Answer:
[tex]Power=11.52hp[/tex]
Explanation:
Given data
[tex]p_{1}=15psia\\p_{2}=70psia\\V_{ol}=0.8ft^{3}/s[/tex]
As
[tex]m=p*V_{ol}[/tex]
Assuming in-compressible flow p is constant
The total change in system mechanical energy calculated as:
Δe=(p₂-p₁)/p
The Power can be calculated as
[tex]P=W\\P=m(delta)e\\P=p*V_{ol}*(p_{2}-p_{1})/p\\ P=V_{ol}*(p_{2}-p_{1})\\P=44(psia.ft^{3}/s )*[(\frac{1Btu}{5.404psia.ft^{3} } )-(\frac{1hp}{0.7068Btu/s } )]\\P=11.52hp[/tex]
The power input required to pump 0.8 ft³/s of water is; P = 11.52 hP
We are given;
Initial pressure; P1 = 15 psia
Final pressure; P2 = 70 psia
Volume flow rate; V' = 0.8 ft³/s
The formula for the mass flow rate is;
m' = ρV'
The total change in the mechanical energy of the system is;
△E = (P2 - P1)/ρ
Now, formula for power is;
P = m' × △E
P = ρV' × (P2 - P1)/ρ
P = V'(P2 - P1)
P = 0.8(70 - 15)
P = 44 psia.ft³/s
Converting to Btu/s gives;
P = 8.142 btu/s
Converting Btu/s to HP from conversion tables gives; P = 11.52 hP
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