Okay, so I\'ve seen how people have responded to this question before but I stil

Question

Okay, so I've seen how people have responded to this question before but I still cannot get the answer. So help would be appreciated! Thank you!

A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. The drawing shows a spherical reservoir that contains 5.90 x 105 kg of water when full. (h = 7.60 m.) The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in each house. Ignore the diameter of the delivery pipes.

A) house a

__________ Pa

B) house b

__________ Pa

Explanation / Answer

The volume of the reservoir can be found from the mass of stored water and water's density: V = m/rho And from this water volume, you can find the diameter, since it is spherical: V = 4/3*Pi*(d/2)^3 V = Pi*d^3/6 d = cbrt(6*V/Pi), cbrt() means cube root In terms of mass: d = cbrt(6*m/(Pi*rho)) At location A, the faucet is a distance d + H from the top of the water surface. To find gauge pressure there, simply use the hydrostatics equation for pressure variation with depth. Neglect flow effects. P_gaugeA = rho*g*(H + d) At location B, the faucet is a distance d + H - h from the top of the water surface. To find gauge pressure there, simply use the hydrostatics equation for pressure variation with depth. Neglect flow effects. P_gaugeB = rho*g*(H + d - h) Make substitutions: P_gaugeA = rho*g*(H + cbrt(6*m/(Pi*rho))) P_gaugeB = rho*g*(H + cbrt(6*m/(Pi*rho)) - h) Data: m:=5.9e5 kg; rho:=1000 kg/m^3; H:=15.0 m; h:=7.60 m; g:=9.8 N/kg; Results (translated from regular Pascals to kPa): P_gaugeA = 1000*9.8*(15+cbrt(6*5.9e5/pi*1000))=248.89kPa P_gaugeB = 1000*9.8*(15+cbrt(6*5.9e5*/pi*1000 - 7.6))= 174.49kPa

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