Steam at 4 MPa and 673 K enter a nozzle steadily with a velocity of 60 m/s and it leaves at 2 MPa and 573 K. The inlet area of the nozzle is 50 cm², and heat is being lost from the nozzle at a rate of 75 kJ/s. Exit velocity in m/s is given by:
(Take specific volume at entry equal to 0.0734310 m³/kg)

Concept:

According to the continuity equation, we have

ṁ_in = ṁ_out 

_{\(\dot m = \rho \times A \times v = \frac{{A \times velocity}}{{specific\;volume}}\;kg/s\)}

The steady flow energy equation

\(\dot m \times \left[ {{h_1} + \frac{{v_1^2}}{{2000}}} \right] - \dot Q = \dot m \times \left[ {{h_2} + \frac{{v_2^2}}{{2000}}} \right]\)

ṁ = mass flow rate in nozzle, h = specific enthalpy, Q̇ = heat lost from the nozzle, v = steam velocity, A = area of nozzle, ρ = density

Calculation:

Given:

A = 50 cm² = 50 × 10^-4 m², Q̇ = 75 kJ/s

At nozzle inlet:

P₁ = 4 MPa, T₁ = 673 K, v₁ = 60 m/s

At nozzle outlet:

P₂ = 2 MPa, T₂ = 573 K

\(m = \rho \times A \times v = \frac{{A \times velocity}}{{specific\;volume}}\;kg/s\)

\(̇̇ m = \frac{{50 \times {{10}^{ - 4}} \times 60}}{{0.0734310}} = 4.085\;kg/s\)

\(\dot m \times \left[ {{h_1} + \frac{{v_1^2}}{{2000}}} \right] - \dot Q = \dot m \times \left[ {{h_2} + \frac{{v_2^2}}{{2000}}} \right]\)
\(4.085 \times \left[ {1.87 \times 673 + \frac{{{{60}^2}}}{{2000}}} \right] - 75 = \;4.085 \times \left[ {1.87 \times 573 + \frac{{v_2^2}}{{2000}}} \right]\)

v₂ = 581.723 m/s.