Fluid Mechanics
Chapter 8 Flow in Pipes 8-76E Solution The flow rate through a piping system between a river and a storage tank is given. The power input to the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The flow is turbulent so that the tabulated value of the loss coefficients can be used (to be verified). 4 The elevation difference between the free surfaces of the tank and the river remains constant. 5 The effect of the kinetic energy correction factor is negligible, = 1. Properties The density and dynamic viscosity of water at 70 F are 6.556 10-4 lbm/ft s. The roughness of galvanized iron pipe is = 0.0005 ft. = 62.30 lbm/ft3 and = 2.360 lbm/ft h =
Analysis The piping system involves 125 ft of 5-in diameter piping, an entrance with negligible loses, 3 standard flanged 90 smooth elbows (KL = 0.3 each), and a sharp-edged exit (KL = 1.0). We choose points 1 and 2 at the free surfaces of the river and the tank, respectively. We note that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm), and the fluid velocity is 6 ft/s at point 1 and zero at point 2 (V1 = 6 ft/s and V2 =0). We take the free surface of the river as the reference level (z1 = 0). Then the energy equation for a control volume between these two points simplifies to
P1 g
1
V12 2g
z1
h pump, u
P2 g
2
V 22 2g
z2
h turbine, e
hL
1
V12 2g
h pump, u
z2
hL
where
1
= 1 and
V2 L h L h L ,total h L ,major h L ,minor f KL 2 D 2g since the diameter of the piping system is constant. The average velocity in the pipe and the Reynolds number are V Re V Ac VD V D2 / 4 1.5 ft 3 /s (5 / 12 ft) 2 / 4
3
2
5 in 125 ft 12 ft
Water tank
11.0 ft/s
1
435,500
River
1.5 ft3/s
(62.3 lbm/ft )(11.0 ft/s)(5/12 ft) 6.556 10
4
lbm/ft s
which is greater than 4000. Therefore, the flow is turbulent. The relative roughness of the
Analysis The piping system involves 125 ft of 5-in diameter piping, an entrance with negligible loses, 3 standard flanged 90 smooth elbows (KL = 0.3 each), and a sharp-edged exit (KL = 1.0). We choose points 1 and 2 at the free surfaces of the river and the tank, respectively. We note that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm), and the fluid velocity is 6 ft/s at point 1 and zero at point 2 (V1 = 6 ft/s and V2 =0). We take the free surface of the river as the reference level (z1 = 0). Then the energy equation for a control volume between these two points simplifies to
P1 g
1
V12 2g
z1
h pump, u
P2 g
2
V 22 2g
z2
h turbine, e
hL
1
V12 2g
h pump, u
z2
hL
where
1
= 1 and
V2 L h L h L ,total h L ,major h L ,minor f KL 2 D 2g since the diameter of the piping system is constant. The average velocity in the pipe and the Reynolds number are V Re V Ac VD V D2 / 4 1.5 ft 3 /s (5 / 12 ft) 2 / 4
3
2
5 in 125 ft 12 ft
Water tank
11.0 ft/s
1
435,500
River
1.5 ft3/s
(62.3 lbm/ft )(11.0 ft/s)(5/12 ft) 6.556 10
4
lbm/ft s
which is greater than 4000. Therefore, the flow is turbulent. The relative roughness of the