The operating conditions of a turbojet aircraft are specified. The temperature and pressure at the turbine exit, the velocity of gases at the nozzle exit, and the propulsive efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The cold-air-standard assumptions are applicable and thus air can be assumed to have constant specific heats at room temperature \left(c_{p}=0.240 \text { Btu/lbm } \cdot{ }^{\circ} F \text { and } k=1.4\right) .
3 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 4 The turbine work output is equal to the compressor work input.
Analysis The T-s diagram of the ideal jet propulsion cycle described is shown in Fig. 9–50. We note that the components involved in the jet-propulsion cycle are steady-flow devices.
(a) Before we can determine the temperature and pressure at the turbine exit, we need to find the temperatures and pressures at other states:
Process 1-2 (isentropic compression of an ideal gas in a diffuser): For convenience, we can assume that the aircraft is stationary and the air is moving toward the aircraft at a velocity of V_{1}=850 ft / s . Ideally, the air exits the diffuser with a negligible velocity \left(V_{2} \cong 0\right) :
h_{2}+\frac{{ V_{2}^{2}}^{\nearrow ^0}}{2}=h_{1}+\frac{V_{1}^{2}}{2}
\begin{aligned} 0 &=c_{p}\left(T_{2}-T_{1}\right)-\frac{V_{1}^{2}}{2} \\ T_{2} &=T_{1}+\frac{V_{1}^{2}}{2 c_{p}} \\ &=420 R +\frac{(850 ft / s )^{2}}{2(0.240 Btu / lbm \cdot R )}\left(\frac{1 Btu / lbm }{25,037 ft ^{2} / s ^{2}}\right) \\ &=480 R \end{aligned}
P_{2}=P_{1}\left(\frac{T_{2}}{T_{1}}\right)^{k /(k-1)}=(5 psia )\left(\frac{480 R }{420 R }\right)^{1.4 /(1.4-1)}=8.0 psia
Process 2-3 (isentropic compression of an ideal gas in a compressor):
\begin{aligned} &P_{3}=\left(r_{p}\right)\left(P_{2}\right)=(10)(8.0 psia )=80 psia \left(=P_{4}\right) \\ &T_{3}=T_{2}\left(\frac{P_{3}}{P_{2}}\right)^{(k-1) / k}=(480 R )(10)^{(1.4-1) / 1.4}=927 R \end{aligned}
Process 4-5 (isentropic expansion of an ideal gas in a turbine): Neglecting the kinetic energy changes across the compressor and the turbine and assuming the turbine work to be equal to the compressor work, we find the temperature and pressure at the turbine exit to be
\begin{aligned} w_{\text {comp }, \text { in }} &=w_{\text {turb,out }} \\ h_{3}-h_{2} &=h_{4}-h_{5} \\ c_{p}\left(T_{3}-T_{2}\right) &=c_{p}\left(T_{4}-T_{5}\right) \\ T_{5} &=T_{4}-T_{3}+T_{2}=2460-927+480=2013 R \\ P_{5} &=P_{4}\left(\frac{T_{5}}{T_{4}}\right)^{k /(k-1)}=(80 psia )\left(\frac{2013 R }{2460 R }\right)^{1.4 /(1.4-1)}=39.7 psia \end{aligned}
(b) To find the air velocity at the nozzle exit, we need to first determine the nozzle exit temperature and then apply the steady-flow energy equation.
Process 5-6 (isentropic expansion of an ideal gas in a nozzle):
T_{6}=T_{5}\left(\frac{P_{6}}{P_{5}}\right)^{(k-1) / k}=(2013 R )\left(\frac{5 psia }{39.7 psia }\right)^{(1.4-1) / 1.4}=1114 R
\begin{aligned} h_{6}+\frac{V_{6}^{2}}{2} &=h_{5}+\frac{{V_{5}^{2}}^{\nearrow ^0} }{2} \\ 0 &=c_{p}\left(T_{6}-T_{5}\right)+\frac{V_{6}^{2}}{2} \\ V_{6} &=\sqrt{2 c_{p}\left(T_{5}-T_{6}\right)} \\ &=\sqrt{2(0.240 Btu / lbm \cdot R )[(2013-1114) R ]\left(\frac{25,037 ft ^{2} / s ^{2}}{1 Btu / lbm }\right)} \\ &=3288 ft / s \end{aligned}
(c) The propulsive efficiency of a turbojet engine is the ratio of the propulsive power developed \dot{W}_{P} to the total heat transfer rate to the working fluid:
\begin{aligned}\dot{W}_{P} &=\dot{m}\left(V_{\text {cxit }}-V_{\text {inlet }}\right) V_{\text {aircraft }} \\&=(100 lbm / s )[(3288-850) ft / s ](850 ft / s )\left(\frac{1 Btu / lbm }{25,037 ft ^{2} / s ^{2}}\right) \\&=8276 Btu / s \quad(\text { or } 11,707 hp ) \\\dot{Q}_{\text {in }} &=\dot{m}\left(h_{4}-h_{3}\right)=\dot{m} c_{p}\left(T_{4}-T_{3}\right) \\&=(100 lbm / s )(0.240 Btu / lbm \cdot R )[(2460-927) R ] \\&=36,794 Btu / s \\\eta_{P} &=\frac{\dot{W}_{P}}{\dot{Q}_{\text {in }}}=\frac{8276 Btu / s }{36,794 Btu / s }=22.5 \%\end{aligned}
That is, 22.5 percent of the energy input is used to propel the aircraft and to overcome the drag force exerted by the atmospheric air.
Discussion For those who are wondering what happened to the rest of the energy, here is a brief account:
\begin{aligned}KE _{\text {out }} &=\dot{m} \frac{V_{g}^{2}}{2}=(100 lbm / s )\left\{\frac{[(3288-850) ft / s ]^{2}}{2}\right\}\left(\frac{1 Btu / lbm }{25,037 ft ^{2} / s ^{2}}\right) \\&=11,867 Btu / s \quad(32.2 \%) \\\dot{Q}_{\text {out }} &=\dot{m}\left(h_{6}-h_{1}\right)=\dot{m} c_{p}\left(T_{6}-T_{1}\right) \\&=(100 lbm / s )(0.24 Btu / lbm \cdot R )[(1114-420) R ] \\&=16,651 Btu / s \quad(45.3 \%)\end{aligned}
Thus, 32.2 percent of the energy shows up as excess kinetic energy (kinetic energy of the gases relative to a fixed point on the ground). Notice that for the highest propulsion efficiency, the velocity of the exhaust gases relative to the ground V_{g} should be zero. That is, the exhaust gases should leave the nozzle at the velocity of the aircraft. The remaining 45.3 percent of the energy shows up as an increase in enthalpy of the gases leaving the engine. These last two forms of energy eventually become part of the internal energy of the atmospheric air (Fig. 9–51).
