Chapter 7, Problem 7.5P

Interpretation Introduction

Interpretation:

The minimum pressure and the 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:

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

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

$$

Minimum pressure = 436.8 kPa

Cross-sectional area at the nozzle exit = $0.0190{\text{ m}}^{\text{2}}$

Given Information:

Inlet pressure of steam, P₁ = 800 kPa

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

Explanation:

The minimum pressure at the exit can be calculated based on the critical pressure ratio:

$\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{ = }{\left(\frac{\text{2}}{\text{γ+1}}\right)}^{\text{γ/γ-1}}\text{ -----(4)}$

Step 1:

Calculate the minimum pressure

Based on equation (4) we have:

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

${\text{where: P}}_{\text{1}}\text{ = 800 kPa and γ = 1}\text{.3 for superheated steam }$

${\text{P}}_{\text{2}}\text{= 800}\times {\left(\frac{\text{2}}{\text{1}\text{.3+1}}\right)}^{\text{1}\text{.3/1}\text{.3-1}}=800\times 0.546=436.8\text{ kPa}$

Step 2:

Calculate the exit velocity, u₂

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

Specific volume of vapor, V_g = V₁= 0.31189 m³/kg

The velocity change is given as:

${\text{Δu}}^2\text{ = }\frac{{\text{2γP}}_{\text{1}}{\text{V}}_{\text{1}}}{\text{γ-1}}\left[\text{1-}{\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)}^{\text{γ-1/γ}}\right]$

$=\frac{\text{2(1}\text{.3)(800)(0}\text{.31189)}}{\text{1}\text{.3-1}}\left[\text{1-}{\left(\frac{436.8}{800}\right)}^{\text{1}\text{.3-1/1}\text{.3}}\right]=281.876{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}$

${\text{Δu}}^2\text{ = }281.876{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}$

${\text{u}}_2^2-{\text{u}}_1^2=281.876$

${\text{u}}_2^2-0=281.876$

${\text{u}}_{\text{2}}=16.79\text{ m/s}$

Step 4:

Calculate the exit cross-sectional area, A

Based on equation (3) we have:

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

where:

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

$\text{V}=\text{ specific volume of vapor = 0}{\text{.42566 m}}^{\text{3}}\text{/kg}$

$\text{A = }\frac{0.75{\text{ kgs}}^{\text{-1}}\times \text{0}{\text{.42566 m}}^{\text{3}}{\text{kg}}^{\text{-1}}}{16.79{\text{ ms}}^{\text{-1}}}=0.0190{\text{ m}}^{\text{2}}$

Thus, minimum pressure = 436.8 kPa

Cross-sectional area at the nozzle exit = $0.0190{\text{ m}}^{\text{2}}$