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PROBLEM ON ENGINE CYCLES Thermal Engineering by RK Rajput - ref page no. 1032 ⁄7)An engine 200mm bore and 300mm stroke works on Otto cycle. The clearance volume is 0.0016m³. The initial temperature and pressure are 1bar and 60°C.If the maximum pressure is limited to 24bar, find : i) The air standard efficiency of the cycle ii) The mean effective pressure of the cycle . Assume ideal conditions. Given : bore diameter d = 0.2m stroke length l = 0.3m clearance volume Vc =0.0016m³ T₁=60+273=333K

P₁= bar Cv =0.718 kJ/kg K

Cp=1.005kJ/kg K

P₃=24 bar To find: (i) η

(ii) mep

Solution: Vs = πd²l/4 = π*0.2²*0.3/4 =0.0094248m³ V = Vc+Vs = 0.0016+0.0094248 = 0.011025 m³ rk = V₂/V₁ = V/Vc = 0.011025/0.0016 = 6.89 ηt = 1- (1/rkγ-1) = 1-(1/6.891.4-1) = 0.5379 = 53.79% T₂/T₁ = (v₁/v₂)γ-1 T₂ = T₁* rkγ-1 = 333 * 6.891.4-1 = 720.66 K P₂/P₁ = (v₁/v₂)γ P₂ = P₁* rkγ = 1* 6.891.4= 14.91 bar P₁v₁ = RT₁

=> v₁=RT₁/P₁ =0.287*333/ 00 = 0.9557 m³/kg

v₂=v₁/6.89 =0.9557/6.89 = 0. 387 m³/kg

Swept volume vs = v₁-v₂ = 0.9557-0.1387 = 0.817 m³/kg P₂/T₂ = P₃/T₃ =>

T₃=P₃T₂/P₂ = 24*720.66/ 4.9 =

Heat supplied Qs = Cv (T₃-T₂)=0.7 8(

60 K

60-720.66) = 315.45 kJ/kg

Work done Wnet = Qs * ηt = 315.45*0.5379=169.68 kJ/kg Mean effective pressure = Work done Wnet / Swept volume vs = 169.68/0.817 = 2.076 bar

Thermal Engineering by RK Rajput - ref page no. 1032 2⁄ )In an ideal Diesel cycle, the temperatures at the beginning of compression, 9 at the end of compression and at the end of heat addition are 97°C,789°C and 1839°C. Find the efficiency of the cycle. Given : T₁=97+273 =370K

T₂=789+273 = 062K

T₃= 839+273 =2

To find : η Solution : T₂/T₁ = (v₁/v₂)γ-1 = rkγ-1 rkγ-1 = 1062/370 => rk = 13.95 T₃/T₂ = v₃/v₂ =rc rc = 2112/1062 = 1.989 T₄ = T₃(rc/rk)γ-1 = 2112*(1.989/13.95)1.4-1 = 968.99 K η = - *(T₄-T₁)/γ(T₃-T₂)+ = 1- {(968.99-370)/1.4(2112-1062)} = 0.5925 =59.25%

2K

Thermal Engineering by RK Rajput - ref page no. 1033 3⁄

) 1 kg of air is taken through a Diesel cycle. Initially the air is at 15°C and

1 ata. The compression ratio is 15 and the heat added is 1850 kJ.Calculate: i) The ideal cycle efficiency

ii) Mean effective pressure

Given: m =1kg

T₁= 5+273=288 K

To find: (i) η

P₁= .0 3 bar

Qs = 1850kJ/kg rk =15

(ii) mep

Solution : T₂/T₁ = (v₁/v₂)γ-1 T₂ = T₁* rkγ-1 = 288 * 151.4-1 = 850.8K P₂/P₁ = (v₁/v₂)γ P₂ = P₁* rkγ = 1.013* 151.4= 44.89 bar Heat supplied Qs = Cp (T₃-T₂) 850= .005(T₃-850.8) T₃/T₂ =v₃/v₂ = rc

=> T₃ =269 .6K => rc =2691.6/850.8 =3.164

η = - [ {rcγ- +/*γ rkγ-1 (rc-1)} ] =1-[{3.1641.4-1}/{1.4*151.4-1*(3.164-1)}] = 0.5513 = 55.13 % Work done Wnet = Qs * ηt = 1850*0.5513=1019.91 kJ/kg P₁v₁ = RT₁

=> v₁=RT₁/P₁ =0.287*850.5/ 0 .3 = 0.8 6 m³/kg

v₂=v₁/ 5 =0.8 6/ 5 = 0.0544 m³/kg Swept volume vs = v₁-v₂ = 0.8 6-0.0544 = 0.7616 m³/kg

Mean effective pressure = Work done Wnet / Swept volume vs = 1019.91/0.7616= 13.39bar Thermal Engineering by RK Rajput - ref page no. 1033 4⁄ ) A compression ignition engine has a stroke 270mm, and a cylinder 4 diameter of 165mm.The clearance volume is 0.000434 m3 and the fuel ignition takes place at constant pressure for 4.5 percent of the stroke. Find the efficiency of the engine assuming it works on the diesel cycle. Given : bore diameter d = 0.27m stroke length l = 0.165m clearance volume Vc =0.000434m³ Heat addition takes place at 4.5% of the stroke. To find : η Solution : Vs = πd²l/4 = π*0.27²*0. 65/4 =0.005773m³ V = Vc+Vs = 0.000434+0.005773 = 0.006207 m³ rk = V₂/V₁ = V/Vc = 0.006207/0.000434 = 14.3 v₃-v₂ = 0.045 (v₂-v₁) = 0.045 ( 4.3v₂-v₂) v₃-v₂ = 0.045* 3.3v₂ =>

v₃/v₂ = rc = 1.5985

η = - [ {rcγ- +/*γ rkγ-1 (rc-1)} ] =1-[{1.59851.4-1}/{1.4*14.31.4-1*(1.5985-1)}] = 0.617 = 61.7 %

Thermal Engineering by RK Rajput - ref page no. 1033

5⁄ ) An air standard dual cycle has a compression ratio of 16 and 8 compression begins at 1 bar and 50°C .The maximum pressure is 70 bar .The heat transferred to air at constant pressure is equal to that at constant volume. Determine : i) The cycle efficiency Take Cv= 0.718 kJ/kg K

ii) Mean effective pressure of the cycle Cp=1.005 kJ/kg K

Given: rk = 16

P₁= bar

T₁=50+273 =323 K

P₃=P₄=70 bar

Qsv =Qsp To find: (i) η

(ii) mep

Solution: T₂/T₁ = (v₁/v₂)γ-1 T₂ = T₁* rkγ-1 = 323 * 161.4-1 = 979K P₂/P₁ = (v₁/v₂)γ P₂ = P₁* rkγ = 1* 161.4= 48.5 bar T₃/T₂ = P₃/P₂ T₃=P₃T₂/P₂ = 70*979/48.5 = 4 3 K Cv (T₃-T₂)= Cp (T₄-T₃) 0.718(1413-979) = .005(T₄-1413) T₄ = 723 K v₄/v₃ =T₄/T₃ = 723/ 4 3 = .2 9 v₅/v₄ = (v₁/v₂)*(v₃/v₄)= 6/ .2 9 = 3. 26 T₅=T₄ *(v₄/v₅)γ-1 = 1723 * (1/13.126)1.4-1 = 615.34 K

Qs = Qsv + Qsp = 2* Qsv Qs = 2* Cv (T₃-T₂) = 2*0.7 8( 4 3-979) = 623.24 kJ/kg Qr = Cv (T₅-T₁) = 0.7 8(6 5.34-323) =209.9 kJ/kg η = -(Qs/Qr) = 1-(209.9/623.24) = 0.6632 =66.32 % Wnet = Qs-Qr = 623.24 - 209.9= 413.32 kJ/kg P₁v₁ = RT₁

=> v₁=RT₁/P₁ =0.287*323/ 00 = 0.927 m³/kg

v₂=v₁/ 6 =0.927/ 6 = 0.0579 m³/kg Swept volume vs = v₁-v₂ = 0.927-0.0579 = 0.8691 m³/kg Mean effective pressure = Work done Wnet / Swept volume vs = 413.32/0.8691 = 4.76 bar