Question

An engineer designing a hydroelectric facility recalls that the flow rate (in gallons per second) is four times the value of the head (in feet) but has forgotten both numbers. If the facility is expected to produce 1.5 MW of power with an efficiency of 88%, determine both the flow rate (in gal/s) and the head (in ft).

Answer 1

Answer: flowrate=505.228

head=101.307ft

Explanation:P=power,m=flowrate,w=work,h=head,g=acceleration due to gravity

P=mw

W=gh

P=1.5×10^6 ×88%

=1320000Watts

g=9.8×3.281 ft/s

=32.154ft/s

m=4h

1320000=32.154×4h^2

4h^2=1320000/32.154

=41052.44

h^2=41052.44/4

=10263.109

h=√10263.109

=101.307ft

m=4h

m=4×101.307

m=405.228gal/s

Answer 2

Flow rate = 86.48 gal/s

Head of water = 21.62 ft

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below.