Worked Example 13.4
A pumping plant delivers 1.3 m^3 s^1 of water into a reservoir with a water
level 75 m above steady state water level in the air vessel. The length of the
supply line from the air vessel to the reservoir is 1500 m, its diameter is
800 mm and the friction coefficient is 0.019. Assuming a head loss coeffi-
cientKs0.3 for flow into the vessel, calculate the minimum necessary
volume of an air vessel required for the protection of the plant if the
maximum permissible head rise at the pump is 40% of the pumping head.
The pressure wave celerity may be assumed as 1000 m s^1. Compare the
result with that of the appropriate simplified method of solution.
Solution
As given above, a1000 m s^1 ,0.019,L1500 m and Q1.3 m^3 s^1.
The steady flow velocity, 0 Q/A1.3/0.5032.59 m s^1. The head loss,
HfLv^20 /2gD12 m. Therefore H 0 75 12 10 97 m (in absolute
units). (Hmax H 0 )/H 0 0.4 (given), so
2 a 0 /gH 0 2.7 (equation (13.21)).From the design graph (Fig. 13.12), 2V 0 a/Q 0 L20, giving V 0 20 m^3. From
the adiabatic gas law it follows that
VmaxV 0 (H 0 /Hmin)1/1.2. (iv)From the design graph (Fig. 13.12), (H 0 Hmin)/H 0 0.37, therefore
Hmin 97 0.37 97 61 m, and from equation (iv), Vmax29.630 m^3.
Therefore, the minimum volume of the vessel30 m^3.
WORKED EXAMPLES 573
Fig. 13.16 Pump–pipeline characteristics