How much of the tank volume is occupied by liquid at the end of the process? 𝑉2L = 𝑀2L 𝑣2L = (1 − 𝑥2 )𝑀𝑣2L = (1 − 0.032) × 13 kg × 0.001708

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m3 = 0.021 m3 . kg 65

c) Sketch the process on the provided 𝑃 − ℎ diagram for ammonia.

EXERCISE 5 Consider the intake stroke of an internal combustion engine under the following idealizations.     

Negligible kinetic and potential energy everywhere Uniform state inside the cylinder at any instant One-dimensional flow Ideal gas Non-adiabatic cylinder walls

a) Make a sketch of the system, including i) its boundary, ii) all the quantities involved in the mass and energy balance, and arrows indicating the positive direction for the energy transfer terms.

Mass balance

Energy balance

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b) Write volume. 𝑀̇1 =

the

mass balance on the control

𝑑𝑀 𝑑𝑡

c) Write the energy balance on the control volume in terms of temperature 𝑇, mass 𝑀, volume 𝑉, energy transfer rate as heat 𝑄̇ and gas constants 𝑐𝑃 and 𝑐𝑣 . 𝑀̇1 ℎ =

𝑑𝑈 + 𝑊̇ + 𝑄̇ 𝑑𝑡

The terms of the energy balance can be written in terms of the specified variables as 𝑀̇1 ℎ =

𝑑𝑀 𝑐 𝑇, 𝑑𝑡 𝑝

𝑑𝑈 𝑑𝑢 𝑑𝑀 𝑑𝑇 𝑑𝑀 =𝑀 +𝑢 = 𝑀𝑐𝑣 + 𝑐𝑣 𝑇 , 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑊̇ =

𝑑𝑊 𝑑 𝑑𝑥 𝑀𝑅𝑇 𝑑𝑉 = (𝑃𝐴𝑑𝑥) = 𝑃𝐴 = , 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑉 𝑑𝑡

giving 𝑑𝑀 𝑑𝑇 𝑑𝑀 𝑀𝑅𝑇 𝑑𝑉 𝑐𝑝 𝑇 = 𝑀𝑐𝑣 + 𝑐𝑣 𝑇 + + 𝑄̇ 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑉 𝑑𝑡 EXERCISE 6 Ten grams of computer chips with a specific heat 𝑐𝑃 = 𝑐𝑣 = 0.3 kJ/(kg ∙ K) are initially at 20 ℃ . These chips are placed in an adiabatic container with 5 grams of saturated liquid R134a at −40 ℃ in order to cool them down. The fluid remains saturated over the process.

Assume that during the cooling process the pressure remains constant. a) Calculate the entropy change of the chips; Application of the 8-step method for energy analysis.

1) Draw a diagram identifying the control volume.

2) Show the forces if the CV moves.

3) Indicate reference frame The reference frame is inertial.

4) List all simplifying idealizations.  the system is closed (perfect seal of the piston);

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67

    

the volume is constant; kinetic and potential energy are negligible if compared to the other energy transfers; Adiabatic container; The fluid remains saturated over the process; The process is isobaric.

5) Indicate all the directions of the energy transfers on the sketch There are no energy transfers 6) Time basis of the energy balance.

Over the period of time in which the cooling occurs 7) Energy and entropy balance. Δ𝑈 = 0,

℘𝑠 = Δ𝑆.

8) Bring in process, state, or other information as necessary to complete the analysis.

The Gibbs equation provides 𝑑𝑠chips =

𝑑𝑢chips 𝑃𝑑𝑣chips 𝑑ℎchips 𝑣chips 𝑑𝑃 + = − 𝑇 𝑇 𝑇 𝑇

Given that the process is isobaric, 𝑑𝑃 = 0 hence 𝑑ℎ = 𝑐𝑝 𝑑𝑇. the Gibbs equation can be integrated resulting in 𝑇2 Δ𝑠chips = 𝑐𝑝 ln ( ) 𝑇1 The R134a fluid is in vapor-liquid equilibrium throughout the process and also once the cooling is completed, thus the fluid is in thermal equilibrium with the chips.at 𝑇2 = −40 ℃.

The entropy change of the chips is (273.15 − 40) K 𝑇2 kJ kJ Δ𝑠chips = 𝑐𝑝 ln ( ) = 0.3 ln [ ] = −0.0687 (273.15 + 20) K 𝑇1 kg ⋅ K kg ⋅ K And in extensive form: Δ𝑆chips = 𝑀chips Δ𝑠chips = 0.010 kg × −0.0687

kJ J = −0.69 kg ⋅ K K

b) Calculate the entropy change of the R134a; The energy balance for the system is ∆𝑈 = 𝑀chips 𝑐𝑣 (𝑇2 − 𝑇1 ) + 𝑀R134a (𝑢2 − 𝑢1 ).

The specific internal energy of the R134a in the final state is therefore 𝑢2 = 𝑢1 −

68

𝑀chips 𝑐𝑣 (𝑇2 − 𝑇1 ) 𝑀R134a

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Initially, R134a is in the saturated liquid state, therefore the specific internal is obtained from Appendix A.4 as 𝑢1 (𝑇 = 233.15 K) ≈ 𝑢L (𝑇 = 232 K) = ℎ1 − 𝑃𝑣1 kJ m3 kJ 3 = −277.95 − (0.04869 × 10 kPa) × 0.000715 = −277.98 . kg kg kg The specific internal energy of R134a in state 2 is kJ 𝑀chips 𝑐𝑣 (𝑇2 − 𝑇1 ) kJ 0.01 kg × 0.3 kg ⋅ K (−40 − 20)K 𝑢2 = 𝑢1 − = −277.98 − 𝑀R134a kg 0.005 kg kJ = −241.98 . kg The quality corresponding to this thermodynamic state is kJ kJ 𝑢2 − 𝑢2L −241.98 kg − (−277.98 kg) 𝑥2 = V = = 0.174. kJ kJ 𝑢2 − 𝑢2L −70.75 − (−277.98 ) kg kg The entropy R134a in state 2 is 𝑠2 = (1 − 𝑥2 )𝑠2L + 𝑥2 𝑠2V = (1 − 0.174) × (−1.1108) = −0.8937

kJ kJ + 0.174 × (−0.1370) kg ⋅ K kg ⋅ K

kJ kg ⋅ K

Noting that 𝑠1 = 𝑠2L , the specific entropy change of the R134a is ∆𝑠 = 𝑠2 − 𝑠1 = 𝑠2 = (1 − 𝑥2 )𝑠2L + 𝑥2 𝑠2V − 𝑠2L = 𝑥2 (𝑠2V − 𝑠2L ) kJ kJ = 0.174[−0.1370 − (−1.1108)] = 0.1694412 kg ⋅ K kg ⋅ K Δ𝑆R134a = 𝑀R134a (𝑠2 − 𝑠1 ) = 0.005 kg × 0.1694412

kJ J = 0.85 kg ⋅ K K

c) Compute the entropy production due to this process; The energy balance is ℘𝑠 = Δ𝑆.

The total change of entropy of the system is Δ𝑆system = Δ𝑆R134a + Δ𝑆chips = 0.85

J J J + (−0.69 ) = 0.16 , K K K

therefore

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℘𝑠 = 0.16

J K

d) state whether this process is possible or not.

The process leads to entropy production, therefore in agreement with the second law of thermodynamics, thus the process is possible.

EXERCISE 7 The majority of current aircraft engines employ gas turbines based on the open Brayton thermodynamic cycle. Hence, being able to analyze the performance of this type of thermal engines is important for the aspiring aerospace engineer, and it is easier to start from the analysis of the ideal Brayton thermodynamic cycle. The sketch of the process flow diagram and of the temperature-entropy diagram for the ideal Brayton cycle are

The corresponding processes are as follows: 1-2: Isentropic compression 2-3: Isobaric transfer of energy as heat 3-4: Isentropic expansion 4-1: Isobaric transfer of energy as heat Assume that air is the working fluid and that it obeys the ideal gas law with constant specific heats (polytropic ideal gas).

Derive the efficiency of an engine operating according to the ideal Brayton cycle, as a 𝑐𝑝 𝑃 function of the cycle pressure ratio Π ≡ 𝑃12 and the ratio of the specific heats 𝛾 ≡ 𝑐 . 𝑣

𝑇

𝑃

2

1

𝛾−1 𝛾

Hint: for a polytropic ideal gas 𝑇1 = (𝑃2 )

The sketches of the components with arrows indicating the positive direction of energy transfers and all the terms of the rate-basis energy balances are

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The energy balances for all the components on a rate basis Compressor: Turbine: Heater: Cooler:

𝑊̇ c = 𝑀̇𝑐𝑝 (𝑇2 − 𝑇1 ) 𝑊̇ t = 𝑀̇𝑐𝑝 (𝑇3 − 𝑇4 ) 𝑄ḣ = 𝑀̇𝑐𝑝 (𝑇3 − 𝑇2 ) 𝑄ċ = 𝑀̇𝑐𝑝 (𝑇4 − 𝑇1 ) Net work output

The definition of efficiency is η ≡ Energy input as heat, and in terms of power and rate of energy input as heat is 𝑇 𝑇1 (𝑇4 − 1) 𝑊̇ t − 𝑊̇ c 𝑀̇𝑐𝑝 (𝑇3 − 𝑇4 ) − 𝑀̇𝑐𝑝 (𝑇2 − 𝑇1 ) 𝑇4 − 𝑇1 1 η= = =1− =1− . 𝑇 ̇ ̇ 𝑇3 − 𝑇2 𝑀𝑐𝑝 (𝑇3 − 𝑇2 ) 3 𝑄h 𝑇2 (𝑇 − 1) 2 For the ideal Brayton cycle whose working fluid is a polytropic ideal gas 𝛾

𝛾

𝑃1 𝑃4 𝑇1 𝛾−1 𝑇4 𝛾−1 𝑇4 𝑇3 = ⇒( ) =( ) ⇒ = , 𝑃2 𝑃3 𝑇2 𝑇3 𝑇1 𝑇2 hence 𝛾−1

𝛾−1 𝑇1 𝑃2 𝛾 η= 1− =1−( ) =1−Π 𝛾 . 𝑇2 𝑃1

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SAMPLE EXAM 6 EXERCISE 1 (TRUE/FALSE QUESTIONS) 1) A modern petrol engine cannot be more efficient than a Carnot engine.

TRUE 2) The value of the internal energy of an ideal gas is always lower than the value of the enthalpy at a given temperature.

TRUE 5) Temperature is an intensive property.

TRUE 4) The specific heat of air does not vary with the temperature.

FALSE 5) In an adiabatic process energy as heat is transferred from high to low temperature.

FALSE 6) A diesel engine with high compression ratio can be more efficient than a Carnot engine.

FALSE 7) An explosion can be regarded as a sudden entropy production.

TRUE 8) By definition 𝑐𝑝 is always smaller than 𝑐𝑣 .

FALSE 9) The production of entropy in the universe is zero.

FALSE 10) In a polytropic process the pressure can change independently of the volume.

FALSE 11) Removing energy as heat from a closed system reduces its entropy.

TRUE 12) A system formed by a simple compressible substance enclosed in a constant volume cannot transfer energy as work.

TRUE 13) In a gas turbine engine the fluid enthalpy at the outlet of the turbine is larger than at the inlet.

FALSE EXERCISE 2 What is the kinetic energy in MJ of an aircraft weighing 150 000 lbm (pound mass) and flying at 880 km/h?

For the conversion of units, use the method of multiplying by suitable units, e.g., 1 kg 1= 2.205 lbm 72

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1 1 1 kg km 1000 𝑚 ℎ 2 2 𝐸𝑘 = 𝑀 𝑉 = × 150 000 lbm × × (880 × × ) 2 2 2.205 lbm ℎ 3600 𝑘𝑚 𝑠 = 2032 MJ EXERCISE 3 Consider the isochoric (constant-volume) compression of a gas from pressure 𝑃1 to 𝑃2 in a piston-cylinder system. The increase of pressure is achieved by controlled energy transfer as heat 𝑄 to the gas, while the piston is kept fixed. a) Make a sketch of the considered system with its control volume, the appropriate nomenclature for all quantities involved in the energy balance, and arrows indicating the positive direction of energy transfers.

The sketch is

Q

U b) Treating the gas as polytropic and ideal, derive the expression for the energy input as heat 𝑄 as a function of the pressure ratio 𝑃2 /𝑃1 .

Accumulation = Input – Output.

The energy balance is therefore ∆𝑈 = 𝑄 − 0.

Recalling that Accumulation = Final − Initial The change of internal energy is therefore ∆𝑈 = 𝑈2 − 𝑈1 For a polytropic gas (constant specific heats) 𝑈2 − 𝑈1 = 𝑀𝑐𝑣 (𝑇2 − 𝑇1 ).

For an ideal gas 𝑃𝑣 = 𝑅𝑇 → 𝑃1 /𝑃2 = 𝑇1 /𝑇2 therefore 𝑃2 𝑄 = ∆𝑈 = 𝑀𝑐𝑣 (𝑇2 − 𝑇1 ) = 𝑀𝑐𝑣 𝑇1 ( − 1) 𝑃1 EXERCISE 4 5 kg of CO2 at its critical state is placed in a perfectly rigid container, which is then cooled until the temperature is 260 K. TESTBANK

73

a) What is the volume occupied by liquid at the end of the process?

STATE 1 From table A.21 in the Appendix, the properties of CO2 at the critical state are ℎ1 = 332.25 kJ/kg, 𝑣1 = 0.002139 m3 / kg,

𝑃1 = 7.3777 MPa

STATE 2 The mass and the volume do not change therefore the overall density does not change, thus 𝑣2 = 𝑣1 .

Again from table A.21 the specific volumes of saturated at 𝑇2 = 260 K are 𝑣 L = 0.001001 𝑚3 /kg,

𝑣 V = 0.0155 𝑚3 /kg.

The quality (or vapor fraction) is 𝑥 =

𝑣2 − 𝑣 L 0.002139 − 0.001001 = = 0.0785 𝑣V − 𝑣L 0.0155 − 0.001001

The specific volume of the liquid in state 2 is therefore 𝑣2𝐿 = (1 − 𝑥)𝑣2 = 0.0009224 𝑚3 /kg.

The volume occupied by the liquid at the end of the process is therefore 𝑉 𝐿 = 𝑣2𝐿 ∙ 𝑀 = 0.0009224 m3 / kg × 5 kg = 0.004612 kg. b) How much energy is transferred as heat during the process?

The energy balance over period of time is ∆𝑈 = 𝑄.

The change in internal energy of the fluid is ( ∆= 𝑓𝑖𝑛𝑎𝑙 – 𝑖𝑛𝑖𝑡𝑖𝑎𝑙) ∆𝑈 = 𝑈2 − 𝑈1 , therefore 𝑄 = 𝑀(𝑢2 − 𝑢1 ).

The values of the specific internal energy can be readily obtained as 𝑢1 = ℎ1 − 𝑃1 𝑣1 = 332.250e3 kJ/kg − (7.377e6 × 0.002139) kJ/kg = 316.4 kJ/kg, and 𝑢2 = ℎ2 − 𝑃2 𝑣2 = (ℎL + 𝑥ℎLV ) − 𝑃2 𝑣2 = (169.44 + 0.0785 × 266.48)e3 kJ/kg − (2.419e6 × 0.002139) kJ/kg = 185.1 kJ/kg.

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Therefore, the energy transferred as heat during the process is 𝑄 = 𝑀(𝑢2 − 𝑢1 ) = 5 𝑘𝑔 (185.1 − 316.4) kJ/kg = −656.5 kJ EXERCISE 5 Consider the simple steam Rankine cycle system operating at steady state, whose process flow diagram is depicted in the figure with the corresponding processes in the T-s diagram.

Water is compressed in a pump (1-2, single point in the 𝑇 − 𝑠 diagram as the temperature and entropy increase are negligible), receives energy as heat, which first increases its temperature and then evaporates it and superheats it (2-3), steam is expanded in a turbine (3-4) and then condensed (4-1). Assume that the flow is one-dimensional at all intercomponent thermodynamic states, that the kinetic energy and potential energy of the fluid at the state points is negligible. Assume in addition that the pump and the turbine are adiabatic.

The following assumptions are valid:     

The work of the pump is negligible (compressing liquid water does not require much work compared to the energy converted into mechanical work by the turbine) The enthalpy of the water at the outlet of the pump is 138 kJ/kg.

The evaporation pressure is 80 bar, the condensation temperature is 30 oC The steam at the inlet of the turbine at 400 oC The isentropic efficiency of the turbine is 0.85

For each of the 4 components, 1) Make a sketch of the pertinent control volume, with the appropriate nomenclature for all quantities involved in the energy balance and arrows indicating the positive direction of energy transfers and derive the steady-state energy balance per unit mass flow of working fluid; Mh1

Pump Energy balance: rate of energy input = rate of energy output 𝑀̇ℎ1 + 𝑊̇Ṗ = 𝑀̇ℎ2

WP

Per unit mass flow of working fluids  𝑤P = ℎ2 − ℎ1 Mh2

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Qin

Boiler 𝑀̇ℎ2 + 𝑄̇in = 𝑀̇ℎ3

Mh2

Mh3

𝑞in = ℎ3 − ℎ2

Turbine 𝑀̇ℎ3 = 𝑀̇ℎ4 + 𝑊̇T 𝑤T = ℎ3 − ℎ4

Condenser 𝑀̇ ℎ4 = 𝑀̇ℎ1 + 𝑄̇out Mh4

Mh1

𝑞out = ℎ4 − ℎ1

Qout

3) Calculate the efficiency of the system.

The properties of the working fluid are obtained from the tables in Appendix A. STATE 3 ℎ3 = ℎ3 (𝑇3 = 673 K, 𝑃3 = 8 MPa) = 3139 𝑠3 = 𝑠3 (𝑇3 = 673 K, 𝑃3 = 8 MPa) = 6.366

kJ kg

kJ Kg K

STATE 4s An interpolation is needed because saturated data are needed at 𝑇 sat (𝑃sat = 0.042 bar) = 303.15 K Saturated data reported for 300 K and 305 K, and for state 4s we need both the enthalpy and the entropy therefore we need to obtain from the tables the values of 𝑠 LV , 𝑠 L , ℎV , ℎL , resulting in

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𝑇 sat = 300 K,

𝑠 LV = 8.1244

kJ , Kg ∙ K

𝑠 L = 0.3931

kJ , Kg ∙ K

kJ , kg

ℎL = 112.6

kJ . kg

kJ , Kg ∙ K

𝑠 L = 0.4622

kJ , Kg ∙ K

kJ , kg

ℎL = 133.5

kJ . kg

ℎV = 2549.9 𝑇 sat = 305 K,

𝑠 LV = 7.9523

ℎV = 2558.9 Linear interpolation gives 𝑇 sat = 303.15 K,

𝑠 LV = 8.059

kJ , Kg ∙ K

ℎV = 2555.6

kJ , kg

𝑠 L = 0.43679 ℎL = 125.7

kJ , Kg ∙ K

kJ . kg

The quality of the isentropic state at 𝑇 sat = 303.15 K knowing its entropy 𝑥 (𝑇 sat = 303.15 K, 𝑠 = 6.37

(𝑠4𝑠 − 𝑠 L ) 6.37 − 0.43679 kJ )= = = 0.74 Kg ∙ K 𝑠 LV 8.059

With the quality, the enthalpy of the isentropic turbine outlet can be computed as ℎ4𝑠 = (1 − 𝑥)ℎL + 𝑥ℎV = 1924 kJ/kg The specific work of the turbine is 𝑤T = 𝜂T 𝑤T,s = 𝜂T (ℎ3 − ℎ4,𝑠 ) = 0.85 (3139 − 1924) = 1033 kJ/kg The specific energy input as heat is 𝑞IN = ℎ3 − ℎ2 = (3139 − 138)

kJ = 3001 kJ/kg kg

The efficiency of the system is 𝜂=

𝑤T = 0.34 𝑞IN

EXERCISE 6 Consider a turbine operating at steady state. The working fluid enters the control volume in state 1 and leaves it in state 2. The boundary of the control volume that separates the turbine from a large environment is at temperature 𝑇0 . The turbine reversibly rejects energy as heat to the environment. Assume that the kinetic and potential energy of the entering and leaving flow can be neglected. a) Derive the expression for the mechanical power output as a function of thermodynamic properties of the inlet and outlet states, of the temperature of the environment and of the rate of entropy production within the process.

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𝑇ℎ𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑖𝑠 𝑀̇ℎ1 = 𝑀̇ℎ2 + 𝑊̇ + 𝑄̇0 .

The entropy balance is ℘̇𝑠 = (𝑀̇ 𝑠2 +

𝑄̇0 ) − 𝑀̇𝑠1 . 𝑇0

Substituting 𝑄̇0 from the energy balance into the entropy balance gives 𝑊̇ = 𝑀̇(ℎ1 − 𝑇0 𝑠1 ) − 𝑇0 𝑀̇(ℎ2 − 𝑇0 𝑠2 ) − 𝑇0 ℘̇𝑠 b) Derive the expression for the maximum mechanical power of the turbine.

The maximum amount of power that can be converted by the turbine is obtained from the expression of the mechanical power in case the rate of entropy production is zero, that is 𝑊̇𝑚𝑎𝑥 = 𝑀̇(ℎ1 − 𝑇0 𝑠1 ) − 𝑇0 𝑀̇(ℎ2 − 𝑇0 𝑠2 ).

EXERCISE 7 Consider a two-part isolated system whereby the total volume of the system and of the parts are constant.

Demonstrate that if the two parts are approaching thermal equilibrium, the energy transfer as heat between the two parts must be from the part at higher temperature to the part at lower temperature. The relation between the thermodynamic temperature and the entropy is 𝜕𝑢 𝑇=( ) 𝜕𝑠 𝑣 The sketch of the system is

The volumes are all constant. The only energy exchange that can occur internally is as heat.

Under the assumption of system isolation 𝑈C = 𝑈A + 𝑈B = const. → 𝑑𝑈A = −𝑑𝑈B The system is approaching thermal equilibrium therefore entropy is at a maximum, hence 𝑑𝑆C = 𝑑𝑆A + 𝑑𝑆B > 0 The entropy function for a simple compressible substance can be expressed as a function of internal energy and volume, therefore 𝑆 = 𝑆(𝑈, 𝑉) → 𝑑𝑆 = (

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𝜕𝑆 𝜕𝑆 ) 𝑑𝑈 + ( ) 𝑑𝑉. 𝜕𝑈 𝑉 𝜕𝑉 𝑈

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The volume is constant, therefore 𝜕𝑆 𝜕𝑆 𝜕𝑆 𝜕𝑆 𝑑𝑆𝐶 = ( ) 𝑑𝑈A + ( ) 𝑑𝑈B = [( ) − ( ) ] 𝑑𝑈A > 0 𝜕𝑈 𝑉,A 𝜕𝑈 𝑉,B 𝜕𝑈 𝑉,A 𝜕𝑈 𝑉,B The definition of thermodynamic temperature is 𝜕𝑈 𝑇≡( ) , 𝜕𝑆 𝑉 therefore we obtain 𝑑𝑆𝐶 = [

1 1 − ] 𝑑𝑈A > 0. 𝑇A 𝑇B

If 𝑑𝑈A is greater than zero, the energy of the molecules of A has increased, and the only mechanism for this to happen is that an infinitesimal amount of energy as heat 𝑑𝑄 has been transferred from B to A. The inequality obtained from the application of the second law of thermodynamics to system C implies that if 𝑑𝑈A > 0 it must be that 1 1 − > 0, 𝑇𝐴 𝑇𝐵 and therefore that 𝑇𝐵 > 𝑇𝐴 , which shows in mathematical terms that if there is an energy transfer as heat from B to A, the internal energy of A increases and the temperature of B must be higher than that of A. Vice versa, if 𝑑𝑈𝐴 < 0, the opposite is true and 𝑇𝐵 < 𝑇𝐴 .

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