Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:
For the pipe 1, the flow velocity is:
Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m
The Reynold´s number is:
Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:
For the pipe 2, the flow velocity is:
The Reynold´s number is:
The head of pipe 1 is:
The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:
The required pumping power is:
Physical movement like running
Answer: a) The technology that deals with the generation, control and transmission of power using pressurized fluids
Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.
Answer: valving the pump discharge to reduce the flow will only result in an increase in the water velocity according to the laws of continuity of flow.
Q = AV = constant.
The pump speed of 150 gpm is, and will remain constant. Valving simply reduces flow area A, which is balanced out by increased velocity V of water through the pipe.
This does not affect the pump speed Q (flow rate) and hence it remains the same.
