Chapter 7, Problem 7.19P

Interpretation Introduction

Interpretation:

The discharge temperature of isobutene and the power output of the turbine under the given conditions need to be calculated

Concept Introduction:

• For an ideal gas under isentropic adiabatic conditions, the temperature and pressure are related as:

$\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}{\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)}^{\text{γ-1/γ}}---(1)$

${\text{where, γ = C}}_{\text{p}}{\text{/C}}_{\text{v}}$

• The isentropic efficiency of a turbine is given as:

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

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

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

$\text{n<mo>˙</mo> = molar flow rate}$

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

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

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

$\text{(or) η }=\frac{{\text{ΔH}}_{\text{actual}}}{{\text{ΔH}}_{\text{adiabatic}}}\text{ ----(5)}$

The discharge temperature of isobutane = $170.3\text{C}$

Power output of turbine = -4250 kW

Given Information:

Inlet pressure, P₁ = 5000 kPa

Inlet Temperature, T₁= $250\text{C}$

Outlet pressure, P₂ = 500 kPa

Molar flow rate = 0.7 kmol/s

Turbine efficiency, ? = 0.80

Explanation:

In this case, isobutane has been assumed to behave as an ideal gas. The discharge temperature can be deduced from equation (1) and the power output can be deduced from equation (4).

The heat capacity ratio for isobutane, i.e. ? = 1.2

Calculation:

Step 1:

Calculate the discharge/final temperature, T₂

Based on equation (1) we have:

$\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}{\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)}^{\text{γ-1/γ}}$

$\frac{{\text{T}}_{\text{2}}}{250}={\left(\frac{500}{5000}\right)}^{\text{1}\text{.2-1/1}\text{.2 }}=0.681$

${\mathrm{T}}_2=170.3C$

Step 2:

Calculate the actual enthalpy change

The adiabatic enthalpy change is related to the change in temperature through the specific heat capacity, C_p

${\text{ΔH}}_{\text{adiabatic}}{\text{ = C}}_{\text{p}}{\text{(T}}_{\text{2}}{\text{-T}}_{\text{1}}\text{) = 0}\text{.0952(170}\text{.3-250) = -7}\text{.59 kJ/mol}$

Based on equation (5) the actual enthalpy change is:-

${\text{ ΔH}}_{\text{actual}}\text{ = η}\times {\text{ΔH}}_{\text{adiabatic}}\text{ = 0}\text{.80}\times \text{(-7}\text{.59) = -6}\text{.07 kJ/mol}$

Step 3:

Calculate the power output of the turbine

Based on equation (4)

$\text{ W<mo>˙</mo> = n<mo>˙</mo>}\Delta {\text{H}}_{\text{actual}}$

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

$\text{i}\text{.e}\text{. W<mo>˙</mo> = Power = 700 mol}{\text{.s}}^{-1}\times (-6.07)\text{ kJ}{\text{.mol}}^{\text{-1}}=\text{ -4250 kJ/s}$

$\text{Power = -4250 kW}$

Thus, the discharge temperature of isobutane = $170.3\text{C}$

Power output of turbine = -4250 kW