Asked by Mohanad
Water towers store water above the level of consumers for times of
heavy use, eliminating the need for high-speed pumps. How high above
a user must the water level be to create a gauge pressure of
3.00×105 N/m2 ?
heavy use, eliminating the need for high-speed pumps. How high above
a user must the water level be to create a gauge pressure of
3.00×105 N/m2 ?
Answers
Answered by
bobpursley
If you had a 1 m^2 pipe, what height would give a weight on the bottom of 3E5N?
weight water= mass*g= density*1m^2*height*9.8Nt/kg
= 1041 *9.8*height N
height= 3E5/(1041*9.8)= 29.4m
Now, to check: you know atomospheric pressure is about 10 m water height, so 300kpa/100kpa= 3 times atm pressure, so height is about 3*10 m
weight water= mass*g= density*1m^2*height*9.8Nt/kg
= 1041 *9.8*height N
height= 3E5/(1041*9.8)= 29.4m
Now, to check: you know atomospheric pressure is about 10 m water height, so 300kpa/100kpa= 3 times atm pressure, so height is about 3*10 m
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