Under ideal conditions, the velocity of steam at the outlet of a nozzle for a drop of 450 kJ/kg from inlet reservoir condition up to the exit is:

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ESE Mechanical 2016 Paper 1: Official Paper
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  1. 649 m/s
  2. 749 m/s
  3. 849 m/s
  4. 949 m/s

Answer (Detailed Solution Below)

Option 4 : 949 m/s
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ST 1: UPSC ESE (IES) Civil - Building Materials
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Detailed Solution

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Concept:

The energy changes in a flow can be analyzed by steady flow energy equation (SFEE)

It is given as

Where h is the enthalpy of the flowing stream and V2/2 represents the kinetic energy and Zg represents the potential energy, Q and W ae the heat and work interaction respectively

Calculation:

Given,

Change in enthalpy h- h= 450 KJ/Kg

Now applying steady flow energy equation by the assumption (W = 0, Q = 0, ΔP.E = 0)

For ideal condition of nozzle

The initial velocity is assumed zero as it is very less compared to the exit velocity

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