A flow at 60 ms^−1 enters a nozzle kept horizontal. The specific enthalpy at the nozzle inlet and exit is 3025 and 2790 kJ kg^−1, respectively. Neglecting the heat loss from the nozzle, calculate (a) the flow velocity at the nozzle exit, (b) the mass flow rate through the nozzle, if the inlet area

Question

A flow at [latex]60 m s ^{-1}[/latex] enters a nozzle kept horizontal. The specific enthalpy at the nozzle inlet and exit is 3025 and [latex]2790 kJ kg ^{-1},[/latex] respectively. Neglecting the heat loss from the nozzle, calculate (a) the flow velocity at the nozzle exit, (b) the mass flow rate through the nozzle, if the inlet area is [latex]0.1 m ^{2}[/latex] and the specific volume at inlet is [latex]0.19 m ^{3} kg ^{-1},[/latex] and (c) the exit area of the nozzle when the specific volume at the exit is [latex]0.5 m ^{3} kg ^{-1}.[/latex]

Accepted Answer

Let the inlet and the exit of the nozzle be denoted by subscripts 1 and 2, respectively. (a) By energy equation, we have

[latex]h_{1}+\frac{V_{1}^{2}}{2}=h_{2}+\frac{V_{2}^{2}}{2}[/latex]

[latex]3025 imes 10^{3}+\frac{60^{2}}{2}=2790 imes 10^{3}+\frac{V_{2}^{2}}{2}[/latex]

[latex]V_{2}^{2}=473600[/latex]

[latex]V_{2}=688.2 ms ^{-1}[/latex]

(b) The mass flowrate is

[latex]\dot{m}= ho A V[/latex]

[latex]=\frac{A_{1} V_{1}}{v_{1}}[/latex]

[latex]=\frac{0.1 imes 60}{0.19}[/latex]

[latex]=31.58 kg s ^{-1}[/latex]

(c) Mass flow rate is also given by

[latex]\dot{m}= ho_{2} A_{2} V_{2}=\frac{A_{2} V_{2}}{v_{2}}[/latex]

Thus,

[latex]A_{2}=\frac{\dot{m} v_{2}}{V_{2}}[/latex]

[latex]=\frac{31.58 imes 0.5}{688.2}[/latex]

[latex]=0.0229 m ^{2}[/latex]

Question 2.2: A flow at 60 ms^−1 enters a nozzle kept horizontal. The spec...

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