Fundamentals of Hydraulic Engineering Systems, 5th edition Robert J. Houghtalen Solution Manual

Type:

Solution Manual

Resource:

Fundamentals of Hydraulic Engineering Systems

Edition:

5th Edition

Author(s):

Robert Houghtalen A. Osman Akan Ned Hwang

Contents 1

FUNDAMENTAL PROPERTIES OF WATER

1

2

WATER PRESSURE AND PRESSURE FORCES

7

3

WATER FLOW IN PIPES

19

4

PIPELINES AND PIPE NETWORKS

31

5

WATER PUMPS

68

6

WATER FLOW IN OPEN CHANNELS

83

7

GROUNDWATER HYDRAULICS

97

8

HYDRAULIC STRUCTURES

109

9

WATER PRESSURE, VELOCITY, AND DISCHARGE MEASUREMENTS

120

10 HYDRAULIC SIMILITUDE AND MODEL STUDIES

126

11 HYDROLOGY FOR HYDRAULIC DESIGN

135

12 STATISTICAL METHODS IN HYDROLOGY

162

Chapter 1 – Problem Solutions 1.2.1

1.2.3

E1 = energy released in lowering steam temperature to 100C from 110°C

E1 = energy needed to vaporize the water E1 = (300 L)(1000 g/L)(597 cal/g)

E1 = (500 L)(1000 g/L)(10C)(0.432 cal/g∙C)

E1 = 1.79x108 cal

6

E1 = 2.16x10 cal The energy remaining (E2) is: E2 = energy released when the steam liquefies E2 = (500 L)(1000 g/L)(597 cal/g) E2 = 2.99x108 cal

E2 = ETotal – E1 E2 = 2.00x108 cal – 1.79x108 cal E2 = 2.10x107 cal

E3 = energy released when the water temperature is lowered from 100°C to 50C E3 = (500 L)(1000 g/L)(50C)(1 cal/g∙C)

The temperature change possible with the remaining energy is: 2.10x107 cal = (300 L)(1000 g/L)(1 cal/g∙C)(T)

7

E3 = 2.50x10 cal;

T = 70C, making the temperature

Thus, the total energy released is: T = 90C when it evaporates. Etotal = E1 + E2 + E3 = 3.26x108 cal _________________________________________

Therefore, based on Table 1.1, P = 0.692 atm _________________________________________

1.2.2 First, convert kPa pressure into atmospheres:

1.2.4 84.6 kPa(1 atm/101.4 kPa) = 0.834 atm E1 = energy required to warm and then melt the ice From Table 1.1, the boiling temperature is 95°C E1 = (10 g)(6C)(0.465 cal/g∙C) + 10g(79.7 cal/g) E1 = energy required to bring the water temperature to 95C from 15°C

E1 = 825 cal. This energy is taken from the water.

E1 = (900 g)(95C - 15C)(1 cal/g∙C)

The resulting temperature of the water will decrease to:

E1 = 7.20x104 cal

825 cal = (0.165 L)(1000 g/L)(20C - T1)(1 cal/g∙C)

E2 = energy required to vaporize the water

T1 = 15.0C. Now we have a mixture of water at 0°C (formerly ice) and the original 165 liters that is now at 15.0 C. The temperature will come to equilibrium at:

E2 = (900 g)(597 cal/g) E2 = 5.37x105 cal

[(0.165 L)(1000 g/L)(15.0C - T2)(1 cal/g∙C)] = 5

Etotal = E1 + E2 = 6.09x10 cal

[(10 g)(T2 - 0C)( 1 cal/g∙C)] ; T2 = 14.1C

1