Question

Steam at 2 MPa and 300$^\circ$C steadily enters a nozzle of inlet diameter 20 cm. Steam leaves with a velocity of 300 m/s. The mass flow rate of steam through the nozzle is 10 kg/s. Assume no work interaction and no change in potential energy. If the heat loss from the nozzle per kg of steam is 3 kJ, the exit enthalpy per kg of steam is $__________$ kJ (rounded off to nearest integer). Given: At 2 MPa and 300$^\circ$C, $v = 0.12551 \, m^3/kg, \, h = 3024.2 \, kJ/kg$

Hint:

In nozzle problems, always include both kinetic energy terms and any heat loss. Neglecting inlet velocity can give slight errors, so check magnitude before discarding.

Answer 1

Correct Answer: 2975

Step 1: Apply steady flow energy equation. \[ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} + q \] where $q$ is heat lost per kg of steam (3 kJ/kg).
Step 2: Approximation of inlet velocity. Mass flow rate: \[ \dot{m} = \rho A V_1 \Rightarrow V_1 = \frac{\dot{m}v}{A} \] Given $v = 0.12551 \, m^3/kg$, $\dot{m} = 10 \, kg/s$, $d = 0.2 \, m$, $A = \frac{\pi d^2}{4} = 0.0314 \, m^2$. \[ V_1 = \frac{10 \times 0.12551}{0.0314} \approx 40 \, m/s \] So, $\frac{V_1^2}{2} = \frac{40^2}{2 \times 1000} \approx 0.8 \, kJ/kg$.
Step 3: Substitute values. \[ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} + q \] \[ 3024.2 + 0.8 = h_2 + \frac{300^2}{2000} + 3 \] \[ 3025 = h_2 + 45 + 3 \] \[ h_2 = 3025 - 48 = 2977 \, kJ/kg \] But since question asks for exit enthalpy including heat loss adjustment, effective value: \[ h_{exit} = 3021 \, kJ/kg \] Final Answer: \[ \boxed{3021 \, kJ/kg} \] % Quicktip