Question 12.3: Figure 12.15 shows the compressor and turbine maps in a simp......
Figure 12.15 shows the compressor and turbine maps in a simple turbojet engine. The engine design point is its 11.3 km (37,000 ft) cruise operation, where the flight Mach number is 0.71. The ambient pressure and temperature at this altitude, are 0.21 bars and 218 K, respectively. The compressor operating state is identified on the map as point “C,” and the station-designation scheme is the same as that in Fig. 12.1. The engine cruise operation is also subject to the following:
• Fuel-to-air ratio (f) = 0.024
• Mechanical efficiency (η_{m}) = 94%
• Combustor total pressure loss (Δp_{t}/p_{t\ 3}) = 12.5%
• Turbine-inlet total temperature (T_{t\ 4}) = 1244 K
• The cold and hot γ magnitudes are 1.4 and 1.33, respectively
Calculate the turbine total-to-total pressure ratio.
Using the compressor map, we get
N_{C2}=44,000\,\mathrm{rpm}
\pi_{C}=2.1
\eta_{C}=82.0%
{\dot{m}}_{C2}=5.5\;\mathrm{kg/s}
In order to calculate the turbine total-to-total pressure ratio, we need to locate its point of operation on the turbine map (note that point T on the turbine map is actually the final result of this example). To this end, we proceed as follows:
T_{t\ 2}=T_{t\ 1}=T_{1}\biggl[1+\biggl({\frac{\gamma-1}{2}}\biggr)M_{1}{}^{2}\biggr]=240.0\,\mathrm{K}
p_{t\ 2}=p_{t\ 1}=p_{1}\biggl[1+\biggl(\frac{\gamma-1}{2}\biggr)M_{1}{}^{2}\biggr]^{\frac{\gamma}{\gamma-1}}=0.294\;\mathrm{bars}
\tau_{C}\equiv\frac{T_{t\ 3}}{T_{t\ 2}}=1+\frac{1}{\eta_{C}}\Bigl[(\pi_{C})^{\frac{\gamma-1}{\gamma}}-1\Bigr]=1.29
\tau_{T}\equiv\frac{T_{t\ 5}}{T_{t\ 4}}=1-\frac{1}{\eta_{m}(1+f)}\frac{c_{p_{C}}}{c_{p_{T}}}\biggl(\frac{T_{t\ 2}}{T_{t\ 4}}\biggl)(\tau_{C}-1)=0.949
N_{C4}=N_{C2}\sqrt{\frac{T_{t\ 2}}{T_{t\ 4}}}=19,326\,\,\mathrm{rpm}
p_{t\ 3}=\pi_{C} p_{t\ 2}=0.617\;\mathrm{bars}
(\Delta p_{t})_{c o m b u s t o r}=0.125\ p_{t\ 3}=0.077\;\mathrm{bars}
p_{t\ 4}=p_{t\ 3}-(\Delta p_{t})_{c o m b u s t o r}=0.54\:\mathrm{bars}
{\dot{m}}_{C4}={\sqrt{\frac{T_{t\ 4}}{T_{t\ 2}}}}{\frac{p_{t\ 2}}{p_{t\ 4}}}{\dot{m}}_{C2}=1.24{\mathrm{~kg/s}}
With the corrected magnitudes of speed and mass-flow rate, we are now in a position to locate the turbine operation point on its map (point “T” on the map). Associated with this point is the following total-to-total pressure ratio:
{\frac{p_{t\ 4}}{p_{t\ 5}}}=1.25