Question #150517

A system contains 0.25 m3 of air pressure of 4.8 bars and 160⁰ C. It is expanded adiabatically till the pressure falls to 1.5 bar. The air is then heated at a constant pressure till its enthalpy increases by 80 kJ. Sketch the process on a P-V diagram and determine the total work done. Use Cp=1.005 kJ/kg.K and Cv=0.714 kJ/kg.K

Expert's answer

Initial stage "P_1=4.8 bars; T_1=160^oC=160+273=433K; V_1=0.25 m^3"

First process; Adiabatic expansion. "P_2=1.5 bar"

Second process: Heat given at constant pressure

"dh=" Change in enthalpy = 80 kJ

So P-V diagram

Let mass of air flowing = m

Now, ideal air has the equation; PV= mRT, where "R=C_p-C_v=1.005-0.714=0.291kJ\/kgK"

Putting all the values; "m=\\frac{P_1V_1}{RT_1}\\frac{4.8*0.25}{0.291*433}=0.00952kg"

Now , during process 1-2

"P_1V_1^r=P_2V_2^r"

"V_2=V_1(\\frac{P_1}{P_2})^{\\frac{1}{r}}" r = 1.4 for air.

"V_2=0.25(\\frac{4.8}{1.5})^{\\frac{1}{1.4}}=0.5738 m^3"

"W_{1-2}=" Work done in adiabatic process.

"W_{1-2}=\\frac{P_1V_1-P_2V_2}{r-1}=\\frac{4.8*0.25-1.5*0.5738}{1.4-1}=0.84825kJ"

Considering process 2-3

First, let us calculate "T_2"

"P_1V_1^r=P_2V_2^r\\implies \\frac{T_1^r}{P_1^{r-1}}= \\frac{T_2^r}{P_2^{r-1}}"

"T_2=T_1(\\frac{P_2}{P_1})^{\\frac{1-r}{r}}=433(\\frac{1.5}{4.8})^{\\frac{1-1.4}{1.4}}=310.5699K"

Now as a change in enthalpy ="mC_pdT"

"80=0.00952(1.005)(T_3-310.5699) \\implies T_3=8672.123K"

By the first law of thermodynamics

"Q=\\triangle U+W"

"80=mC_vdT+W_{2-3}"

"80=0.00952*0.714(8672.123-310.5699)+W_{2-3}"

"W_{2-3}=23.16418 kJ"

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