Question 12.6: Figure 12.18 shows the compressor and turbine maps for a tur......

OUTER LOOP
Step 1: Let us arbitrarily pick a point (not shown on the compressor map) where

\frac{p_{t\ 3}}{p_{t\ 2}}=2.2

(\dot m)_{C2}=7.0\mathrm{~kg/s}

{ N}_{C2}=58,000\mathrm{\ rpm}

\eta_{C}=75.0%

Step 2: Let us pick a T_{t\ 4}/T_{t\ 2} magnitude of 3.5. Now the turbine corrected speed can be computed:

N_{C4}=N_{C2}\sqrt{\frac{T_{t\ 2}}{T_{t\ 4}}}=31,000\;\mathrm{rpm}

INNER LOOP
1) Let (p_{t4}/p_{t5}) be 2.4:

(\dot m)_{C4}=4.3\,\mathrm{kg/s}\,(\mathrm{from~the~turbine~map})

({\dot{m}})_{C2}={\frac{(\dot m)_{C4}}{(1+f)}}{\sqrt{\frac{T_{t\ 2}}{T_{t\ 4}}}}{\Biggl({\frac{p_{t\ 4}}{p_{t\ 3}}}\Biggr)}{\Biggl({\frac{p_{t\ 3}}{p_{t\ 2}}}\Biggr)}=4.64\;\mathrm{kg/s}\ <\;{\mathrm{7.0~k g/s}}\;{\mathrm{(the~target~value)}}

2) Let (p_{t4}/p_{t5}) be 2.6:

(\dot m)_{C4}=6.5\;{\mathrm{kg/s}}

(\dot m)_{C2}=7.02\,\mathrm{kg/s}\approx7.0 (the target value)

BACK TO THE OUTER LOOP
With the preceding results, let us now recalculate the previously assumed temperature ratio (T_{t\ 4}/T_{t\ 2}) by substituting in equation (12.41), which yields

{\frac{T_{t\ 4}}{T_{t\ 2}}}=\left[{\frac{1}{(1+f)\eta_{m}\eta_{T}\eta_{C}}}\right]\left({\frac{c_{p_{C}}}{c_{p_{T}}}}\right){\frac{\left[\left({\frac{p_{t\ 3}}{p_{t\ 2}}}\right)^{\frac{\gamma_{C}-1}{{\gamma_{C}}}}-1\right]}{\left[1-\left({\frac{p_{t\ 5}}{p_{t\ 4}}}\right)^{\frac{\gamma_{T}-1}{{\gamma_{T}}}}\right]}}   (12.41)

{\frac{T_{t\ 4}}{T_{t\ 2}}}=1.74\neq3.5 (the targeted temperature-ratio magnitude)

Now, we have no choice but to complete the entire (double-loop) procedure with a new magnitude of the temperature ratio.
Let us now pick a point on the compressor map where p_{t\ 3}/p_{t\ 2}=2.05,(\dot m)_{C2}= {\mathrm{6.5\ k g/s,\ a n d}}\ N_{C2}=50,000\mathrm{~rpm}

OUTER LOOP
Let us select the temperature ratio (T_{t\ 4}/T_{t\ 2}) to be \underline{1.74} this time. In doing so, note that:

1) The problem statement does not confine us to a specific value of T_{t\ 4}/T_{t\ 2}.
2) The vast difference in the assumed and then computed values of T_{t\ 4}/T_{t\ 2} may make it harder to match the turbine to the compressor under such a magnitude.
3) The newly selected temperature ratio is excessively low. A comment at the end of the solution will shed some light on this particular choice.

N_{C4}=37,900\,\mathrm{rpm}

INNER LOOP
1) Let (p_{t4}/p_{t5}) be 2.8:

(\dot m)_{C4}=6.0\,\mathrm{kg/s}

(\dot m)_{C2}=8.56\,\mathrm{kg/s}\neq6.5\,\mathrm{kg/s} (our current target value)

2) Let (p_{t4}/p_{t5}) be 2.7:

(\dot m)_{C4}=5.0\,\mathrm{kg/s}

(\dot m)_{C2}=7.13\,\mathrm{kg/s}\neq6.5\,\mathrm{kg/s}

3) Let (p_{t4}/p_{t5}) be 2.63:

(\dot m)_{C4}=4.55\;\mathrm{kg/s}

(\dot m)_{C2}=6.5\;\mathrm{kg/s} (identical to the target value)

BACK TO THE OUTER LOOP
Using equation (12.41), we update the assumed temperature-ratio magnitude:

{\frac{T_{t\ 4}}{T_{t\ 2}}}=1.91\not=1.74 (the target temperature-ratio magnitude)

NOTE
As is clear by now, we have to reexecute the double-loop procedure, taking the temperature ratio (T_{t\ 4}/T_{t\ 2}) to be the newly computed magnitude or, generally speaking, any other magnitude we wish, knowing that our choice may not be exactly consistent with our wish to shorten this iterative procedure.
It seems appropriate, however, to skip all of these intermediate steps indicating, only the final results. The “finality” here simply means that the magnitudes of temperature ratio (T_{t\ 4}/T_{t\ 2}) at the beginning of the outer loop and at the end of the inner loop are sufficiently close to one another.

FINAL RESULTS
With a temperature ratio of 1.78 (at both the outer-loop beginning and inner-loop end), we get the two points “C” and “T” on the compressor and turbine maps, respectively. The operational points “C” and “T” shown on the corresponding maps in Fig. 12.18 both correspond to a (T_{t_{4}}/T_{t_{2}} ) of 1.78, and are defined as follows:

Point C:

{\frac{p_{t\ 3}}{p_{t\ 2}}}=2.05

(\dot m)_{C2}=6.2\mathrm{~kg/s}

N_{C2}=45,500\;\mathrm{rpm}

Point T:

{\frac{p_{t\ 4}}{p_{t\ 5}}}=2.55

(\dot m)_{C4}=4.40\mathrm{~kg/s}

{ N}_{C4}=34,100\;\mathrm{rpm}

COMMENTS
This example is meant to provide a closer look at the tedious procedure to produce the engine operating lines on (particularly) the compressor map (Fig. 12.10). It is clear that our chosen T_{t\ 4}/T_{t\ 2} magnitude is excessively small. To make some practical “sense” out of the results, let us compute the combustor temperature ratio (T_{t\ 4}/T_{t\ 2}) to at least prove it to be above unity:

{\frac{T_{t\ 4}}{T_{t\ 2}}}=1+{\frac{1}{\eta_{C}}}(\pi_{C}{}^{\frac{\gamma-1}{\gamma}}-1)=1.27

The combustor total-to-total temperature ratio can now be computed:

{\frac{T_{t\ 4}}{T_{t\ 3}}}={\frac{T_{t\ 4}}{T_{t\ 2}}}{\frac{T_{t\ 2}}{T_{t\ 3}}}={\frac{1.78}{1.27}}=1.40\gt 1.0

Asuitable description therefore is that the chosen magnitude of T_{t\ 4}/T_{t\ 2} represents a far off-design engine-operation mode.