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jonny [76]
3 years ago
9

g A pump is required to deliver 100 gpm at a head of 100 ft, but the pump rated capacity is 150 gpm at a head of 100 ft. If the

pump speed is constant, what will be the effect of valving the pump discharge to reduce flow
Engineering
1 answer:
Thepotemich [5.8K]3 years ago
8 0

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.

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3 years ago
A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is
jonny [76]

Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}

a = -0.185\,\frac{ft}{s^{2}}

The time required to travel is:

t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }

t = 277.477\,s\,(4.625\min)

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

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