Known An ideal turbojet engine operates at steady state. Key operating conditions are specified.
Find Determine the velocity at the nozzle exit, in ft/s, and the pressure, in lbf / in .{ }^{2}, at each principal state.
Schematic and Given Data:
Engineering Model
1. Each component is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines.
2. The diffuser, compressor, turbine, and nozzle processes are isentropic.
3. There is no pressure drop for flow through the combustor.
4. The turbine work output equals the work required to drive the compressor.
5. Except at the inlet and exit of the engine, kinetic energy effects can be ignored. Potential energy effects are negligible throughout.
6. The working fluid is air modeled as an ideal gas.
Analysis To determine the velocity at the exit to the nozzle, the mass and energy rate balances for a control volume enclosing this component reduce at steady state to give
0=\dot{Q}_{ cv }^{\nearrow0}-\dot{W}_{ cv }^{\nearrow0}+\dot{m}\left[\left(h_{4}-h_{5}\right)+\left(\frac{ y _{4}^{2^{\nearrow0}}- V _{5}^{2}}{2}\right)+g\left(z_{4}-{z}_{5}\right)^{\nearrow0}\right]
where \dot{m} is the mass flow rate. The inlet kinetic energy is dropped by assumption 5. Solving for V _{5}
V _{5}=\sqrt{2\left(h_{4}-h_{5}\right)}
This expression requires values for the specific enthalpies h_{4} \text { and } h_{5} at the nozzle inlet and exit, respectively. With the operating parameters specified, the determination of these enthalpy values is accomplished by analyzing each component in turn, beginning with the diffuser. The pressure at each principal state can be evaluated as a part of the analyses required to find the enthalpies h_{4} \text { and } h_{5}.
Mass and energy rate balances for a control volume enclosing the diffuser reduce to give
h_{1}=h_{ a }+\frac{ V _{ a }^{2}}{2}
With h_{ a } from Table A-22E and the given value of V_{ a }
1 h_{1}=102.7 Btu / lb +\left[\frac{(909.3)^{2}}{2}\right]\left(\frac{ ft ^{2}}{ s ^{2}}\right)\left|\frac{1 lbf }{32.2 lb \cdot ft / s ^{2}}\right| \left|\frac{1 Btu }{778 ft \cdot lbf }\right|
= 119.2 Btu/lb
Interpolating in Table A-22E givesp_{ r 1}=1.051. The flow through the diffuser is isentropic, so pressure p_{1} is
p_{1}=\frac{p_{ r 1}}{p_{ ra }} p_{ a }
With p_{ r } data from Table A-22E and the known value of p_{ a }
p_{1}=\frac{1.051}{0.6268}\left(11.8 lbf / in .^{2}\right)=19.79 lbf / in .^{2}
Using the given compressor pressure ratio, the pressure at state 2 is p_{2}=8\left(19.79 lbf / in .{ }^{2}\right)=158.3 lbf / in .{ }^{2}
The flow through the compressor is also isentropic. Thus,
p_{ r 2}=p_{ r 1} \frac{p_{2}}{p_{1}}=1.051(8)=8.408
Interpolating in Table A-22E, we get h_{2}=216.2 Btu/lb.
At state 3 the temperature is given as T_{3}=2150^{\circ} R. From Table A-22E, h_{3}=546.54 Btu/lb. By assumption 3, p_{3}=p_{2}. The work developed by the turbine is just sufficient to drive the compressor (assumption 4). That is,
\frac{\dot{W}_{ t }}{\dot{m}}=\frac{\dot{W}_{ c }}{\dot{m}}
or
h_{3}-h_{4}=h_{2}-h_{1}
Solving for h_{4}
h_{4}=h_{3}+h_{1}-h_{2}=546.54+119.2-216.2
= 449.5 Btu/lb
Interpolating in Table A-22E with h_{4} \text { gives } p_{ r 4}=113.8.
The expansion through the turbine is isentropic, so
p_{4}=p_{3} \frac{p_{ r 4}}{p_{ r 3}}
With p_{3}=p_{2} \text { and } p_{ r } data from Table A-22E
p_{4}=\left(158.3 lbf / in .^{2}\right) \frac{113.8}{233.5}=77.2 lbf / in .{ }^{2}
The expansion through the nozzle is isentropic to p_{5}=11.8 lbf / in .{ }^{2} Thus,
p_{ r 5}=p_{ r 4} \frac{p_{5}}{p_{4}}=(113.8) \frac{11.8}{77.2}=17.39
From Table A-22E, h_{5}=265.8 Btu/lb, which is the remaining specific enthalpy value required to determine the velocity at the nozzle exit.
Using the values for h_{4} \text { and } h_{5} determined above, the velocity at the nozzle exit is
V _{5}=\sqrt{2\left(h_{4}-h_{5}\right)}
=\sqrt{2(449.5-265.8) \frac{ Btu }{ lb }\left|\frac{32.2 lb \cdot ft / s ^{2}}{1 lbf }\right| \left| \frac{778 ft \cdot lbf }{1 Btu }\right|}
= 3034 ft/s (2069 mi/h)
1 Note the unit conversions required here and in the calculation of V _{5}.
2 The increase in the velocity of the air as it passes through the engine gives rise to the thrust produced by the engine. A detailed analysis of the forces acting on the engine requires Newton’s second law of motion in a form suitable for control volumes (see Sec. 9.12.1).
Skills Developed
Ability to…
• sketch the schematic of the turbojet engine and the T–s diagram for the corresponding air-standard cycle.
• evaluate temperatures and pressures at each principal state and retrieve necessary property data.
• apply mass, energy, and entropy principles.
• calculate the nozzle exit velocity.
Quick Quiz
Using Eq. 6.47, the isentropic nozzle efficiency, what is the nozzle exit velocity, in ft/s, if the efficiency is 90%? Ans. 2878 ft/s.
\eta_{\text {nozzle }}=\frac{V_{2}^{2} / 2}{\left( V _{2}^{2} / 2\right)_{ s }} (6.47)
