Question

5 kg of air initially at 8 bar and 380 K expands polytropically (pv1.2 = 

constant) until the pressure is reduced to one-fifth value. Calculate: (i) 

Final specific volume and temperature .

(ii) Change of internal energy, work done and heat interaction .

Take: R = 0.287 kJ/kg K and γ= 1.4.

Answer 1

Solution.

$m=5kg; p_1=8bar; T_1=380K; pv^{1.2}=const; p_2=\dfrac{p_1}{5}=1.6bar;$

$R= 0.287 kJ/kg K; \gamma=1.4;$

$i)p_1v_1=RT_1\implies v_1=\dfrac{RT_1}{p_1};$

$v_1=\dfrac{0.287\sdot10^3\sdot380}{8\sdot10^5}=0.1363m^3/kg;$

$v_2=v_1(\dfrac{p_1}{p_2})^{\dfrac{1}{n}};$

$v_2=0.1363\sdot(\dfrac{8}{1.6})^{\dfrac{1}{1.2}}=0.5211m^3/kg;$

$\dfrac{T_2}{T_1}=(\dfrac{p_2}{p_1})^{\dfrac{n-1}{n}};$

$T_2=T_1(\dfrac{p_2}{p_1})^{\dfrac{n-1}{n}};$

$T_2=190.6K;$

$ii)u_2-u_1=c_v(T_2-T_1)=\dfrac{R}{\gamma-1}(T_2-T_1);$

$u_2-u_1=\dfrac{287}{1.4-1}(190.6-380)=-64110J/kg;$

$w_{1-2}=\dfrac{R(T_1-T_2)}{n-1};$

$w_{1-2}=128290J/kg;$

$Q_{1-2}=\dfrac{\gamma-n}{\gamma-1}w;$

$Q_{1-2}=64150J/kg;$

Answer:$i)v_2=0.5211m^3/kg;T_2=190.6K;$

$ii)u_2-u_1=-64110J/kg; w_{1-2}=128290J/kg; Q_{1-2}=64150J/kg.$