1.) Water at 20°C is pumped at a constant rate of 9 m^3/h from a large reservoir resting On the floor to the open top of an experimental absorption tower. The point of discharge is 5 m above the floor, and frictional losses in the 50-mm pipe from the reservoir to the tower amount to 2.5 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only 0.1 kW?
1.) Water at 20°C is pumped at a constant rate of 9 m^3/h from a large reservoir resting On the floor to the open top of an experimental absorption tower. The point of discharge is 5 m above the floor, and frictional losses in the 50-mm pipe from the reservoir to the tower amount to 2.5 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only 0.1 kW?
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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required drawing of problem and show your complete solution and formula provided in pic thanks.

Transcribed Image Text:1.) Water at 20°C is pumped at a constant rate of 9
m^3/h from a large reservoir resting On the floor to the
open top of an experimental absorption tower. The
point of discharge is 5 m above the floor, and frictional
losses in the 50-mm pipe from the reservoir to the
tower amount to 2.5 J/kg. At what height in the
reservoir must the water level be kept if the pump can
develop only 0.1 kW?

Transcribed Image Text:P₂-P₁
1
(v² − v ² ) + ²/1 (2₂ − 2₁) + ²² − ²₁ + ΣF + W₁ = 0 (English)
2agc
gc
Unir for formula english
z=ft, F=Ws= ft lbf / lbm
v=ft/s, g= 32.174ft/s^2, p = lbm/ft3, gc= 31.174 lbm ft/
lbf s^2, P= lbf/ft^2= 144lbf/in^2(psia),
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