Question

A steam enters in an insulated nozzle with a velocity of 40 m/s. The initial pressure and temperature of the steam is 6.87 bar and ${225}^{o}C$. It leaves at a pressure of 2.1 bar and a velocity of 450 m/s. The final enthalpy of the steam is
(At 6.87 bar and ${225}^{o}C$, h = 2850 kJ/kg).

• A. 2250 kJ/kg
• B. 2698 kJ/kg
• C. 3000 kJ/kg
• D. 2749.55 kJ/kg

Answer 1

The correct option is D 2749.55 kJ/kg


The steady flow energy equation,

$h_1+\frac{c_1^2}{2}+g{z}_1+Q=h_2+\frac{c_2^2}{2}+g{z}_2+W$

$\because W=0,Q=0,g(z_2-z_1)=0$

$h_2-h_1=\frac{c_1^2-c_2^2}{2}$

$h_2-2850=\frac{(40{)}^2-(450{)}^2}{2\times {10}^3}$

$\Rightarrow h_2=2749.55kJ/kg$