Question

The air in a steady flow enters the system at a pressure of 8 bar ${180}^{\circ }C$, with a velocity of 80 m/s and leaves at 1.4 bar, ${20}^{\circ }C$ with a velocity of 40 m/s. The temperature of the surroundings is ${20}^{\circ }C$ and pressure is 1 bar.

What is the maximum theoretic work which must be available?

• A. $160.8kJ/kg$
• B. $163.2kJ/kg$
• C. $170.8kJ/kg$
• D. $181.49kJ/kg$

Answer 1

The correct option is D $181.49kJ/kg$

$P_1=8bar,T_1=180+273=453K$

$P_2=1.4bar,T_2=20+273=293K$

$P_0=1bar$

$C_1=80m/s,C_2=40m/s$

Availabaility of the air at the inlet

$=(h_1-h_0)-T_0(s_1-s_0)+\frac{c_1^2}{2}$

$=c_p(T_1-T_0)-T_0(s_1-s_0)+\frac{c_1^2}{2}$

$(s_1-s_0)=c_{p}lo{g}_e\left(\frac{T_1}{T_0}\right)-Rlo{g}_e\left(\frac{P_1}{P_0}\right)$

$=1.005lo{g}_e\left(\frac{453}{293}\right)-0.287lo{g}_e\left(\frac{8}{1}\right)$

$=0.437-0.596=-0.159kJ/kgK$

Availbaility of the air at the inlet

$=1.005(453-293)-293(-0.159)+\frac{{80}^2}{2\times 1000}=210.58kJ$

Availbaility at the exit

$=(h_1-h_0)-T_0(s_2-s_0)+\frac{c_2^2}{2}$

$=-T_0(s_2-s_0)+\frac{c_1^2}{2}[h_2-h_0,since,T_2-T_0]$

$s_2-s_0=-Rlo{g}_e\left(\frac{P_2}{P_0}\right)=-0.287lo{g}_e\left(\frac{1.4}{1}\right)$

$=-0.9656kJ/kgK$

$\therefore \text{Availability at the exit}=-293-(-0.9656)+\frac{{40}^2}{2\times 1000}=29.09kJ/kg$

Theoretical work which must be available = availability at the inlet - availability at the exit

$=210.58-29.09=181.49kJ/kg$