Asked by Eileen
What must the water pressure be to supply water to the third flor of a buiking 35.0 ft up with a pressure of 40.0 lb/in(2) at that level? I have the answer,but I need to know how to get the answer. thanks
Answers
Answered by
bobpursley
Find the weight of a column of water A in area, h high.
weight= densitywater*g*area*height
but pressure= weight/area= density*g*h
but one wants an additional 40psi, so that has to be added..
pressure=density*g*h+ 40lbs/in^2*
so solve. Watch units. I would work it in English, density*g= 0.0361273lbs/in^3
h= 35ft= 35*12inches
weight= densitywater*g*area*height
but pressure= weight/area= density*g*h
but one wants an additional 40psi, so that has to be added..
pressure=density*g*h+ 40lbs/in^2*
so solve. Watch units. I would work it in English, density*g= 0.0361273lbs/in^3
h= 35ft= 35*12inches
Answered by
Bradley
Find required water pressure to supply water at h = 35.0 ft and P = 40.0 Lb/in^2.
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Working equation: P = hDw where
P = pressure, h = height, Dw = weight density (Dw for water is 62.4 Lb/ft^2)
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P at bottom = hDw = (35.0 ft)(62.4 Lb/ft^3) = 2184 Lb/ft^2
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Convert to Lb/in^2 to match question: 2184 Lb/[ ft^2(144 in^2/ft^2)] = 15.2 Lb/in^2
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We want 40.0 Lb/in^2. To this we add (P)ressure just calculated above:
Pressure required = 40.0 Lb/in^2 + 15.2 Lb/in^2 = 55.2 Lb/in^2.
-----------------------------------
Working equation: P = hDw where
P = pressure, h = height, Dw = weight density (Dw for water is 62.4 Lb/ft^2)
-----------------------------------
P at bottom = hDw = (35.0 ft)(62.4 Lb/ft^3) = 2184 Lb/ft^2
-----------------------------------
Convert to Lb/in^2 to match question: 2184 Lb/[ ft^2(144 in^2/ft^2)] = 15.2 Lb/in^2
-----------------------------------
We want 40.0 Lb/in^2. To this we add (P)ressure just calculated above:
Pressure required = 40.0 Lb/in^2 + 15.2 Lb/in^2 = 55.2 Lb/in^2.
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