- #1
WateryFungi
- 1
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- Homework Statement
- Air with a volume of 1.59 m^3 at 2.06267 atm and 298.7 K is compressed by a compressor, which applies 260 kJ/kg of flow work to increase the pressure to 2,094 kPa. Find the final temperature in Kelvin.
- Relevant Equations
- First Law of Thermodynamics
Energy Balance Equation for Open Systems
Ideal Gas Equation
h = Pv + u
Hi
I'd like to know if my current approach to the problem has any issues.
My attempt:
Balancing the equation:
Qin + Win +mΘout = Qout + Wout + mΘout
Qin+m(hin + v2in/2 + gzin) = m(hout + v2out/2 + gzout)
Wflow + ΔU + m(hin) = m(hout)
I factored out the mass:
wflow + Δu + hin = hout
And I used h = Pv +u where h = specific enthalpy, P=pressure, v=specific volume, u=specific internal energy, to get:
wflow = P2v2 - P1v1
I then used the ideal gas equation Pv = RT, with R = 0.287 kJ/kg*K and P = 2.06267 atm = 209 kPa to get v1
v1 = (0.287 kJ/kg*K)(298.7 K) / (209 kPa) = 0.4102 m3/kg
And plugged it in the previous equation for v2
v2 = (wflow + P1v1) / P2
v2 = [260 kJ/kg + (209 kPa)(0.4102)]/(2,094 kPa) = 0.1651 m3/kg
I then used the ideal gas equation again to solve for T2.
Any insight would be greatly appreciated.
I'd like to know if my current approach to the problem has any issues.
My attempt:
Balancing the equation:
Qin + Win +mΘout = Qout + Wout + mΘout
Qin+m(hin + v2in/2 + gzin) = m(hout + v2out/2 + gzout)
Wflow + ΔU + m(hin) = m(hout)
I factored out the mass:
wflow + Δu + hin = hout
And I used h = Pv +u where h = specific enthalpy, P=pressure, v=specific volume, u=specific internal energy, to get:
wflow = P2v2 - P1v1
I then used the ideal gas equation Pv = RT, with R = 0.287 kJ/kg*K and P = 2.06267 atm = 209 kPa to get v1
v1 = (0.287 kJ/kg*K)(298.7 K) / (209 kPa) = 0.4102 m3/kg
And plugged it in the previous equation for v2
v2 = (wflow + P1v1) / P2
v2 = [260 kJ/kg + (209 kPa)(0.4102)]/(2,094 kPa) = 0.1651 m3/kg
I then used the ideal gas equation again to solve for T2.
Any insight would be greatly appreciated.