Chapter 7, Problem 7.4P

Interpretation Introduction

Interpretation:

The exit velocity and cross-sectional area at the nozzle exit 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:

${\left({\text{H}}_{\text{1}}\text{+}\frac{{\text{u}}_{\text{1}}^{\text{2}}}{\text{2}}\right)}_{\text{inlet}}\text{= }{\left({\text{H}}_{\text{2}}\text{+}\frac{{\text{u}}_{\text{2}}^{\text{2}}}{\text{2}}\right)}_{\text{outlet}}$

$\text{(or)}$

$\text{ΔH + }\frac{{\text{Δu}}^{\text{2}}}{\text{2}}\text{ = 0 ----(1)}$

• For a process that takes place at constant entropy i.e. isentropic, the change in entropy is zero. In other words, the entropy in the final state (S₂) is equal to that in the initial state (S₁). The change in entropy is given as:

${\text{ΔS = S}}_{\text{2}}{\text{-S}}_{\text{1}}\text{ -----(2)}$

$\text{When, }\Delta \text{S = 0 }$

${\text{S}}_{\text{2}}{\text{ = S}}_{\text{1}}$

The rate of mass flow $\text{(m<mo>˙</mo>)}$ is related to the cross-sectional area (A) of a nozzle through the following expression:

$\text{m<mo>˙</mo> = }\frac{\text{uA}}{\text{V}}\text{ ----(3)}$

$\text{where u = velocity and V = specific volume}$

$$

Exit velocity = 534.4 m/s

Cross-sectional area at the nozzle exit = $6.05\times {10}^{-4}{\text{ m}}^{\text{2}}$

Given Information:

Inlet pressure of steam, P₁ = 800 kPa

Inlet Temperature of steam T₁= $280\text{C}$

Outlet pressure P₂ = 525 kPa

Explanation:

Since this is a constant enthalpy process, S₂ = S₁

From the final state entropy, the exit temperature and the exit enthalpy (H₂) can be deduced. The change in enthalpy can be used to calculate the exit velocity based on equation (1). Finally, from the calculated value of exit velocity, the exit cross-sectional area can be calculated using equation (3).

Calculation:

Step 1:

Calculate the final state entropy, S₂

Based on the steam tables for inlet conditions, P₁ = 800 kPa and T₁= $280\text{C}$

Specific enthalpy of vapor, H_g = H₁= 3014.5 kJ/kg

Specific enthalpy of vapor, S_g = S₁= 7.1593 kJ/kg-K

Since, S₂ = S₁

We have, S₂= 7.1593 kJ/kg-K

Step 2:

Calculate the final state enthalpy, H₂

The exit temperature, T₂corresponding to the exit pressure P₂ = 525 kPa and S₂= 7.1593 kJ/kg-K, can be calculated by interpolation:

At T = $220\text{C}$, S = 7.1236 kJ/kg-K, H = 2853.8 kJ/kg

At T = $240\text{C}$, S = 7.2078 kJ/kg-K, H = 2896.8 kJ/kg

${\text{T}}_{\text{2}}\text{ = 220 + }\frac{\text{7}\text{.1593-7}\text{.1236}}{\text{7}\text{.2078-7}\text{.1236}}\text{(240-220)}=\text{ 228}\text{.34}\text{C}$

Similarly, H₂ at T₂ = $228.34\text{C}$ can be calculated by interpolation of the specific enthalpy values:

${\text{H}}_{\text{2}}\text{ = 2853}\text{.8 + }\frac{\text{228}\text{.34-220}}{\text{240-220}}\text{(2896}\text{.8-2853}\text{.8)}=2871.7\text{ kJ/kg}$

Step 3:

Calculate the exit velocity, u₂

Based on equation (3) we have:

$\begin{array}{l}{\text{H}}_{\text{2}}-{\text{H}}_{\text{1}}\text{ + }\frac{{\text{u}}_2^2-{\text{u}}_1^2}{\text{2}}\text{ = 0 }\\ (2871.7-3014.5)\text{ + }\frac{{\text{u}}_2^2-0}{\text{2}}=0\end{array}$

${\text{u}}_2^2=285.6\text{ kJ/kg}$

$\begin{array}{l}{\text{Now, 1 kJ/kg = 1000 m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{u}}_2^2=285600{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{u}}_2=534.4\text{ m/s}\end{array}$

Step 3:

Calculate the exit cross-sectional area, A

Based on equation (3) we have:

$\text{A = }\frac{\text{m<mo>˙</mo>V}}{\text{u}}$

Where:

$\text{m<mo>˙</mo> = exit mass flow rate = 0}\text{.75kg/s}$

$\text{V}=\text{ specific volume of vapor = 0}{\text{.43113 m}}^{\text{3}}\text{/kg}$

$\text{A = }\frac{0.75{\text{ kgs}}^{\text{-1}}\times \text{0}{\text{.43113 m}}^{\text{3}}{\text{kg}}^{\text{-1}}}{534.4{\text{ ms}}^{\text{-1}}}=6.05\times {10}^{-4}{\text{ m}}^{\text{2}}$

Thus, exit velocity = 534.4 m/s

The cross-sectional area at the exit = $6.05\times {10}^{-4}{\text{ m}}^{\text{2}}$