AE 2010 Thermodynamics (Problem Set #3 Gas State Equations, work and phase change) NEW VERSION UPDAT

AE 2010 Thermodynamics (Problem Set #3: Gas State Equations, work and phase change) NEW VERSION UPDATE WITH SOLUTION Georgia Institute Of Technology • Always indicate any assumptions you make. If you use any results or equations from the class notes or text in you solutions, please note and reference them (but you better be sure they are applicable to the problem at hand).

• Show all your work, no credit for just answers. When applicable, try to solve the problem algebraically first. Only use numbers/values in the final steps of your solution – and be sure to include units when you insert numbers.

• If the problem statement is given in ENGLISH units, the answer must also be in English units; if the problem statement is in SI units, the answer must be in SI units.

1. Gas Storage A small satellite being designed requires a nitrogen storage tank to store propellant for the cold gas thruster used to maintain the satellite’s orientation. The requirements call for 50.0 kg of N2 to be stored in a rigid container at 5°C and 65.0 bar when the satellite is deployed in space. Assuming that nitrogen is an ideal gas, what is the minimum required size (i.e., volume) of the storage tank?

While the satellite awaits launch, it is estimated that the temperature in the tank could reach as high as 45°C. If this happens, what will be the maximum internal pressure that the storage vessel must be able to handle (again, assume nitrogen is an ideal gas)?

2. Work and Energy A closed vessel contains 10 lbm of an ideal gas with a molecular weight of 60.1 lbm/lbmole and a specific heat ratio of 1.23. The gas is initially at a pressure of 2.0 atm and a temperature of 140°F. The gas pressure is then tripled without changing the density of the gas. Find the changes in internal energy, U, and enthalpy, H, of the gas, during this process AND the amount of work done to the gas. You MUST include a sketch of the process on a p-v diagram.

3. Ideal Gases – Internal Energy and Enthalpy Consider a process that takes an ideal gas between an initial state (350.K, 1.0 atm) and a final state (1000. K, 10.0 atm). Calculate the specific enthalpy and internal energy changes (i.e., per unit mass) for such a process for two substances: H 2 and H2O. Calculate your answer using the following three different approaches: (a) assume the specific heats are constants (i.e., the calorically perfect gas assumption) - and use a ROOM TEMPERATURE value for the specific heats; (b) allow the specific heat to vary with temperature and use the polynomial expressions for specific heats found in Table D.11;

(c) use the tabulated values for u and h of the gases found in the appropriate Tables in your text (D.4 and D.7)

4. Water - Energy Change and Work 0.50 lbm of water is contained in a closed piston-cylinder system initially at 300. F and 500. psia. If the water is heated at constant pressure: (a) what is the final temperature if the final volume is 0.40 ft3; (b) what is the enthalpy change (i.e., final minus initial) of the water if the final volume is 0.40 ft3; (c) what is the enthalpy change if the final temperature is 600. F; (d) how much work is done if the final temperature is 600. F.

Problem 1. Gas Storage Given: 50.0 kg of N2 initially in closed, rigid container that eventually is heated to conditions shown in figure.

T2=45C=318K

Find: (a) V1min (b) p2 (c) Estimate of V error due to ideal gas assumption Assume: ideal gas for parts (a) and (b), at Q equilibrium Analysis: a) Volume We know our initial state because we know (p,T), so we can use a state equation to find other state variables. In this case, use ideal gas state eqn. pV = mRT

Nm 8 314 R RTNm kmolK V

= 50 .0 kg p

p

278K

m 0 1 605 bar

28.01kg kmol



635m3 = 77 8

kg m3

50.0kg .

0

bar

b) p2 Since V=const. and the system is closed (m=const), we know density (or specific volume) also does not change, so state 2 is also known (v2, T2). Therefore, we can again use a state equation to find the other state variables.

 p1   mR   T2 =  T2  V   T1 

2m RT

V2

since (mR/V ) is a constant in this problem. So  605 bar  = (318K ) = 74 bar   2  278K  c) V1 error

Videal − real

First let’s define the fractional error

Verror =

1− = Vreal

Videal Vreal

We need some “real” gas EOS information about N2 at our conditions (278K, 65bar). There are a number of possible options. We could find a new EOS for N2, or we could estimate the compressibility using the generalized compressible chart to find Zpv/RT. Using the latter approach, pm( RT ) 1 Verror = −1 1 ( mRT Z p) To find Z, we first need to find the critical temperature and pressure for N2.

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From Appendix H of Black and Hartley or from http://encyclopedia.airliquide.com, we get Tcrit=126K, pcrit=34 bar.

the

web,

e.g.,

So we have for the reduced temperature and pressure: 65bar 278 K .2 pr .91 r 126 K 34bar Examining Figure E.3 of the text (or from many places on the web), we find Z0.98.

1 = − So V 1 .02 or the ideal-gas-based volume will be a few % LARGER than the error .098 necessary volume.

ALTERNATIVELY, We can calculate the N2 specific volume (or density from an EOS like the BeattieBridgeman EOS given in Table H.2 in the textbook – or using an online calculator like http://www.peacesoftware.de/einigewerte/stickstoff_e.html. Using the latter, we find

 = 79.99kg m3 which means our ideal gas density was 1.6% too low (or – again - our volume calculated from the ideal gas EOS was about 2% too high).

Implications: 1) If our storage tank was a sphere, it would have a diameter of ~1 m. So it takes a pretty big tank to hold 50 kg of a gas, even at high pressure.

2) Even though the gas is being stored at a pressure almost twice the critical pressure, the fact that the temperature is well above the critical temperature means the gas is still nearly ideal.

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−

Problem 2. Work & Energy Change Given: 10.lbm of ideal gas with MW=60.1,  =1.23

p1=2.0 atm T1=140.F=600.R m1=10. lbm

Find: (a) U12 (kJ) (b) H12 (kJ) (b) W12

p2=3 p1 v2= v1 m2=m1

p

Assume: • calorically perfect gas (since one value of  given) • no work but compression/expansion work • volume stays constant during the ENTIRE process Analysis:

v

a) U12 Using the caloric equation of state of an ideal gas AND assuming cal. perfect T2

U =



m u12

T1

R MW

)

= mc (T vT vdT



( − )

−1

12

We know everything in this equation except T2. To get the final temperature, we can use the ideal gas eq’n of state since we know two intensive variables at state 2, i.e., p2 and v2 (actually we don’t know v2, but we know it is the same as v1 which we can get from a state equation too).

v2 R

p1 R

v1

p  RT    = 3T1 R  p1 

or

T2

T1

T1

=0.033 BTU/lbmR =  S o

U12

2(600.R) = 1730BTU

10.lbm

b) H12

U m H + 



U12 + ( RT )12 =  12 H12

12

mR T12

 BTU  1200R = 2120BTU 1730BTU +10lbm  033 lb m R  

or we could have gotten there even more simply for our TPG/CPG …

)=

12

.123(1730BTU ) = 2120BTU

c) With the standard expression for compression/expansion work

=  12 + (pV )12 =

-5-

2

 pdV

0 since there is no change in volume in our assumed path. There could also

1

be other kinds of work being done, but we assumed there was none.

Implications: 1) In this case the internal energy increases without NET work being done to the system – this suggests there must have been some heat transfer to raise the temperature 2) Note that if we had not had V stay constant during the complete process (for example look at the path shown in the figure), then there WOULD p have been net compression work done to the gas in the vessel. (In the example path shown, the work done in the third part of the path to the gas would be greater than the work done by the gas as it expanded in the first part of the path.)

3) So if we had NOT assumed a fixed volume for the whole path and NOT assumed that v compression work was the only possible work, then we could not have provided an answer for part c. Without knowing the process involved in going from state 1 to state 2, you can’t figure out how much work was required.

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Problem 3. Ideal Gases – U and H

T

T2=1000. K p2=10.0 atm

Given: Simple compressible substance, H2 or H2O undergoing change shown in figure.

Find: h12 and u12 for T1=350. K (a) calor. perfect (with room T values) p1=1.0 atm (b) polynomial fits from D.11 (c) tables (D.4 and D.7) Assume: Ideal gases Analysis: We can use the following for thermally perfect gases:

ch = 

12

cu = 

12

 p dT and  vdT

(1)

MW

(3)

p

(2)

p

then

or  hu−

R

12 =  12

MW

(

)

(4)

a) For calor. =perfect specific So eq (1) gives: dTassumption, c ( ) and uheatsc are ( constants. ) 212

 p

p

212 1

v

From Table D.9 cp,H2=14.31 kJ/kgK, cp,H2O =1.865 kJ/kgK, cv,H2=10.19 kJ/kgK, cv,H2O =1.404 kJ/kgK

−

)=650K.

12,HH2

kg O12,

kg

2

Similarly from eq (2)

12,HH2

kg

12, O

2

kg

b) From Table D.11, (T in Kelvin) a cp b = a + bT + cT 2 + dT 3 + eT 4 with R c d e

H2 3.057 2.677 10−3 − 5.180 10−6 5.52110−9 − 1.812 10−12

(

T2

H 2O 4.07 − 1.108 10−3 4.152 10−6 − 2.964 10−9 8.07 10−13

)

2 3 4 From eq (1), h12 = c p dT =  R a + bT + cT + dT + eT dT T1

1000 R  T2 T3 T4 T5  + c + d + e h12 =  aT + b MW  2 5  350 3 4

So using the values from above h

and from eq (4), u

= h 12,H2

−

12,H2

= 10340

kJ

8.314 kJ kmolK

= 1350

, h

650K = 10340

2.016 kg kmol

kJ

kJ

kJ − 4.124(650) kg kg

u12,H 2 = 7660 kJ kg Similarly

u

= 1350 − 12,H2O

c) From Table D.4,

8.314 18.02

650.

kJ kg

= 1050

kJ kg

h12,H 2 = h(1000K )− h(350K ) = (14459 − 9945)

kJ

= 9510

kJ

kg kJ = 1350 = h ( 1000K ) − h ( 350K ) = ( 1993.0 − 6436.74 ) h from Table D.7, 12,H2O kg kg

Similarly, and

kg kJ

and

kJ kJ = (10334 − 3502) = 6830 and 12,H2 kg kg Values given to the same kJ kJ u = (1531.5 − 485.2) = 1050 ~number of significant figures as 12,H2O in part (b) kg kg u

For H2O, these are essentially the same numbers obtained from the polynomial fits to cp in part b, BUT NOT for the H2 results.

Implications: 1) The results for H2 using the table data are close (within ~2-3%) to the ones we found using the simpler calorically perfect gas assumption (case a) meaning H2 is close to being calorically perfect from 350-1000K. 2) HOWEVER for H2O, the change in the values are somewhat larger (~10-13%) comparing the (room T) constant specific heat calculation to the more accurate table data. Therefore, we can conclude that the calorically perfect assumption is less valid for H2O from 350-1000K. 3) The polynomial fit data for the H2O specific heat provides essentially the same answer as the table data, BUT the polynomial fit given for H 2 produces significantly different results from the table data. This suggests the H2 polynomial may be a poor fit to its actual cp/R values.

For example, another source (NASA tables) suggests the following polynomial values for H2’s cp/R: a 3.337279

B -4.94024710-05

c 4.99456810-07

d -1.79566410-10

e 2.00255410-14

Problem 4. Water - Energy Change and Work Given: Closed piston/cylinder (Vconst) undergoing p=constant process and initial conditions shown Find: (a) T2 if V2=0.40 ft3 (b) h2-h1 if V2=0.40 ft3 (c) h2-h1 if T2=600.F (d) W12 if T2=600.F Assume: process slow enough to be quasiequil.

Analysis: a) T2

p2=p1

m=0.50 lbm T1=300. F p1=500. psia

p

At state 2, we know p2=p1=500.psia and v2=V2/m=0.4 ft3/0.50lbm=0.8 ft3/lbm we have from B.1E that at 500 psia vf=0.01975 ft3/lbm and vg=0.92806 ft3/lbm.

Since our specific volume is between these two  saturated mixture

2 1

2 T

v

467 F b) h2-h1

 T ofind h1, we need to find the initial state. Since T1 = 300 F

Tsat (500 psia) , the initial

state is a compressed liquid. From Table B.4E, 2

(500 sia,300.)

70 58 BTU lbm

To find h2, we can find the quality at state 2. From Table B.2E at 500 psia,

v2 −

f

x = 2

.(0 =

800−

019749) ft3 lb m 3

v fg

.090831 ft

=.0859

lbm

Then fg

−

(10958

2

449.47 BTU lbm + 755.05.( 859)BTU lbm = 1095 8

270.58)BTU lbm = 828 BTU lbm

This process is shown on the p-v state diagram above.

BTU lbm

c) h2-h1 for T2=600F At our new state 2 (let’s call it 2), we now have p2=500.psia and T2=600.F.

467 F , state 2 is a superheated vapor. From Table B.3E, h(500 sia,600 ' 2

2

(1297.9

−

) = 1297

BTU lbm

70 58)BTU lbm = 1027 BTU lbm

This process is also shown on the p-v diagram above.

d) W12 To calculate the work in the expansion process 1→2 (with our quasi-equil. assumption) we can use 2 '

1

since pressure is constant during the whole process.

F = p mTable B.4E, v ro 1 H O

F = p m Table B.3E, ro ' 2 H O

(500 sia,300 (500 sia,600

= v 2

v =v

) .0 17415 ft3 lb m ) .1580 ft3 lb m

2

So

pm(v2 v ) lb f 144in 2 (.05 0 l b )(.1580 = 500.

2 m 2 in ft f

−.0 1 7 4 1 5 )

ft3

lbm 5 = 58 2 BTU (= 5.7

f

kJ )

778 ftlb f

Implications: 1) Work is done BY the water as it expands from a liquid to a gas, and since most of the volume change occurs as it evaporates (liquid to gas), that is where most of the work is done 2) Comparing the work done in this process (52.8 BTU or 105.6 BTU/lb m) to the enthalpy change (1027 BTU/lbm), we get the idea that the energy change in the process is MUCH greater than the work done (this implies the process requires a lot of heat transfer into the water, much more than the work done).