In a jet propulsion unit air is drawn into the rotary compressor at 15°C and 1.01 bar and delivered at 4.04 bar. The isentropic efficiency of compression is 82% and the compression is uncooled. After delivery the air is heated at constant pressure until the temperature reaches 750°C.

Question

In a jet propulsion unit air is drawn into the rotary compressor at 15°C and 1.01 bar and delivered at 4.04 bar. The isentropic efficiency of compression is 82% and the compression is uncooled. After delivery the air is heated at constant pressure until the temperature reaches 750°C. The air then passes through a turbine unit which drives the compressor only and has an isentropic efficiency of 78% before passing through the nozzle and expanding to atmospheric pressure of 1.01 bar with an efficiency of 88%. Neglecting any mass increase due to the weight of the fuel and assuming that R and γ are unchanged by combustion, determine :
(i) The power required to drive the compressor.
(ii) The air-fuel ratio if the fuel has a calorific value of 42000 kJ/kg.
(iii) The pressure of the gases leaving the turbine.
(iv) The thrust per kg of air per second.
Neglect any effect of the velocity of approach.
Assume for air : R = 0.287 kJ/kg K, γ = 1.4.

Accepted Answer

Given : [latex] T_{1} [/latex] = 15 + 273 = 288 K,
[latex]p_{1} = 1.01 bar, p_{2} = 4.04 bar, T_{3} [/latex]= 750 + 273 = 1023 K,
[latex]η _{compressor}[/latex] = 82%, [latex]η_{turbine} [/latex] = 78%, [latex]η _{nozzle} [/latex] = 88%,
[latex]R_{air} = 0.287 kJ/kg K, γ_{air} [/latex] = 1.4.

Refer Fig. 40.

[latex]\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ - 1}{γ }} = (\frac{4.04}{1.01}) ^{\frac{1.4 - 1}{1.4 }} [/latex] = 1.486

∴ [latex] T_{2}[/latex] = 2.88 × 1.486 = 428 K

[latex] η_{compressor} = \frac{T_{2} - T_{1}}{T_{2}′ - T_{1}} i.e., 0.82 = \frac{428 - 288}{T_{2}′ - 288} [/latex]

∴ [latex] T_{2}′ = ( \frac{428 - 288}{0.82}) [/latex] + 288 = 458.7 K

[latex] c_{p} = R × (\frac{γ }{γ - 1}) = 0.287 × \frac{1.4 }{(1.4 - 1)}[/latex] = 1.004 kJ/kg K

(i) Power required to drive the compressor :
Power required to drive the compressor (per kg of air/sec.)

= [latex] c_{p} (T_{2}′ – T_{1}) [/latex] = 1.004(458.7 – 288) = 171.38 kW.

(ii) Air-fuel ratio :
[latex]m_{f} × C = (m_{a} + m_{f}) × c_{p} × (T_{3} – T_{2}′)[/latex]
where, [latex]m_{a} [/latex] = Mass of air per kg of fuel, and
= Air-fuel ratio.

∴ [latex]m_{a} = \frac{m_{f} × C }{c_{p} (T_{3} – T_{2}′)} - m_{f}[/latex]

∴ [latex] \frac{m_{a}}{m_{f} } = \frac{C}{c_{p} (T_{3} – T_{2}′)} - 1 [/latex]

= [latex] \frac{42000}{1.004 (1023 - 458.7)} [/latex] – 1 = 73.1

i.e., Air-fuel ratio = 73.1 : 1.

(iii) Pressure of the gases leaving the turbine, [latex]p_{4}[/latex] :
Neglecting effect of fuel on mass flow,
Actual turbine work = actual compressor work
i.e., [latex] c_{p} (T_{2}′ – T_{1}) = c_{p} (T_{3} – T_{4}′)[/latex]
or [latex] T_{2}′ – T_{1} = T_{3} – T_{4}′[/latex]
∴ 458.7 – 288 = 1023 – [latex]T_{4}′[/latex]
∴ [latex]T_{4}′ [/latex]= 852.3 K

Also, [latex]η_{ turbine } = \frac{ T_{3} – T_{4}′}{ T_{3} – T_{4}} [/latex]

0.78 = [latex] \frac{ 1023 – 852.3}{ 1023 – T_{4}} [/latex]

∴ [latex]T_{4} = 1023 – (\frac{ 1023 – 852.3}{0.78})[/latex] = 804 K

∴ [latex]\frac{ T_{4} }{ T_{3} } = (\frac{p_{4}}{p_{3}})^{\frac{γ - 1}{γ }} [/latex]

or [latex] \frac{p_{4}}{p_{3}} = (\frac{ T_{4} }{ T_{3} })^{\frac{γ }{γ - 1}} = (\frac{ 804 }{ 1023 })^{\frac{1.4 }{1.4 - 1}} [/latex] = 0.43

or [latex]p_{4} [/latex] = 4.04 × 0.43 = 1.74 bar.

(iv) Thrust per kg of air per second :

[latex]\frac{ T_{4}′ }{ T_{5} } = (\frac{p_{4}}{p_{5}})^{\frac{γ - 1}{γ }} = (\frac{1.74}{1.01})^{\frac{1.4 - 1}{1.4 }} [/latex] = 1.168

∴ [latex]T_{5} = \frac{ T_{4}′ }{1.168} = \frac{ 852.3 }{1.168} [/latex] = 729.7 K

[latex]η_{nozzle} = \frac{ T_{4}′ - T_{5}′}{T_{4}′ - T_{5}} [/latex]

0.88 = [latex]\frac{ 852 - T_{5}′}{852.3 - 729.7} [/latex]

∴ [latex]T_{5} ′[/latex] = 852.3 – 0.88 (852.3 – 729.7) = 744.4 K

If [latex] C_{j}[/latex] is the jet velocity, then

[latex] \frac{C_{j}²}{2} = c_{p} (T_{4}′ – T_{5}′)[/latex]

∴ [latex] C_{j} = \sqrt{2 × c_{p} (T_{4}′ – T_{5}′)}[/latex].

= [latex]\sqrt{2 × 1.004 (852.3 - 744.4 ) × 1000}[/latex] = 465.5 m/s

∴ Thrust per kg per second = 1 × 465.5 = 465.5 N.

Question 11.21: In a jet propulsion unit air is drawn into the rotary compre...

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