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IRISSAK [1]
3 years ago
7

Air at 7 deg Celcius enters a turbojet engine at a rate of 16 kg/s and at a velocity of 300 m/s (relative to engine). Air is hea

ted in the combustion chamber and it leaves the engine at 427 deg Celcius. Determine the thrust produced by this turbojet engine.
Engineering
1 answer:
Savatey [412]3 years ago
4 0

To solve this problem we will start using the given temperature values and then transform them to the Kelvin scale. From there through the properties of the tables, described for the Air, we will find the entropy. With these three data we can perform energy balance and find the speed of the fluid at the exit, which will finally help us find the total force:

V_i = 300m/s

T_1 = 7\°C = 280K

T_2 = 427\°C = 700K

\dot{m} 16kg/s

\dot{Q}_{in} = 15000kJ/s

Using the tables for gas properties of air we can find the enthalpy in this two states, then

T_1 = 280K \rightarrow h_1 = 280.13kJ/Kg\cdot K

T_2 = 700K \rightarrow h_2 = 713.27kJ/Kg\cdot K

Applying energy equation to the entire engine we have that

\frac{\dot{Q}_{in}}{\dot{m}}+h_1 +\frac{1}{2} V^2_{in} = h_2 + \frac{1}{2} V^2_{out}

\frac{15000}{16}*10^3+280.13*10^3+\frac{1}{2} 300^2 = 713.27*10^3+\frac{1}{2} V^2_{out}

V_{out} = 1048.198m/s

Finally the force in terms of mass flow and velocity is

F = \dot{m} (V_{out}-V_{in})

F = 16(1048.198-300)

F= 11971.168N

Therefore the thrust produced by this turbojet engine is  11971.168N

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Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr
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Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

P _1= 15 psia

P _1= 100 psia

We know that work for isothermal process  

W=mRT\ln \dfrac{P_1}{P_2}

Lets take mass is 1 kg.

So work per unit mass

W=RT\ln \dfrac{P_1}{P_2}

We know that for air R=0.287KJ/kg.K

W=RT\ln \dfrac{P_1}{P_2}

W=0.287\times 310.92\ln \dfrac{15}{100}

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

\Delta S=-R\ln \dfrac{P_2}{P_1}

\Delta S=-0.287\ln \dfrac{100}{15}

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