Dry saturated steam enters a frictionless adiabatic nozzle with negligible velocity at a temperature of 300 °C [h₁ = 2751 kJ/kg]. It is expanded to a pressure of 5 MPa isentropically [h₂ = 2651 kJ/kg]. What will be the exit velocity of steam?

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SJVN ET Mechanical 2019 Official Paper

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Answer 1

Concept:

The nozzle is a duct of varying cross-sectional area in which velocity increase with a corresponding drop in pressure.

The steady flow energy equation for a nozzle can be written as,

\(h_1+\frac{C_1^2}{2000}+Q = h_2+\frac{C_2^2}{2000}+ W\;\)

And for adiabatic and frictionless flow, Q = 0

and also there is no shaft work in the nozzle, so, W = 0

h1 = Specific enthalpy at the inlet of nozzle,

h2 = Specific enthalpy at the outlet of nozzle,

C1, and C2 are velocities at the inlet and outlet of the nozzle respectively.

For nozzle, SFEE can be written as,

\(h_1+\frac{C_1^2}{2000}= h_2+\frac{C_2^2}{2000}\;\)

Or velocity at the outlet of the nozzle can be written as,

\(C_2= \sqrt{C_1^{2}+ 2000(h_1-h_2)}\;\)

Calculation:

Given:

h1 = 2751 kJ/kg, h2= 2651 kJ/kg

Isentropic enthalpy drop, h1 - h2 = 100 kJIkg

To find -

Outlet velocity, C2 = ?

So now on putting the given values in the formula,

\(C_2= \sqrt{2000(h_1-h_2)}\;\)

\(C_2= \sqrt{2000 \times 100} =447.21\;m/s\;\)