Chapter 5.5, Problem 33P

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 375°C and 400 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm², determine the velocity and the volume flow rate of the steam at the nozzle exit.

FIGURE P5–33

To determine

The velocity and volume flow rate of the steam at the nozzle exit.

Answer to Problem 33P

The exit volume flow rate of the steam is $1.5472{\text{m}}^3\text{/s}$.

The exit velocity of the steam is $259.0027\text{m/s}$.

Explanation of Solution

Write the energy rate balance equation for one inlet and one outlet system.

  $\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$    …… (I)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

Consider the steam flows at steady state through the nozzle. Hence, the rate of change in net energy of the system becomes zero.

  $\Delta {\dot{E}}_{\text{system}}=0$

Here, the heat loss from the nozzle is estimated as $25\text{kW}$.

  ${\dot{Q}}_{\text{2}}=25\text{kW}$

There is no heat transfer at inlet i.e. ${\dot{Q}}_1=0$ and no work done in nozzle i.e. $\dot{W}=0$. The inlet, outlet are at same elevation, the potential energy becomes negligible i.e. $gz=0$.

The Equations (I) reduced as follows to obtain the exit velocity.

  $\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[0+{\dot{Q}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\left[{\dot{Q}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)={\dot{Q}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-{\dot{Q}}_2=\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\end{array}$

  $\begin{array}{l}{h}_1+\frac{V_1{}^2}{2}-\frac{{\dot{Q}}_2}{\dot{m}}=h_2+\frac{V_2{}^2}{2}\\ V_2{}^2=2\left[h_1-h_2+\frac{V_1{}^2}{2}-\frac{{\dot{Q}}_2}{\dot{m}}\right]\\ V_2=\sqrt{2\left[h_1-h_2+\frac{V_1{}^2}{2}-\frac{{\dot{Q}}_2}{\dot{m}}\right]}\end{array}$                                                                        …… (II)

At steady flow, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for inlet mass flow rate.

  $\dot{m}=\frac{A_1{V}_1}{v_1}$                                                                                                    …… (III)

Here, the cross-sectional area is $A$, the velocity is $V$ and the specific volume is $v$; the suffix 1 indicates the inlet condition.

Write the formula for exit volumetric flow rate.

  ${\dot{V}}_2=\dot{m}{v}_2$                                                                                                     …… (IV)

At both the inlet and exit of the nozzle, the steam is at the state of superheated condition.

Refer Table A-6, “Superheated water”.

At inlet:

Obtain the following properties corresponding to the temperature of $400°\text{C}$ and the pressure of $800\text{kPa}\left(0.8\text{MPa}\right)$.

  $\begin{array}{l}{v}_1=0.38429{\text{m}}^3\text{/kg}\\ h_1=3267.7\text{kJ/kg}\end{array}$

At outlet:

Obtain the following properties corresponding to the temperature of $375°\text{C}$ and the pressure of $400\text{kPa}\left(0.4\text{MPa}\right)$- using linear interpolation.

Write the formula of interpolation method of two variables.

  $y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$                                                                          …… (V)

Show the temperature and enthalpy values from the Table A-6 as in below table.

S.No.	x	y
	Temperature $\left(T\right)$,in $\text{°C}$	Enthalpy $\left(h\right)$, in $\text{kJ/kg}$
1	300	3067.1
2	375	?
3	400	3273.9

Substitute $300$ for $x_1$, $375$ for $x_2$, $400$ for $x_3$, $3067.1$ for $y_1$, and $3273.9$ for $y_3$ in Equation (V).

  $\begin{array}{c}{y}_2=\frac{\left(375-300\right)\left(3273.9-3067.1\right)}{\left(400-300\right)}+3067.1\\ =3222.2\text{kJ/kg}\end{array}$

Thus, the enthalpy $\left(h_2\right)$ corresponding to the temperature of $375°\text{C}$ and the pressure of $400\text{kPa}\left(0.4\text{MPa}\right)$is $3222.2\text{kJ/kg}$.

Similarly, the exit specific volume $\left(v_2\right)$ corresponding to the temperature of $375°\text{C}$ and the pressure of $400\text{kPa}\left(0.4\text{MPa}\right)$-using linear interpolation method is $0.74321{\text{m}}^3\text{/kg}$.

Conclusion:

Substitute $800{\text{cm}}^2$ for $A_1$, $10\text{m/s}$ for $V_1$ and $0.38429{\text{m}}^3\text{/kg}$ for $v_1$ in Equation (III).

  $\begin{array}{c}\dot{m}=\frac{\left(800{\text{cm}}^2\right)\left(10\text{m/s}\right)}{0.38429{\text{m}}^3\text{/kg}}\\ =\frac{\left(800{\text{cm}}^2\times \frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\left(10\text{m/s}\right)}{0.38429{\text{m}}^3\text{/kg}}\\ =\frac{0.8{\text{m}}^3\text{/s}}{0.38429{\text{m}}^3\text{/kg}}\\ =2.0818\text{kg/s}\end{array}$

Substitute $2.0818\text{kg/s}$ for $\dot{m}$ and $0.74321{\text{m}}^3\text{/kg}$ for $v_2$ in Equation (IV).

  $\begin{array}{c}{\dot{V}}_2=\left(2.0818\text{kg/s}\right)\left(0.74321{\text{m}}^3\text{/kg}\right)\\ =1.5472{\text{m}}^3\text{/s}\end{array}$

Thus, the exit volume flow rate of the steam is $1.5472{\text{m}}^3\text{/s}$.

Substitute $3267.7\text{kJ/kg}$ for $h_1$, $3222.2\text{kJ/kg}$ for $h_2$, $10\text{m/s}$ for $V_1$, $25\text{kW}$ for ${\dot{Q}}_2$ and $2.0818\text{kg/s}$ for $\dot{m}$ in Equation (II).

  $\begin{array}{c}{V}_2=\sqrt{2\left[3267.7\text{kJ/kg}-3222.2\text{kJ/kg}+\frac{{\left(10\text{m/s}\right)}^2}{2}-\frac{\left(-25\text{kW}\right)}{2.0818\text{kg/s}}\right]}\\ =\sqrt{2\left[\left(45.5\text{kJ/kg}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kJ/kg}}\right)+50{\text{m}}^2{\text{/s}}^2-\frac{\left(25\text{kW}\times \frac{1\text{J/s}}{1\text{W}}\right)}{2.0818\text{kg/s}}\right]}\\ =\sqrt{2\left[\left(45.5\times {10}^3{\text{m}}^2{\text{/s}}^2\right)+50{\text{m}}^2{\text{/s}}^2-\left(12.0088\text{kJ/kg}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kJ/kg}}\right)\right]}\\ =\sqrt{2\left[\left(45.5\times {10}^3{\text{m}}^2{\text{/s}}^2\right)+50{\text{m}}^2{\text{/s}}^2-12.0088\times {10}^3{\text{m}}^2{\text{/s}}^2\right]}\end{array}$

  $\begin{array}{c}=\sqrt{67082.4{\text{m}}^2{\text{/s}}^2}\\ =259.0027\text{m/s}\end{array}$

Thus, the exit velocity of the steam is $259.0027\text{m/s}$.