In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference be
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Answer:
a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s
b) The pressure difference across the pump is approximately 293.118 kPa
Explanation:
The efficiency of the pump = 78%
The power of the pump = 5 -kW
The height of the pool above the underground water, h = 30 m
The diameter of the pipe on the intake side = 7 cm
The diameter of the pipe on the discharge side = 5 cm
a) The maximum flowrate of the pump is given as follows;
Where;
P = The power of the pump
Q = The flowrate of the pump
ρ = The density of the fluid = 997 kg/m³
h = The head of the pump = 30 m
g = The acceleration due to gravity ≈ 9.8 m/s²
= The efficiency of the pump = 78%
= 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s
The maximum flowrate of the pump ≈ 0.013305 m³/s = 13,305.22 cm³/s
b) The pressure difference across the pump, ΔP = ρ·g·h
∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa
The pressure difference across the pump, ΔP ≈ 293.118 kPa
Answer:
Mechanical power of pump is 74.07%.
Explanation:
Power of motor = 15 KW
Efficiency of motor= 90%
So the actual power(P) supplied by motor = 0.9 x 15 KW
P=13.5 KW
Water flow rate = 50 L/s
Volume flow rate = 50 L/s
We know that
So
We know that pump is an open system and work input for open system can be calculated as
W=VΔP
ΔP is the pressure difference
V is the volume flow rate
So by putting the values
W=0.05 (300-100) (here ΔP=300 - 100=200 KPa)
W=10 KW
So mechanical power of pump
η =0.7407
Mechanical power of pump is 74.07%.