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)

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BHEL ET Mechanical Held on May 2019

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Answer 1

Concept:

According to the continuity equation, we have

ṁ_in = ṁ_out

The steady flow energy equation

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

v₂ = 581.723 m/s.

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