Page 133 - Water Engineering Hydraulics, Distribution and Treatment

◦
T = absolute temperature = 82 + 460 = 542 R.
a
◦
◦
2
3
= P /[R T ] = (34.5 × 144 lb/ft )/[(53.3 ft/ R)(542 R)] = 0.172 lb/ft .
g a
a
a
3
W = 0.55 lb/s = Q = (0.172 lb/ft )(Q).
a
3
Q = 3.19 ft /s = Av = A(19 ft/s).
2
2
A = 3.19∕19 = 0.168 ft = D × 0.785.
0.5
D = (0.168∕0.785)
= 0.46 ft = 5.55 in.
Solution 2 (SI System):
2
P = 239.4 kPa = 239.4 kN/m .
a
R = 0.287 kJ/kg-K = 29.26 m/K.
g
2
= P /[R T ] = (239.4 kN/m )/[(29.26 m/K)(27.8 + 273)K]. 5.1 Fluid Mechanics, Hydraulics, and Water Transmission 111
g a
a
a
3
3
3
= 0.027 kN/m = 0.027 × 1,000∕9.81 kg/m = 2.752 kg/m .
a
Here 1 kg = 9.81 N.
3
W = 0.25 kg/s = Q = (2.752 kg/m )Q.
a
3
Q = 0.09084 m /s = Av = A(5.8 m/s).
2
2
A = 0.01788 m = D × 0.785.
D = (0.01788∕0.785) 0.5 = 0.15 m = 15 cm.
EXAMPLE 5.3 COMPRESSIBLE CARBON DIOXIDE GAS FLOW CALCULATIONS
Carbon dioxide passes point A in a 3.5 in. (0.0889 m) pipe at a velocity of 16 ft/s (4.88 m/s). The pressure at point A is 32 psi (222
◦
◦
◦
kPa) and the temperature is 70 F(21 C). At point B downstream the pressure is 22 psi (152.7 kPa) and the temperature is 85 F
◦
(29.4 C). For a barometric reading of 14.7 psi (102 kPa), determine the gas velocity at point B, and the gas flows at points A and B.
◦
The gas constant of carbon dioxide is 38.86 ft/ R (0.209 kJ/kg-K).
Solution 1 (US Customary System):
P = absolute pressure = P gauge + P atm = 32 + 14.7 = 46.7 psi.
a
2
2
2
2
P = (46.7 lb/in. ) (144 in. /ft ) = 6,724.8 lb/ft .
a
◦
T = absolute temperature = T + 460 = 70 + 460 = 530 R.
a
= P ∕[R T ] = [(32 + 14.7) × 144]∕[38.86(70 + 460)].
A
g a
a
◦
◦
3
2
= 6,724.8 lb/ft ∕[(38.86 ft/ R)(530 R)] = 0.326 lb/ft .
A
Similarly can be determined:
B
3
= P /[R T ] = [(22 + 14.7) × 144]∕[38.86(85 + 460)] = 0.249 lb/ft .
B a g a
W (lb/s) = ( )(A )(v ) = ( )(A )(v ) = ( )(Q ) = ( )(Q ).
A A A B B B A A B B
Since A = A ,then( )(v ) = ( )(v ).
A B A A B B
3
3
(0.326 lb/ft )(16 ft/s) = (0.249 lb/ft )(v ).
B
v = 20.9 ft/s.
B
2
2
A = A = (3.5∕12) (3.14∕4) = 0.0668 ft .
A
B
2
3
Q = (A )(v ) = (0.0668 ft )(16 ft/s) = 1.068 ft /s.
A
A
A
3
2
Q = (A )(v ) = (0.0668 ft )(20.9 ft/s) = 1.396 ft /s.
B
B
B
Solution 2 (SI System):
2
P = absolute pressure = P gauge + P atm = 222 + 102 = 324 kPa = 324 kN/m .
a
◦
R = 0.209 kJ/kg-K = 209 J/kg-K = 209 × 0.101972 kg-m/kg-K = 21.31 m/K = 38.86 ft-lb/lb- R.
g
T = absolute temperature = T + 273 = 21 + 273 = 294 K.
a
3
2
= P /[R T ] = (324 kN/m )/[(21.31 m/K)(294 K)] = 0.0517 kN/m .
g a
a
A
3
3
= (0.0517 × 1,000)∕9.81 kg/m = 5.27 kg/m .
A