INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
8th Edition
ISBN: 9781260940961
Author: SMITH
Publisher: INTER MCG
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Chapter 7, Problem 7.15P
Interpretation Introduction

Interpretation:

The steam rate through the turbine and the turbine efficiency under the given conditions needs to be calculated

Concept Introduction:

  • The energy balance equation that relates the inlet and outlet enthalpies and velocities is given based on the first law of thermodynamics as:
  • (H1+ u 1 2 2)inlet(H2+ u 2 2 2)outlet(or)ΔH + Δu22 = 0 ----(1)

  • For an adiabatic process, there is no loss or gain of heat i.e. q = 0
  • ΔS =qT + SGΔS=S2-S1 = SG-----(2)

The process occurs at constant enthalpy, i.e. isentropic

  • The isentropic efficiency of a turbine is given as:
  • η=Actual work done by turbine Isentropic Work =W˙aW˙s ----(3)

    where, the rate of work done is given as:

    W˙ = m˙ΔH -----(4)

    m˙ = rate of mass flow

    ΔH = change in enthalpy

    Thus, turbine efficiency is given as:

    η =W˙aW˙s=H2a-H1H2s-H1 -----(5)

Steam rate = 4.1 kg/s

Turbine efficiency = 0.82

Given Information:

Inlet pressure of steam, P1 = 2400 kPa

Inlet Temperature of steam T1= 500C0

Outlet pressure of steam, P2 = 20 kPa

Power output = 3500 KW

Explanation:

Based on the steam tables, for inlet conditions: P1 = 2400 kPa, T1= 500C0

Specific enthalpy of vapor H1= 3462.9 kJ/kg

Specific entropy of vapor, S1=7.3439 kJ/kg-K

For outlet at P2 = 20 kPa

Specific enthalpy of vapor H2= 2609.9 kJ/kg

Step 1:

Calculate the steam rate through the turbine

Based on equation (1) we have:

m˙ = W˙ΔH 

Now, Power = Energy/time = Rate of work done 

i.e. W˙ = -3500 kJ/s

m˙ = W˙ΔH=W˙H2-H1=35002609.93462.9=4.1 kg/s

Step 2:

Calculate the adiabatic (isentropic) enthalpy at the exit

Since this is an isentropic process, S2 = S1 = 7.3429 kJ/kg-K

Based on the steam tables at outlet, P2 = 20 kPa the value of S2 = 7.3429 kJ/kg-K represents ‘wet’ steam.

At 20 kPa: Sf = 0.8321 kJ/kg-k and Sg = 7.9094 kJ/kg-k

S2 = Sf + x(Sg-Sf)where x = steam quality7.3429=0.8321+x(7.90940.8321)=0.92

For an adiabatic (isentropic) process at P = 20 kPa:H2s = Hf + x(Hg-Hf) =251.5 + 0.92(2609.9-251.5) = 2421.23 kJ/kg

Step 3:

Calculate the turbine efficiency

Based on equation (5) we have:

η =W˙aW˙s=H2a-H1H2s-H1=2609.93462.92421.23462.9=0.82

Thus, steam rate = 4.1 kg/s

Turbine efficiency = 0.82

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