Chapter 7, Problem 7.9P

Interpretation Introduction

Interpretation:

The mass flow rate and the state of steam 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:

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

(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:

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

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

$$

The state of steam at exit is superheated vapor with a mass flow rate of 2.78 kg/s

Given Information:

Inlet pressure of steam, P₁ = 1400 kPa

Inlet Temperature of steam T₁= $325\text{C}$

Outlet pressure of steam, P₂ = 140 kPa

Cross-sectional area at the throat of the nozzle = 6 cm² = $6\times {10}^{-4}{\text{ m}}^2$

Explanation:

For a converging-diverging nozzle, mass flow rate $\text{(m<mo>˙</mo>)}$ is related to the critical pressure ratio (P₂/P₁) as:

$\text{m<mo>˙</mo> =}\frac{{\text{γP}}_{\text{1}}\text{A}}{\sqrt{{\text{γRT}}_{\text{1}}}}\text{M}{\left(\text{1+}\frac{\text{γ-1}}{\text{2}}{\text{M}}^2\right)}^{\text{2-γ/2}}\text{ -----(4)}$

$\text{where: M is the Mach number which is related to the critical pressure ratio as:}$

$\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{ = }{\left(\text{1+}\frac{\text{γ-1}}{\text{2}}{\text{M}}^{\text{2}}\right)}^{\text{γ/γ-1}}\text{ ----(5)}$

Step 1:

Calculate the Mach number at exit

Based on equation (5) we have:

$\frac{\text{1400}}{\text{140}}\text{ = }{\left(\text{1+}\frac{\text{1}\text{.3-1}}{\text{2}}{\text{M}}^{\text{2}}\right)}^{\text{1}\text{.3/1}\text{.3-1}}$

$\text{M = 3}\text{.37}$

Step 2:

Calculate the mass flow rate

Based on equation (4) we have:

$\text{m<mo>˙</mo> =}\frac{\text{(1}\text{.3)(1400)(6}\times {\text{10}}^{-4})}{\sqrt{\text{(1}\text{.3)(0}\text{.008314)(325)}}}(3.37){\left(\text{1+}\frac{\text{1}\text{.3-1}}{\text{2}}{(3.37)}^2\right)}^{\text{2-1}\text{.3/2}}\text{ = 2}\text{.78 kg/s}$

Thus, the state of steam at exit is superheated vapor with a mass flow rate of 2.78 kg/s.