Question 12.1: The cruise operation of a single-spool turbojet engine (Fig.......

Part Ia:

T_{t\ 2}=T_{1}\biggl[1+\biggl(\frac{\gamma-1}{2}\biggr)M_{1}{}^{2}\biggr]=288.0\,\mathrm{K}

p_{t\ 2}=p_{1}\Big[1+\Big({\frac{\gamma-1}{2}}\Big)M_{1}{}^{2}\Big]^{\frac{\gamma}{\gamma-1}}=0.722\,\mathrm{bars}

N_{C2}=\frac{N}{\sqrt{\theta_{2}}}=34,000\,\mathrm{rpm}

N_{C4}={\frac{N}{\sqrt{\theta_{4}}}}=17,000\,\,\mathrm{rpm}

Because the turbine is choked,

\dot m_{C4}=2.0\,\mathrm{kg/s}

Let us now consider the following equality:

\dot{m}_{C4}=\dot{m}_{C2}\sqrt{\frac{T_{t\ 4}}{T_{t\ 2}}}\biggl(\frac{p_{t\ 2}}{p_{t\ 4}}\biggr)

In this equality, both \dot m_{C2} and p_{t\ 4} are unknown. Knowing that N_{C2} will always be 34,000 rpm (θ_{2} = 1.0), we can proceed to solve this equality through a trial-and-error procedure, as follows:

First attempt – Set \dot m_{C2} to 13.0 kg/s: Now

{\dot{m}}_{C4}=2.0\,\mathrm{kg/s}={\dot{m}}_{C2}{\sqrt{\frac{T_{t\ 4}}{T_{t\ 2}}}}\biggl({\frac{p_{t\ 2}}{p_{t\ 4}}}\biggr)

This will provide us with a magnitude of 9.39 bars for p_{t\ 4}.
On the other hand, we can compute the same variable (say p_{t\ 4}{}^{\prime}), where

p_{t\ 4}{}^{\prime}={\frac{p_{t\ 4}}{p_{t\ 3}}}{\frac{p_{t\ 3}}{p_{t\ 2}}}p_{t\ 2}=8.06\;\mathrm{bars}

Comparing the two magnitudes of p_{t\ 4}, this attempt has produced an error of 15.2%, which is unacceptable.

Second attempt – Set “\dot m_{C2}” to 12.0 kg/s: Following the same approach as on the first attempt, we get

p_{t\ 4}=8.66\;\mathrm{bars}

p_{t\ 4}{}^{\prime}=8.72\mathrm{~bars}

This yields an error of 0.7%, which is acceptable. Let us now proceed with an average p_{t\ 4} magnitude of 8.69 bars.
At this point, the compressor operating conditions can be read-off the compressor map as follows:

\dot m_{C2}=12.0\,\mathrm{kg/s}

\pi_{C}={\frac{p_{t\ 3}}{p_{t\ 2}}}=13.0

N_{C2}=34,000\,\mathrm{rpm}

\eta_{C}=80%

The compressor total-to-total temperature ratio (\tau_{C}) can be computed as

\tau_{C}=1+\frac{1}{\eta_{C}}\biggl[(\pi_{C})^{\frac{\gamma_{C}-1}{\gamma_{C}}}-1\biggr]=2.351

With the fuel-to-air ratio f\approx0, we can compute the turbine total-to-total temperature ratio (\tau_{T}) as

\tau_{T}=1-\left[\frac{1}{\eta_{m}}\frac{c_{p_{C}}}{c_{p_{T}}}\frac{T_{t\ 2}}{T_{t\ 4}}(\tau_C -1)\right]=0.701

Just as easily, we can calculate the turbine total-to-total pressure ratio (π_{T}) as

\pi_{T}=\left[1-\frac{1}{\eta_{T}}(1-\tau_{T})\right]^{\frac{\gamma_{T}}{\gamma_{T}-1}}=0.169

It follows that

p_{t\ 5}=\pi_{T}p_{t\ 4}=1.472\;\mathrm{bars}

Part Ib:

T_{t\ 5}=\tau_{T}T_{t\ 4}=807.6\,\mathrm{K}

Part Ic:

{\dot{m}}_{2}={\dot{m}}_{C2}{\frac{\delta_{2}}{\sqrt{\theta_{2}}}}=8.66\;{\mathrm{kg/s}}

Part Id:

p_{t\ 5}=p_{t\ 6}=p_{6}\biggl[1+\biggl(\frac{\gamma_{T}-1}{2}\biggr)M_{6}{}^{2}\biggr]^{\frac{\gamma_{T}}{\gamma_{T}-1}}

which, upon substitution, yields

M_{6}=1.307

Noting that T_{t\ 5} is equal to T_{t\ 6}, we can calculate the exit static temperature as follows:

T_{6}=\frac{T_{t\ 6}}{\left[1+(\frac{\gamma_{T}-1}{2})M_{6}{}^{2}\right]}=630.0\mathrm{\,K}

We can now proceed to calculate the engine specific thrust as follows:

V_{6}=M_{6}a_{6}=M_{6}\sqrt{\gamma_{T}R T_{6}}=641.0\,\mathrm{m/s}

V_{1}=M_{1}a_{1}=M_{1}\sqrt{\gamma_{C}R T_{1}}=270.3\,\mathrm{m/s}

F/\dot m=V_{6}-V_{1}=370.7\ \mathrm{N/kg}

Part Ie: We now calculate the accurate value of the fuel-to-air ratio (by applying the energy equation to the combustor) as follows:

f=\frac{(c_{p_{T}}\,T_{t\ 4}-c_{p_{C}}\,T_{t\ 3})}{(H-c_{p_{T}}\,T_{t\ 4})}=0.0152

Part If: The specific fuel consumption is now easy to calculate:

S F C={\frac{f}{F/\dot m_{2}}}=4.11\times10^{-5}{\mathrm{~kg/N}}

Part II: Following a constant -π_{C} (i.e., horizontal) line on the compressor map until we get to the surge line, the compressor instability prevails at the point where

(N_{C})_{m i n.}=33{,}700\;\mathrm{rpm}

(\dot m_{C})_{m i n.}=10.7\mathrm{~kg/s}

which, in terms of physical variables, translate into

(N)_{m i n.}=\sqrt{\theta_{2}}(N_{C})_{m i n.}=33,700\;\mathrm{rpm}

(\dot m_{2})_{m i n.}=\frac{\delta_{2}}{\sqrt{\theta_{2}}}(\dot m_{C})_{m i n.}=7.73\mathrm{~kg/s}