Question

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:

1. 62.425
2. 58.899
3. 624.25
4. 581.723

Answer 1

Correct Answer - Option 4 : 581.723

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

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

\(̇̇ m = \frac{{50 \times {{10}^{ - 4}} \times 60}}{{0.077395}} = 3.877\;kg/s\)

\(\dot m \times \left[ {{h_1} + \frac{{v_1^2}}{{2000}}} \right] - \dot Q = \dot m \times \left[ {{h_2} + \frac{{v_2^2}}{{2000}}} \right]\)
\(3.877 \times \left[ {1.87 \times 673 + \frac{{{{60}^2}}}{{2000}}} \right] - 75 = \;3.877 \times \left[ {1.87 \times 573 + \frac{{v_2^2}}{{2000}}} \right]\)

v₂ = 581.723 m/s.

The value of specific volume has not been given in the question. It has been taken from steam table based on the value of pressure and temperature. As it is official question of BHEL Management Trainee, we cannot change the language of the question.