The following data pertain to a jet engine flying at an altitude of 9000 metres with a speed of 215 m/s. Thrust power developed 750 kW Inlet pressure and temperature 0.32 bar, – 42°C Temperature of gases leaving the combustion chamber 690°C Pressure ratio 5.2 Calorific value of fuel 42500

Question

The following data pertain to a jet engine flying at an altitude of 9000 metres with a speed of 215 m/s.
Thrust power developed 750 kW
Inlet pressure and temperature 0.32 bar, – 42°C
Temperature of gases leaving the combustion chamber 690°C
Pressure ratio 5.2
Calorific value of fuel 42500 kJ/kg
Velocity in ducts (constant) 195 m/s
Internal efficiency of turbine 86%
Efficiency of compressor 86%
Efficiency of jet tube 90%
For air : [latex] c_{p} [/latex] = 1.005, γ = 1.4, R = 0.287
For combustion gases, [latex] c_{p} [/latex] = 1.087
For gases during expansion, γ = 1.33.
Calculate the following :
(i) Overall thermal efficiency of the unit ;
(ii) Rate of air consumption ;
(iii) Power developed by the turbine ;
(iv) The outlet area of jet tube ;
(v) Specific fuel consumption is kg per kg of thrust.

Accepted Answer

Refer Fig. 42.
Given : T.P. = 750 kW ; [latex]p _{1} = 0.32 bar, T_{1} = – 42 + 273 = 231 K ; T_{3} = 690 + 273 = 963 K ; r_{pc} = 5.2 ; C = 42500 kJ/kg ; C_{a} = 215 m/s, C_{4} ^{′ } = 19.5 m/s, η_{c} = 0.86 ; η_{t} = 0.86 ; η_{jt} [/latex] = 0.9.
Refer Fig. 42.
Let [latex]m_{f} [/latex]= kg of fuel required per kg of air
Then, heat supplied per kg of air
= 42500 [latex] m_{f } = (1 + m_{f}) × 1.087 (T_{3} – T_{2} ^{′ }) [/latex] ...(i)

Now, [latex]\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ - 1}{γ }} = (5.2)^{\frac{1.4 - 1}{1.4}} = (5.2)^{0.2857}[/latex] = 1.60

or [latex]T_{2} [/latex] = 231 × 1.60 = 369.6 K

Also, [latex]η_{c} = \frac{ T_{2} - T_{1} }{T_{2} ^{′ } - T_{1} } or T_{2} ^{′ } = T_{1} + \frac{ T_{2} - T_{1} }{η_{c} } = 231 + \frac{369.6 - 231 }{0.86} [/latex] = 392.2 K

Substituting the value of [latex]T_{2} ^{′ } [/latex] in eqn. (i), we get
42500 [latex]m_{f} = (1 + m_{f}) × 1.087 (963 – 392.2) = 620.46 (1 + m_{f})[/latex]
or 42500 [latex] m_{f } = 620.46 + 620.46 m_{f} [/latex]

or [latex] m_{f } = \frac{620.46 }{(42500 - 620.46)}[/latex] = 0.0148 = fuel-air ratio

∴ Air-fuel ratio [latex] = \frac{ 1}{0.0148}[/latex] = 67.56 : 1
The discharge velocity [latex] C_{j} = C_{5} ^{′ } [/latex] cannot be determined from the thrust equation because the rate of air-flow is not known. It may be determined from the expression of jet efficiency.

Jet efficiency, [latex] η_{jet} = \frac{Final kinetic energy in the jet}{Isentropic heat drop in the jet pipe + Carry-over from the turbine } [/latex]

or [latex] η_{jet} = \frac{C_{j}² / 2}{c_{pg} (T_{4} ^{′ } – T_{5}) + C_{4} ^{′ } ² / 2} (where C_{4} ^{′ } = 195 m/s ) [/latex] ...(ii)

Since the turbine’s work is to drive the compressor only, therefore,

[latex] c_{pa} (T_{2} ^{′ } – T_{1}) = c_{pg} (1 + \frac{m_{f }}{m_{a}} ) (T_{3} - T_{4} ^{′ } ) [/latex]

or 1.005 (392.2 – 231) = 1.087(1 + 0.0148) (963 – [latex]T_{4} ^{′ } ) [/latex]

or [latex]T_{4} ^{′ } = 963 - \frac{1.005 (392.2 - 231)}{1.087 (1 + 0.0148)} [/latex] = 816.13 K

Let [latex] r_{pt} = expansion pressure ratio in turbine i.e., r_{pt} = \frac{p_{3}}{p_{4}}[/latex]

[latex] r_{pj} = expansion pressure ratio in jet tube i.e., r_{pj} = \frac{p_{4}}{p_{5}}[/latex]

∴ [latex] r_{pt} × r_{pj} = \frac{p_{3}}{p_{4}} × \frac{p_{4}}{p_{5}} \simeq[/latex] 5.2

Now, [latex] η_{t} = \frac{T_{3} - T_{4} ^{′ } }{T_{3} - T_{4} } or T_{4} = T_{3} - \frac{T_{3} - T_{4} ^{′ } }{η_{t}} [/latex]

= 963 – [latex] \frac{963 - 816.13}{0.86} [/latex] = 792.2 K

Also, [latex]\frac{ T_{3} }{ T_{4} } = (\frac{p_{3}}{p_{4}})^{\frac{γ - 1}{γ }} = (\frac{p_{3}}{p_{4}})^{\frac{1.33 - 1}{1.33 }} or \frac{963}{792.2} = ( r_{pt})^{0.248}[/latex]

or [latex] r_{pt} = ( \frac{963}{792.2})^{\frac{1}{0.248}} [/latex] = 2.197

∴ [latex] r_{pj} = \frac{p_{4}}{p_{5}} = \frac{5.2}{2.197} [/latex] = 2.366

Thus, [latex]\frac{ T_{4}′ }{ T_{5} } = (r_{pj})^{\frac{γ - 1}{γ }} = ( 2.366)^{\frac{1.33 - 1}{1.33 }} [/latex] = 1.238

or [latex]T_{5} = \frac{ T_{4} ^{′ } }{ 1.238} = \frac{ 816.13 }{ 1.238} [/latex] = 659.23 K

Substituting the values in eqn. (ii), we get

0.9 = [latex] \frac{C_{j}² / 2}{1.087 × 1000 (816.13 - 659.23) + 195² / 2} = \frac{C_{j}² / 2}{189562.8} [/latex]

∴ [latex]C_{j} = \sqrt{0.9 × 189562.8 × 2} [/latex] = 584.13 m/s

(i) Overall efficiency, [latex]η_{0}[/latex] :

[latex]η_{0} = \frac{[(1 + \frac{m_{f }}{m_{a}} ) C_{j} - C_{a} ] C_{a}}{(\frac{m_{f }}{m_{a}} ) × C} = \frac{[(1 + 0.0148) × 584.13 - 215] 215}{1000 × 0.0148 × 42500} [/latex]

= 0.1291 or 12.91%.

(ii) Rate of air consumption, [latex]\dot{m}_{a}[/latex] :

Thrust power = Thrust × Velocity of the unit

750 = [latex] [\frac{\left\{ (1 + \frac{m_{f }}{m_{a}} ) C_{j} - C_{a} \right\} \dot{m}_{a}}{1000}] C_{a}[/latex]

or 750 = [latex] \frac{\left\{(1 + 0.0148) × 584.13 - 215 \right\} × \dot{m}_{a}}{1000} × 215 = 81.22 \dot{m}_{a}[/latex]

or [latex]\dot{m}_{a} = \frac{750}{81.22}[/latex] = 9.234 kg/s.

(iii) Power developed by the turbine, [latex]P_{t}[/latex] :

[latex]P_{t} = \dot{m}_{a} (1 + \frac{m_{f }}{m_{a}} ) c_{pg} (T_{3} – T_{4} ^{′ }) [/latex]

= 9.234(1 + 0.0148) × 1.087(963 – 816.13) = 1496 kW.

(iv) The outlet area of jet tube, [latex]A_{jt} [/latex]:

Now, [latex] \frac{C_{j}² - C_{4} ^{′ }²}{2} = c_{pg} (T_{4} ^{′ } – T_{5} ^{′ }) [/latex]

or [latex] T_{5} ^{′ } = T_{4} ^{′ } - \frac{C_{j}² - C_{4} ^{′ }²}{2 × c_{pg}}[/latex]

= 816.13 – [latex] \frac{(584.13² - 195²)}{2 × 1.087 × 1000} [/latex] = 676.67 K

Assume the exit pressure of the gases be equal to atmospheric pressure i.e., 0.32 bar.

Density of exhaust gases, ρ = [latex] \frac{p_{5} ^{′ } }{R T _{5} ^{′ } } = \frac{0.32 × 10^{5}}{0.29 × 1000 × 676.67} [/latex] = 0.163 m³/kg
(Assuming R = 0.29 for the gases)

Also, discharge of jet area = [latex]A_{jt} × C_{j} × ρ = \dot{m}_{a} (1 + \frac{m_{f }}{m_{a}} )[/latex]

or [latex]A_{jt} [/latex] × 584.13 × 0.163 = 9.234 (1 + 0.0148)

or [latex]A_{jt} [/latex] = 0.0984 m².

(v) Specific fuel consumption in kg per kg of thrust :

Specific fuel consumption = [latex] \frac{0.0148 × 9.234 × 3600}{1000 × (750/215)} [/latex]

= 0.141 kg/thrust-hour.

Question 11.23: The following data pertain to a jet engine flying at an alti...

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