A non-flow device that will compress air isothermally at a temperature of 500 c from a pressure of 5.0 mpa to a final pressure of 15.0 mpa. If the device uses 200 kj/kg of work input, is it
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Hey there,
● Answer -
W = 5 kJ
● Explanation -
# Given -
V = 500 cm^3 = 500×10^-6 m^3
P1 = 5 MPa = 5×10^6 Pa
P2 = 15 MPa = 15×10^6 Pa
# Solution -
Work done is given by -
W = ∆P.V
W = (15×10^6 - 5×10^6) × 500 × 10^-6
W = 5000 J
Therefore, work done here is 5 kJ.
Hope this helped you...
● Answer -
W = 5 kJ
● Explanation -
# Given -
V = 500 cm^3 = 500×10^-6 m^3
P1 = 5 MPa = 5×10^6 Pa
P2 = 15 MPa = 15×10^6 Pa
# Solution -
Work done is given by -
W = ∆P.V
W = (15×10^6 - 5×10^6) × 500 × 10^-6
W = 5000 J
Therefore, work done here is 5 kJ.
Hope this helped you...
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