Answer:
Explanation:
Calculating the exit temperature for K = 1.4
The value of [tex]c_p[/tex] is determined via the expression:
[tex]c_p = \frac{KR}{K_1}[/tex]
where ;
R = universal gas constant = [tex]\frac{8.314 \ J}{28.97 \ kg.K}[/tex]
k = constant = 1.4
[tex]c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}[/tex]
[tex]c_p= 1.004 \ kJ/kg.K[/tex]
The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :
[tex]0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}[/tex] ------ equation(1)
we can rewrite the above equation as :
[tex]0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}[/tex]
[tex]T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}[/tex]
where:
[tex]T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K[/tex]
[tex]T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}[/tex]
[tex]T_2 = 402.36 \ K[/tex]
Thus, the exit temperature = 402.36 K
The exit pressure is determined by using the relation:[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}[/tex]
[tex]P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}[/tex]
[tex]P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}[/tex]
[tex]P_2 = 17.79 \ bar[/tex]
Therefore, the exit pressure is 17.79 bar
