Question

Steam at 5 MPa and 400 C enters a nozzle steadily with a velocity of 80 m/s, and it leavesat 2 MPa and 300 C. The inlet area of the nozzle is 50 cm2, and heat is being lost at a rateof 120 kJ/s. Determine the following:
a) the mass flow rateof the steam.
b) the exit velocity of the steam.
c) the exitarea of the nozzle.

Answer 1

Answer:

a) the mass flow rate of the steam is  $\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam  is $\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle is  $A_2$ = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

$P_1 =$ 5 MPa

$T_1$ = 400° C

Velocity V = 80 m/s

Exit:

$P_2 =$ 2 MPa

$T_2$ = 300° C

From the properties of steam tables  at $P_1 =$ 5 MPa and $T_1$ = 400° C we obtain the following properties for enthalpy h and the speed v

$h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg$

From the properties of steam tables  at $P_2 =$ 2 MPa and $T_1$ = 300° C we obtain the following properties for enthalpy h and the speed v

$h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg$

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

$m_1=m_2=m_3$

Thus

$m_1 =\dfrac{V_1 \times A_1}{v_1}$

$m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}$

$m_1 =\dfrac{0.4 }{0.057838 }$

$\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam.

Using Energy Balance equation:

$\Delta E _{system} = E_{in}-E_{out}$

In a steady flow process;

$\Delta E _{system} = 0$

$E_{in} = E_{out}$

$m(h_1 + \dfrac{V_1^2}{2})$ $= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})$

$- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})$

$- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$V_2^2 = 316631.29 \ m/s$

$V_2 = \sqrt{316631.29 \ m/s$

$\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

$m = \dfrac{V_2A_2}{v_2}$

making $A_2$ the subject of the formula ; we have:

$A_2 = \dfrac{ m \times v_2}{V_2}$

$A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}$

$A_2$ = 0.0015435 m²