Chapter 17.7, Problem 110P

Steam enters a converging nozzle at 5.0 MPa and 400°C with a negligible velocity, and it exits at 3.0 MPa. For a nozzle exit area of 75 cm², determine the exit velocity, mass flow rate, and exit Mach number if the nozzle (a) is isentropic and (b) has an efficiency of 94 percent.

To determine

The exit velocity, mass flow rate, and exit Mach number if the nozzle is isentropic.

Answer to Problem 110P

The exit velocity of the stream is $528.5\text{m}/\text{s}$.

The mass flow rate is $46.08\text{kg}/\text{s}$.

The Mach number at the exit of nozzle is $0.917$.

Explanation of Solution

For isentropic,

The flow of steam through the nozzle is steady and isentropic.

Write the expression of energy balance equation for the converging-diverging nozzle.

$h_1+V_1^2/2=h_2+V_2^2/2$

Inlet velocity is equal to zero $V_1=0$.

$V_2=\sqrt{2\left(h_1-h_2\right)}$ (I)

Here, enthalpy at exit is $h_2$, enthalpy at inlet is $h_1$, velocity of steam at the inlet of nozzle is $V_1$, and velocity of steam at exit of nozzle is $V_2$ .

Write the expression to calculate the exit area of the nozzle.

$\begin{array}{l}{A}_2=\frac{\dot{m}{v}_2}{V_2}\\ \dot{m}=\frac{A_2{V}_2}{v_2}\end{array}$ (II)

Here, mass flow rate of steam is $\dot{m}$ , specific volume of the steam at the exit is $v_2$ .

Write the expression to calculate the velocity of sound through the steam at the exit of nozzle.

$c_2={\left(\frac{\partial P}{\partial \rho }\right)}_s^{1/2}={\left(\frac{\Delta P}{\Delta \left(1/v\right)}\right)}_s^{1/2}$ (III)

Here, pressure drop in the nozzle is $\Delta P$, and drop in the specific volume of the steam is $\Delta \left(1/v\right)$.

Write the expression to calculate the Mach number for the steam at the exit of nozzle.

${\text{Ma}}_2=\frac{V_2}{c_2}$ (IV)

Here, Mach number of the steam at the exit is ${\text{Ma}}_2$ .

Conclusion:

Refer Table A-6, “Superheated water”, obtain the values of $h_{01}$, and $s_{2s}$ at inlet pressure of $5\text{MPa}$ and temperature of $400°\text{C}$ as $\text{3196}\text{.7}\text{kJ}/\text{kg}$, and $6.6483\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Here, stagnation enthalpy at the inlet is $h_{01}$, at superheated condition the entropy of saturated steam is $s_{2s}$.

Refer Table A-6, “Superheated water”, obtain the values of $h_2$ at an entropy of $6.6483\text{kJ}/\text{kg}\cdot \text{K}$ and exit pressure of $3\text{MPa}$ as $\text{3057}\text{kJ}/\text{kg}$.

The stagnation enthalpy of steam at the inlet is equal to the actual enthalpy

at the inlet $\left(h_{01}=h_1\right)$.

Substitute $\text{3196}\text{.7}\text{kJ}/\text{kg}$ for $h_1$ and $\text{3057}\text{kJ}/\text{kg}$ for $h_2$ in Equation (II).

$\begin{array}{c}{V}_2=\sqrt{2\left(\text{3196}\text{.7}\text{kJ}/\text{kg}-\text{3057}\text{kJ}/\text{kg}\right)}\\ =\sqrt{2\left(\text{3196}\text{.7}\text{kJ}/\text{kg}-\text{3057}\text{kJ}/\text{kg}\right)\left(\frac{1000{\text{m}}^2/{\text{s}}^2}{1\text{kJ}/\text{kg}}\right)}\\ =528.5\text{m}/\text{s}\end{array}$

Thus, the exit velocity of the stream is $528.5\text{m}/\text{s}$.

Refer Table A-6, “Superheated water”, obtain the value of $v_2$ at the entropy of $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ and a exit pressure of $3\text{MPa}$ as $0.08601{\text{m}}^3/\text{kg}$.

Substitute $75{\text{cm}}^2$ for $A$, $0.08601{\text{m}}^3/\text{kg}$ for $v_2$, and $528.5\text{m}/\text{s}$  for $V_2$ in Equation (III).

$\begin{array}{c}\dot{m}=\frac{75{\text{cm}}^2\times 528.5\text{m}/\text{s}}{0.08601{\text{m}}^3/\text{kg}}\\ =\frac{75{\text{cm}}^2\left(\frac{1{\text{m}}^2}{\text{10000}{\text{cm}}^2}\right)\times 528.5\text{m}/\text{s}}{0.08601{\text{m}}^3/\text{kg}}\\ =46.08\text{kg}/\text{s}\end{array}$

Therefore, the mass flow rate is $46.08\text{kg}/\text{s}$.

Refer Table A-6, “Superheated water”, obtain the value of specific volume of steam at the entropy of $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ at pressure just below and above the specified pressure of $2.5\text{MPa}$ and $3.5\text{MPa}$ as $0.09906{\text{m}}^3/\text{kg}$ and $0.07632{\text{m}}^3/\text{kg}$ respectively.

Substitute $\left(3500-2500\right)\text{kPa}$ for $\Delta P$, and $\left(\frac{1}{0.09906}-\frac{1}{0.07632}\right)\text{kg}/{\text{m}}^{\text{3}}$ for $\Delta \left(1/v\right)$ in Equation (IV).

$\begin{array}{c}{c}_2={\left(\frac{\left(3500-2500\right)\text{kPa}}{\left(\frac{1}{0.09906}-\frac{1}{0.07632}\right)\text{kg}/{\text{m}}^{\text{3}}}\right)}^{1/2}\\ ={\left(\frac{\left(3500-2500\right)\text{kPa}}{\left(\frac{1}{0.09906}-\frac{1}{0.07632}\right)\text{kg}/{\text{m}}^{\text{3}}}\left(\frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kPa}\cdot {\text{m}}^3}\right)\right)}^{1/2}\\ =576.6\text{m}/\text{s}\end{array}$

Substitute $576.6\text{m}/\text{s}$ for $c_2$, and $528.5\text{m}/\text{s}$ for $V_2$ in Equation (V).

$\begin{array}{c}{\text{Ma}}_2=\frac{528.5\text{m}/\text{s}}{576.6\text{m}/\text{s}}\\ =0.917\end{array}$

Hence, the Mach number at the exit of nozzle is $0.917$.

To determine

The exit velocity, mass flow rate, and exit Mach number if the has an efficiency of 94 percent.

Answer to Problem 110P

The exit velocity of the stream is $512.4\text{m}/\text{s}$.

The mass flow rate is $44.34\text{kg}/\text{s}$.

The Mach number at the exit of nozzle is $0.885$.

Explanation of Solution

Nozzle has an efficiency of 94 percent:

Write the expression for the efficiency of nozzle.

${\eta }_N=\frac{h_{01}-h_2}{h_{01}-h_{2s}}$ (V)

Here, efficiency of nozzle is ${\eta }_N$, stagnation enthalpy at the inlet is $h_{01}$, enthalpy at exit is $h_2$, and superheated enthalpy at exit is $h_{2s}$.

Write the expression of energy balance equation for the converging-diverging nozzle.

$h_1+V_1^2/2=h_2+V_2^2/2$

Inlet velocity is equal to zero $V_1=0$.

$V_2=\sqrt{2\left(h_1-h_2\right)}$ (VI)

Here, velocity of steam at the inlet of nozzle is $V_1$, and velocity of steam at exit of nozzle is $V_2$ .

Write the expression to calculate the exit area of the nozzle.

$A_2=\frac{\dot{m}{v}_2}{V_2}$ (VII)

Here, mass flow rate of steam is $\dot{m}$ , specific volume of the steam at the exit is $v_2$ .

Write the expression to calculate the velocity of sound through the steam at the exit of nozzle.

$c_2={\left(\frac{\partial P}{\partial \rho }\right)}_s^{1/2}={\left(\frac{\Delta P}{\Delta \left(1/v\right)}\right)}_s^{1/2}$ (VIII)

Here, pressure drop in the nozzle is $\Delta P$, and drop in the specific volume of the steam is $\Delta \left(1/v\right)$.

Write the expression to calculate the Mach number for the steam at the exit of nozzle.

${\text{Ma}}_2=\frac{V_2}{c_2}$ (IX)

Here, Mach number of the steam at the exit is ${\text{Ma}}_2$ .

Conclusion:

Refer Table A-6, “Superheated water”, obtain the values of $h_{01}$, and $s_{2s}$ at inlet pressure of $5\text{MPa}$ and temperature of $400°\text{C}$ as $\text{3196}\text{.7}\text{kJ}/\text{kg}$, and $6.6483\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Here, at superheated condition the entropy of saturated steam is $s_{2s}$.

Refer Table A-6, “Superheated water”, obtain the values of $h_{2s}$ at an entropy of $6.6483\text{kJ}/\text{kg}\cdot \text{K}$ and a pressure of $3\text{MPa}$ as $\text{3057}\text{kJ}/\text{kg}$.

Substitute $\text{3057}\text{kJ}/\text{kg}$ for $h_{2s}$, $\text{3196}\text{.7}\text{kJ}/\text{kg}$ for $h_{01}$, and 0.90 for ${\eta }_N$ in Equation (V).

$\begin{array}{l}94\%=\frac{\text{3196}\text{.7}\text{kJ}/\text{kg}-h_2}{\text{3196}\text{.7}\text{kJ}/\text{kg}-\text{3057}\text{kJ}/\text{kg}}\\ 94\left(\frac{1}{100}\right)=\frac{\text{3196}\text{.7}\text{kJ}/\text{kg}-h_2}{\text{3196}\text{.7}\text{kJ}/\text{kg}-\text{3057}\text{kJ}/\text{kg}}\\ h_2=3065.4\text{kJ}/\text{kg}\end{array}$

The stagnation enthalpy of steam at the inlet is equal to the actual enthalpy

at the inlet $\left(h_{01}=h_1\right)$.

Substitute $\text{3196}\text{.7}\text{kJ}/\text{kg}$ for $h_1$ and $3065.4\text{kJ}/\text{kg}$ for $h_2$ in Equation (VI).

$\begin{array}{c}{V}_2=\sqrt{2\left(\text{3196}\text{.7}\text{kJ}/\text{kg}-3065.4\text{kJ}/\text{kg}\right)}\\ =\sqrt{2\left(\text{3196}\text{.7}\text{kJ}/\text{kg}-3065.4\text{kJ}/\text{kg}\right)\left(\frac{1000{\text{m}}^2/{\text{s}}^2}{1\text{kJ}/\text{kg}}\right)}\\ =512.4\text{m}/\text{s}\end{array}$

Thus, the exit velocity of the stream is $512.4\text{m}/\text{s}$.

Refer Table A-6, “Superheated water”, obtain the value of $v_2$ at the entropy of $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ and a exit pressure of $3\text{MPa}$ as $0.08666{\text{m}}^3/\text{kg}$.

Substitute $75{\text{cm}}^2$ for $A_2$, $0.08666{\text{m}}^3/\text{kg}$ for $v_2$, and $512.4\text{m}/\text{s}$ for $V_2$ in Equation (VII).

$\begin{array}{c}\dot{m}=\frac{75{\text{cm}}^2\times 512.4\text{m}/\text{s}}{0.08666{\text{m}}^3/\text{kg}}\\ =\frac{75{\text{cm}}^2\left(\frac{1{\text{m}}^2}{\text{10000}{\text{cm}}^2}\right)\times 512.4\text{m}/\text{s}}{0.08666{\text{m}}^3/\text{kg}}\\ =44.34\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate is $44.34\text{kg}/\text{s}$.

Refer Table A-6, “Superheated water”, obtain the value of specific volume of steam $\Delta \left(1/v\right)$ at the entropy of $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ at pressure just below and above the specified pressure of $2.5\text{MPa}$ and $3.5\text{MPa}$ as $0.09981{\text{m}}^3/\text{kg}$ and $0.07689{\text{m}}^3/\text{kg}$ respectively.

Substitute $\left(3500-2500\right)\text{kPa}$ for $\Delta P$, and $\left(\frac{1}{0.09981}-\frac{1}{0.07689}\right)\text{kg}/{\text{m}}^{\text{3}}$ for $\Delta \left(1/v\right)$ in Equation (VIII).

$\begin{array}{c}{c}_2={\left(\frac{\left(3500-2500\right)\text{kPa}}{\left(\frac{1}{0.09981}-\frac{1}{0.07689}\right)\text{kg}/{\text{m}}^{\text{3}}}\right)}^{1/2}\\ ={\left(\frac{\left(3500-2500\right)\text{kPa}}{\left(\frac{1}{0.09981}-\frac{1}{0.07689}\right)\text{kg}/{\text{m}}^{\text{3}}}\left(\frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kPa}\cdot {\text{m}}^3}\right)\right)}^{1/2}\\ =578.7\text{m}/\text{s}\end{array}$

Substitute $578.7\text{m}/\text{s}$ for $c_2$, and $512.4\text{m}/\text{s}$ for $V_2$ in Equation (IX).

$\begin{array}{c}{\text{Ma}}_2=\frac{512.4\text{m}/\text{s}}{578.7\text{m}/\text{s}}\\ =0.885\end{array}$

Thus, the Mach number at the exit of nozzle is $0.885$.