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The steady flow energy equation:

Q = m(h2 – h1) is applicable for?

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SSC JE ME Previous Paper 8 (Held on: 27 Sep 2019 Morning)
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  1. Nozzle
  2. Turbine
  3. Compressor
  4. Boiler

Answer (Detailed Solution Below)

Option 4 : Boiler
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Detailed Solution

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Explanation:

Steady Flow Energy Equation (S.F.E.E.)

Now,

Boiler:

  • Ẇ  = 0, no work is done by the boiler,
  • Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0)

∴ Q̇ = ṁ(h2 - h1)

Nozzle:

  • Q̇ = 0, as the nozzle is perfectly insulated.
  • Ẇ = 0, no work is done by the nozzle, v1 <<< v2 and potential energy change is neglected

∴  

Turbine:

  • Q = 0, for the adiabatic or perfectly insulated turbine,
  • Change in kinetic and potential energy is neglected

∴  Ẇ = ṁ(h1 – h2)

Compressor:

  • Q̇ = 0, for the adiabatic or perfectly insulated turbine,
  • Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0)

∴ Ẇ = ṁ(h2 - h1)

Points to remember:

Device

S.F.E.E

Nozzle

Turbine

Ẇ = ṁ(h1 – h2)

Compressor

Ẇ = ṁ(h2 - h1)

Boiler

Q = ṁ(h2 - h1)

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