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:

$\begin{array}{l}{\left({\text{H}}_{\text{1}}\text{+}\frac{{\text{u}}_{\text{1}}^{\text{2}}}{\text{2}}\right)}_{\text{inlet}}\text{= }{\left({\text{H}}_{\text{2}}\text{+}\frac{{\text{u}}_{\text{2}}^{\text{2}}}{\text{2}}\right)}_{\text{outlet}}\\ \text{(or)}\\ \text{ΔH + }\frac{{\text{Δu}}^{\text{2}}}{\text{2}}\text{ = 0 ----(1)}\end{array}$

• For an adiabatic process, there is no loss or gain of heat i.e. q = 0
$\begin{array}{l}\text{ΔS =}\frac{\text{q}}{\text{T}}{\text{ + S}}_{\text{G}}\\ \text{ΔS}={\text{S}}_{\text{2}}{\text{-S}}_{\text{1}}{\text{ = S}}_{\text{G}}\text{-----(2)}\end{array}$

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

• The isentropic efficiency of a turbine is given as:

$\text{η}=\frac{\text{Actual work done by turbine}}{\text{ Isentropic Work }}=\frac{{\text{W<mo>˙</mo>}}_{\text{a}}}{{\text{W<mo>˙</mo>}}_{\text{s}}}\text{ ----(3)}$

$\text{where, the rate of work done is given as:}$

$\text{W<mo>˙</mo> = m<mo>˙</mo>}\Delta \text{H -----(4)}$

$\text{m<mo>˙</mo> = rate of mass flow}$

$\Delta \text{H = change in enthalpy}$

$\text{Thus, turbine efficiency is given as:}$

$\text{η =}\frac{{\text{W<mo>˙</mo>}}_{\text{a}}}{{\text{W<mo>˙</mo>}}_{\text{s}}}=\frac{{\text{H}}_{\text{2a}}{\text{-H}}_{\text{1}}}{{\text{H}}_{\text{2s}}{\text{-H}}_{\text{1}}}\text{ -----(5)}$

$$

Steam rate = 4.1 kg/s

Turbine efficiency = 0.82

Given Information:

Inlet pressure of steam, P₁ = 2400 kPa

Inlet Temperature of steam T₁= $500\text{C}$

Outlet pressure of steam, P₂ = 20 kPa

Power output = 3500 KW

Explanation:

Based on the steam tables, for inlet conditions: P₁ = 2400 kPa, T₁= $500\text{C}$

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:

$\text{m<mo>˙</mo> = }\frac{\text{W<mo>˙</mo>}}{\Delta \text{H}}$

$\text{Now, Power = Energy/time = Rate of work done }$

$\text{i}\text{.e}\text{. W<mo>˙</mo> = -3500 kJ/s}$

$\text{m<mo>˙</mo> = }\frac{\text{W<mo>˙</mo>}}{\Delta \text{H}}=\frac{\text{W<mo>˙</mo>}}{{\text{H}}_{\text{2}}{\text{-H}}_{\text{1}}}=-\frac{3500}{2609.9-3462.9}=4.1\text{ kg/s}$

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: S_f = 0.8321 kJ/kg-k and S_g = 7.9094 kJ/kg-k

$\begin{array}{l}{\text{S}}_{\text{2}}{\text{ = S}}_{\text{f}}{\text{ + x(S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)}\\ \text{where x = steam quality}\\ 7.3429=0.8321+\text{x}(7.9094-0.8321)=0.92\end{array}$

$\begin{array}{l}\text{For an adiabatic (isentropic) process at P = 20 kPa:}\\ {\text{H}}_{\text{2s}}{\text{ = H}}_{\text{f}}{\text{ + x(H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{) =251}\text{.5 + 0}\text{.92(2609}\text{.9-251}\text{.5) = 2421}\text{.23 kJ/kg}\end{array}$

Step 3:

Calculate the turbine efficiency

Based on equation (5) we have:

$\text{η =}\frac{{\text{W<mo>˙</mo>}}_{\text{a}}}{{\text{W<mo>˙</mo>}}_{\text{s}}}=\frac{{\text{H}}_{\text{2a}}{\text{-H}}_{\text{1}}}{{\text{H}}_{\text{2s}}{\text{-H}}_{\text{1}}}=\frac{2609.9-3462.9}{2421.2-3462.9}=0.82$

Thus, steam rate = 4.1 kg/s

Turbine efficiency = 0.82