Page 172 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
EXAMPLE 5.45 MASS FLOW OF INCOMPRESSIBLE FLUID
Determine the mass-flow rate at section B of a pipe line if the static pressure at section A is 102 psig (707.9 kPa = 707.0 kN/m )and
the 4-in. (101.6-mm)-diameter pipe is flowing full with turbulent flow at an average velocity of 32 ft/s (9.75 m/s).
Solution 1 (US Customary System):
Q = Av.
Q = Av = mass flow
3
= (62.4lb∕ft )[( ∕4)(4∕12) ft ](32 ft∕s).
Mass flow = 174 lb/s.
Solution 2 (SI System): 2 2 2
Q = Av.
Q = Av = mass flow
3 2 2
= (9.8kN∕m )[( ∕4)(0.1016) m ](9.75 m∕s).
Mass flow = 0.77 kN/s.
Mass flow rate is the same at all sections. Mass flow at Section A equals mass flow at Section B
EXAMPLE 5.46 MASS FLOW OF COMPRESSIBLE FLUID
Determine the mass-flow rate of air traveling in a long length of 1-in. (25.4-mm)-diameter pipe. At section A, the pressure is 31
◦
◦
2
psia (215.14 kPa = 215.14 kN/m ), the temperature is 300 F (148.88 C), and the air velocity is 32 ft/s (9.75 m/s). At the downflow
2
section B, the pressure has been reduced by friction and heat loss to 21 psig (145.7 kPa = 145.7 kN/m )(R , the gas constant of air =
g
◦
53.34 ft-lb/lb- R (286.9 J/kg-K); 1 J = 0.101972 kg-m; and 1 N = 0.101972 kg.)
Solution 1 (US Customary System):
R = PV/T.
g
◦
◦
2
53.4 ft-lb/lb- R = (31 × 144 lb/ft )(V)/(300 + 460 R).
3
V = 9.081 ft /lb.
3
3
= 1∕V = 1∕9.081 ft /lb = 0.11 lb/ft .
Mass flow = Q = Av
2
3
2
= (0.11 lb∕ft )[( ∕4)(1∕12) ft ](32 ft∕s)
= 0.019 lb∕s.
Solution 2 (SI System):
R = PV/T.
g
2
286.9 J/kg-K × (0.101972 kg-m/J) = (215.14 kN/m )V/(148.88 +275)K.
3
V = 57.36 m /kN.
3
= 1∕V = 1∕57.46 = 0.0174 kN/m .
Mass flow = Q = Av
3
2
2
= (0.0174 kN∕m )[( ∕4)(0.0254) m ](9.75 m∕s)
= 0.0000859 kN∕s
= 0.00875 kg∕s
= 8.75 g∕s.