TITLE.PM5

618 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-2.pm5 or 0.72 (T 3 – 689.1) = 1500 or T 3 = 1500 072. 689.1 = 2772.4 K. (Ans ...

GAS POWER CYCLES 619 dharm \M-therm\Th13-2.pm5 For the constant volume process 2-3 : p T p T p pT T 2 2 3 3 3 23 2 12 3^1842 614 ...

620 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-2.pm5 Fig. 13.9. Atkinson cycle. Now, T T p p 5 3 5 3 1 = F HG I KJ −γ γ or T ...

GAS POWER CYCLES 621 dharm \M-therm\Th13-2.pm5 p (bar) V(m )^3 Adiabatic compression Heat addition at constant volume 3 7 2 1 1 ...

622 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-2.pm5 (i)Compression ratio. (ii)Thermal efficiency of the cycle. (iii)Work do ...

GAS POWER CYCLES 623 dharm \M-therm\Th13-2.pm5 (iii)Work done : Again, for adiabatic compression 1-2, T T 2 1 = V V 1 2 F^1 HG I ...

624 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-2.pm5 p (bar) 3 (^112) 1 4 1 V (m )^3 Adiabatics V = 0.45 m 1 3 21.48 Fig. 13 ...

GAS POWER CYCLES 625 dharm \M-therm\Th13-2.pm5 For the adiabatic (or isentropic) process 3-4 p 3 V 3 γ = p 4 V 4 γ p 4 = p 3 × V ...

626 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-2.pm5 (b) Determine the air-standard efficiency of the cycle when the cycle d ...

GAS POWER CYCLES 627 dharm \M-therm\Th13-2.pm5 (b) Change in efficiency : For air γ = 1.4 ∴ r = T T 3 1 F 12 14 1 HG I KJ /(.−) ...

628 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-2.pm5 ∴ T 2 = T 4 = TT 13. Proved. (b) Power developed, P : T T m 1 3 310 145 ...

GAS POWER CYCLES 629 dharm \M-therm\Th13-2.pm5 Putting the value of ρ in ηDiesel, we get ηDiesel = 1 –^11 1 1 1 1 1 ()r r r r r ...

630 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-3.pm5 s T p = Const. 3 4 2 1 v = Const. v p 4 1 2 3 Adiabatic Adiabatic Fig. ...

GAS POWER CYCLES 631 dharm \M-therm\Th13-3.pm5 = r ρ F γ HG I KJ − 1 Q v v v v v v v v 4 r 3 1 3 1 2 2 3 ==×= F HG I ρKJ ∴ T 4 = ...

632 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-3.pm5 Mean effective pressure pm is given by : pm = pvr r v r r 11 11 1 11 1 ...

GAS POWER CYCLES 633 dharm \M-therm\Th13-3.pm5 = 0.06 (V 1 – V 2 ) = 0.06 (15 V 2 – V 2 ) = 0.84 V 2 or V 3 = 1.84 V 2 ∴ρ = V V ...

634 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-3.pm5 = 1 – 0.248 × 1.563 = 0.612 or 61.2% When the fuel is cut-off at 8%, we ...

GAS POWER CYCLES 635 dharm \M-therm\Th13-3.pm5 Solution. Refer Fig. 13.17. p (bar) 2 3 Vs Vs 8 100 Adiabatics 4 1 1 (27°C) Vc V( ...

636 ENGINEERING THERMODYNAMICS dharm \M-therm\Th13-3.pm5 For the adiabatic (or isentropic) process 1-2 p 1 V 1 γ = p 2 V 2 γ or ...

GAS POWER CYCLES 637 dharm \M-therm\Th13-3.pm5 (iii)Mean effective pressure, pm : Mean effective pressure of Diesel cycle is giv ...