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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.

1 Answer

3 votes

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

Step-by-step 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 =(V_1 * A_1)/(v_1)


m_1 =(80 \ m/s * 50 * 10 ^(-4) \ m^2)/(0.057838 \ m^3/kg)


m_1 =(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 + (V_1^2)/(2))
= Q_(out) + m (h_2 + (V_2^2)/(2))


- Q_(out) = m (h_2 - h_1 + (V_2^2-V^2_1)/(2))


- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + (V_2^2- 80 m/s^2)/(2)) * ((1 \ kJ/kg)/(1000 \ m^2/s^2))


- 120 kJ/s = 6.92 \ kg/s (-172.5 + (V_2^2- 80 m/s^2)/(2)) * ((1 \ kJ/kg)/(1000 \ m^2/s^2))


- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( (V_2^2- 80 m/s^2)/(2)) * ((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 = (V_2A_2)/(v_2)

making
A_2 the subject of the formula ; we have:


A_2 = ( m * v_2)/(V_2)


A_2 = ( 6.92 * 0.12551)/(562.7)


A_2 = 0.0015435 m²

User Ahmed Ziani
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