Question 17.16: Steam Flow through a Converging–Diverging Nozzle Steam enter...

Steam enters a converging–diverging nozzle with a low velocity. The throat and exit areas and the Mach number are to be determined.
Assumptions 1 Flow through the nozzle is one-dimensional. 2 The flow is isentropic between the inlet and the throat, and is adiabatic and irreversible between the throat and the exit. 3 The inlet velocity is negligible.

Analysis We denote the entrance, throat, and exit states by 1, t, and 2, respectively, as shown in Fig. 17–60.
(a) Since the inlet velocity is negligible, the inlet stagnation and static states are identical. The ratio of the exit-to-inlet stagnation pressure is

\frac{P_{2}}{P_{01}}=\frac{300 kPa }{2000 kPa }=0.15

It is much smaller than the critical-pressure ratio, which is taken to be P^{*} / P_{01}=0.546  since the steam is superheated at the nozzle inlet. Therefore, the flow surely is supersonic at the exit. Then the velocity at the throat is the sonic velocity, and the throat pressure is

P_{t}=0.546 P_{01}=(0.546)(2 MPa )=1.09 MPa

At the inlet,

\left.\begin{array}{l}P_{1}=P_{01}=2 MPa \\T_{1}=T_{01}=400^{\circ} C\end{array}\right\} \quad \begin{aligned}&h_{1}=h_{01}=3248.4 kJ / kg \\&s_{1}=s_{t}=s_{2 s}=7.1292 kJ / kg \cdot K\end{aligned}

Also, at the throat,

\left.\begin{array}{rl}P_{t} & =1.09 MPa \\s_{t} & =7.1292 kJ / kg \cdot K\end{array}\right\} \quad \begin{aligned}&h_{t}=3076.8 kJ / kg \\&v_{t}=0.24196 m ^{3} / kg\end{aligned}

Then the throat velocity is determined from Eq. 17–3 to be

h_{01}=h_{02}

V_{t}=\sqrt{2\left(h_{01}-h_{t}\right)}=\sqrt{[2(3248.4-3076.8) kJ / kg ]\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)}=585.8 m / s

The flow area at the throat is determined from the mass flow rate relation:

A_{t}=\frac{\dot{m} v_{t}}{V_{t}}=\frac{(2.5 kg / s )\left(0.2420 m ^{3} / kg \right)}{585.8 m / s }=10.33 \times 10^{-4} m ^{2}=10.33 cm ^{2}

At state 2s,

\left.\begin{array}{rl} P_{2 s} & =P_{2}=300 kPa \\ s_{2 s} & =s_{1}=7.1292 kJ / kg \cdot K \end{array}\right\} \quad h_{2 s}=2783.6 kJ / kg

The enthalpy of the steam at the actual exit state is (see Chap. 7)

\begin{aligned}\eta_{N} &=\frac{h_{01}-h_{2}}{h_{01}-h_{2 s}} \\0.93 &=\frac{3248.4-h_{2}}{3248.4-2783.6} \longrightarrow h_{2}=2816.1 kJ / kg\end{aligned}

Therefore,

\left.\begin{array}{l}P_{2}=300 kPa \\h_{2}=2816.1 kJ / kg\end{array}\right\} \quad \begin{aligned}&v_{2}=0.67723 m ^{3} / kg \\&s_{2}=7.2019 kJ / kg \cdot K\end{aligned}

Then the exit velocity and the exit area become

\begin{aligned}&V_{2}=\sqrt{2\left(h_{01}-h_{2}\right)}=\sqrt{[2(3248.4-2816.1) kJ / kg ]\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)}=929.8 m / s \\&A_{2}=\frac{\dot{m} v_{2}}{V_{2}}=\frac{(2.5 kg / s )\left(0.67723 m ^{3} / kg \right)}{929.8 m / s }=18.21 \times 10^{-4} m ^{2}=18.21 cm ^{2}\end{aligned}

(b) The velocity of sound and the Mach numbers at the throat and the exit of the nozzle are determined by replacing differential quantities with differences,

c=\left(\frac{\partial P}{\partial \rho}\right)_{s}^{1 / 2} \cong\left[\frac{\Delta P}{\Delta(1 / v)}\right]_{s}^{1 / 2}

The velocity of sound at the throat is determined by evaluating the specific volume at s_{t}=7.1292 kJ / kg \cdot K and at pressures of 1.115 and 1.065 MPa \left(P_{t} \pm 25 kPa \right): :

c=\sqrt{\frac{(1115-1065) kPa }{(1 / 0.23776-1 / 0.24633) kg / m ^{3}}\left(\frac{1000 m ^{2} / s ^{2}}{1 kPa \cdot m ^{3} / kg }\right)}=584.6 m / s

The Mach number at the throat is determined from Eq. 17–12 to be

Ma =\frac{V}{c}

Ma =\frac{V}{c}=\frac{585.8 m / s }{584.6 m / s }= 1 . 0 0 2

Thus, the flow at the throat is sonic, as expected. The slight deviation of the Mach number from unity is due to replacing the derivatives by differences.
The velocity of sound and the Mach number at the nozzle exit are determined by evaluating the specific volume at s_{2}=7.2019 kJ / kg \cdot K and at pressures of 325 and 275 kPa \left(P_{2} \pm 25 kPa \right) :

c=\sqrt{\frac{(325-275) kPa }{(1 / 0.63596-1 / 0.72245) kg / m ^{3}}\left(\frac{1000 m ^{2} / s ^{2}}{1 kPa \cdot m ^{3} / kg }\right)}=515.4 m / s

and

Ma =\frac{V}{c}=\frac{929.8 m / s }{515.4 m / s }=1.804

Thus the flow of steam at the nozzle exit is supersonic.