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:
[tex]V_i = 300m/s[/tex]
[tex]T_1 = 7\°C = 280K[/tex]
[tex]T_2 = 427\°C = 700K[/tex]
[tex]\dot{m} 16kg/s[/tex]
[tex]\dot{Q}_{in} = 15000kJ/s[/tex]
Using the tables for gas properties of air we can find the enthalpy in this two states, then
[tex]T_1 = 280K \rightarrow h_1 = 280.13kJ/Kg\cdot K[/tex]
[tex]T_2 = 700K \rightarrow h_2 = 713.27kJ/Kg\cdot K[/tex]
Applying energy equation to the entire engine we have that
[tex]\frac{\dot{Q}_{in}}{\dot{m}}+h_1 +\frac{1}{2} V^2_{in} = h_2 + \frac{1}{2} V^2_{out}[/tex]
[tex]\frac{15000}{16}*10^3+280.13*10^3+\frac{1}{2} 300^2 = 713.27*10^3+\frac{1}{2} V^2_{out}[/tex]
[tex]V_{out} = 1048.198m/s[/tex]
Finally the force in terms of mass flow and velocity is
[tex]F = \dot{m} (V_{out}-V_{in})[/tex]
[tex]F = 16(1048.198-300)[/tex]
[tex]F= 11971.168N[/tex]
Therefore the thrust produced by this turbojet engine is 11971.168N
