Question

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

Answer 1

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

$Q=\triangle U+W$

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

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

$W_{2-3}=23.16418 kJ$