Answer:64.10 Btu/lbm
Explanation:
Work done in an isothermally compressed steady flow device is expressed as
Work done = P₁V₁ In { P₁/ P₂}
Work done=RT In { P₁/ P₂}
where P₁=13 psia
P₂= 80 psia
Temperature =°F Temperature is convert to °R
T(°R) = T(°F) + 459.67
T(°R) = 55°F+ 459.67
=514.67T(°R)
According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm
Work = RT In { P₁/ P₂}
0.06855 x 514.67 In { 13/ 80}
=0.06855 x 514.67 In {0.1625}
= 0.06855 x 514.67 x -1.817
=- 64.10Btu/lbm
The required work therefore for this isothermal compression is 64.10 Btu/lbm
Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:

where;
m = mass flow rate = 120 kg/min
the pressure at the inlet
= 96 kPa
the pressure at the exit
= ???
the pressure
= 1000 kg/m³
∴




400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
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Answer:
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