Chapter 7, Problem 7.11P

Interpretation Introduction

Interpretation:

The temperature of steam at the nozzle entrance 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
$\begin{array}{l}\text{ΔS =}\frac{\text{q}}{\text{T}}\\ \text{ΔS}={\text{S}}_{\text{2}}{\text{-S}}_{\text{1}}\text{ = 0-----(2)}\end{array}$

  $$

Temperature at the nozzle inlet $182.5\text{C}$

Given Information:

Outlet Temperature of air T₂= $15\text{C}$

Inlet velocity, u₁ = negligible = 0 m/s

Outlet velocity, u₂ = 580 m/s

Explanation:

Step 1:

Calculate the change in enthalpy

Based on equation (1) we have:

  $\begin{array}{l}\Delta \text{H = }-\frac{{\text{u}}_2^2-{\text{u}}_1^2}{\text{2}}\\ \text{ΔH=}-\frac{{(580)}^2-{(0)}^2}{2}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{ = }-168200{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{Since, 1 kJ/kg = 1000 m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{ΔH = }-\text{168}\text{.2kJ/kg}\\ \text{Molar mass of air = 0}\text{.02897 kg/mol}\\ \text{Thus, ΔH = }-\text{168}\text{.2kJ/kg}\times 0.02897\text{ kg/mol = -4}\text{.873 kJ/mol}\end{array}$

Step 2:

Calculate the inlet temperature, T₁

For an ideal gas the change in enthalpy is related to the change in temperature through the specific heat C_p;

  $\begin{array}{l}{\text{ΔH = C}}_{\text{p}}{\text{(T}}_{\text{2}}{\text{-T}}_{\text{1}}\text{)}\\ \text{where Cp = 7/2 R = 7/2}\times \text{0}\text{.008314 kJ/mol-K = 0}\text{.0291 kJ/mol-k}\\ {\text{T}}_{\text{1}}{\text{ = T}}_{\text{2}}\text{ - }\frac{\text{ΔH}}{{\text{C}}_{\text{p}}}\text{ = (15 +273)K -}\frac{\text{(-4}\text{.873 kJ/mol)}}{0.0291\text{ kJ/mol}\text{.K}}\text{ = 455}\text{.46 K}\\ \text{In terms of degrees C we have:}\\ {\text{T}}_{\text{1}}\text{ = 455}\text{.46-273 = 182}\text{.5}\text{C}\end{array}$

Thus, the temperature at the nozzle inlet $182.5\text{C}$