Page 134 - Water Engineering Hydraulics, Distribution and Treatment

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Water Hydraulics, Transmission, and Appurtenances
Chapter 5
Similarly can also be determined:
B
= P /[R T ] = (152.7 + 102)/[21.31(29.4 + 273)] = 0.0395 kN/m .
g a
a
B
3
3
= (0.0395 × 1,000)∕9.81 kg/m = 4.029 kg/m .
B
W (kg/s) = ( )(A )(v ) = ( )(A )(v ) = ( )(Q ) = ( )(Q ).
A
A
B
B
A
A
A
B
Since A = A ,then( )(v ) = ( )(v ).
A
B
A
A
B
B
3
3
(5.27 kg/m )(4.88 m/s) = (4.029 kg/m )(v ).
B
v = 6.38 m/s.
B
2
2
A = A = (0.0889) (3.14∕4) = 0.0062 m .
A
B
3
2
Q = (A )(v ) = (0.0062 m )(4.88 m/s) = 0.03 m /s.
A
A
A
2
3
Q = (A )(v ) = (0.0062 m )(6.38 m/s) = 0.04 m /s. B B 3
B
B
B
EXAMPLE 5.4 COMPRESSIBLE AIR FLOW CALCULATIONS
◦
◦
Air flows in a 6.5 in. (165 mm) pipe at a pressure of 35 psi (242.9 kPa) gauge and a temperature of 100 F (37.78 C). If the barometric
pressure is 14.7 psi (102 kPa) and the velocity is 10.8 ft/s (3.29 m/s), determine the air mass flow (lb/s and kg/s) when air is flowing.
◦
Gas constant for air is 53.3 ft/ R absolute (0.287 kJ/kg-K).
Solution 1 (US Customary System):
The gas laws require absolute units for temperature and pressure, when using Eq. (5.2).
P = absolute pressure = P gauge + P atm = 35 + 14.7 = 49.7 psi.
a
2
2
2
2
P = (49.7 lb/in. ) (144 in. /ft ) = 7,156.8 lb/ft .
a
◦
R = 53.3 ft/ R absolute.
g
◦
T = absolute temperature = T + 460 = 100 + 460 = 560 R absolute.
a
◦
◦
3
2
= P /[R T ] = (7,156.8 lb/ft )/[(53.3 ft/ R)(560 R)] = 0.24 lb/ft for air.
a
g a
W = mass flow = ( )(A)(v) = ( )(Q).
2
2
3
W = (0.24 lb/ft )[(6.5∕12) (3.14∕4) ft ](10.8 ft/s) = 0.597 lb/s.
Solution 2 (SI System):
2
P = absolute pressure = P gauge + P atm = 242.9 + 102 = 344.9 kPa = 344.9 kN/m .
a
◦
R = 53.3 ft/ R absolute = 0.287 kJ/kg-K = 287 × 0.101972 kg-m/kg-K = 29.26 m/K.
g
Here 1 J = 0.101972 kg-m.
T = absolute temperature = T + 273 = 37.78 + 273 = 310.8 K absolute.
a
2
3
3
= P /[R T ] = (344.9 kN/m )/[(29.26 m/K)(310.8 K)] = 0.0379 kN/m for air = 3.87 kg/m .
g a
a
Here 1 kg = 9.81 N.
W = mass flow = ( )(A)(v) = ( )(Q).
3
2
W = (3.87 kg/m )[(0.165 m) (3.14∕4)](3.29 m/s) = 0.272 kg/s.
EXAMPLE 5.5 CALCULATION OF VELOCITIES IN WATER PIPES
When 550 gpm (2,082 Lpm) water flows through a 12 in. (30.48 cm) pipe, which later reduces to a 6 in. (15.24 cm) pipe, calculate
the average velocities in the two pipes.
Solution 1 (US Customary System):
3
Q = (A )(v ) = (A )(v ) = 550 gpm = 1.224 ft /s.
1
1
2
2
2
2
2
A = (12∕12) × 0.785 ft = 0.785 ft .
1