Ice Lecture 2a

ILUSTRATIVE EXAMPLES-1 1.1. Otto Cycle Engine Vs; Vc Distinguish between the swept and clearance volumes of a reciprocating engine. Define compression ratio. The engine of the Ford Zephyr car has six cylinders of 82.55 mm bore and 79.5 mm stroke. The compression ratio is 7.8. Determine the cubic capacity of the engine and the clearance volume of each cylinder. π  2 π 2 Swept volume of one cylinder =  d  =   x 8.255 x 7.95 4 4     = 425.5 cm3 (or cc) Cubic capacity of the engine = total swept volume of all cylinders = 425.5 x 6 = 2553 cm3 Compression ratio, r = Or 7.8 = 1 + ∴

total volume clearance volume + swept volume = clearance volume clearence volume

swept volume clearance volume

swept volume = r − 1 = 7 .8 − 1 = 6 .8 clearance volume

Thus, clearance volume of each cylinder clearencevolume =

swept volume 425.5 = = 62.58 cm3 6.8 6.8

1.2. Bhp; fhp. A certain engine produces 10 ihp. Its mechanical efficiency is 80 per cent. Find the bhp delivered and friction horsepower (fhp). Mechanical efficiency (ηm) =

bhp ihp

bhp = ηm x ihp = 0.80 x 10 = 8 fhp = ihp – bhp = 10 – 8 = 2 1.3. Mechanical efficiency at various loads A certain engine at full load delivers 100 bhp. It requires 25 fhp to rotate it without fuel at the same speed. Find its mechanical efficiency. Assuming that the mechanical losses remain constant what will be the mechanical efficiency at (a) half load, (b) quarter load. Mechanical efficiency ηm =

bhp 100 100 = = = 0.80 or 80% bhp + fhp 100 + 25 125 13

(a) At half load Mechanical efficiency η m =

bhp 50 50 = = = 0.667 or 66.7% bhp + fhp 50 + 25 75

(b) At quarter load Mechanical efficiency η m =

bhp 25 25 = = = 0.5 or 50 % bhp + fhp 25 + 25 50

1.3. Petrol Engine: ihp; fhp; ηth; ηt; fuel and air consumption A four-stroke petrol engine delivers 48 bhp with a mechanical efficiency of 80 per cent. The fuel consumption of the engine is 0.3 kg per bhp-hr and the air-fuel ratio is 14:1. The heating value of the fuel is 10000 kcal/kg. Find (a) ihp, (b) fhp, (c) brake thermal efficiency (d) indicated thermal efficiency, (e) fuel consumption per hour, and (f) air consumption/hr. (a) Mechanical efficiency, η m = ∴ ihp =

bhp ihp

bhp 48 = = 60 η m 0 .8

(b) fhp = ihp – bhp = 60 – 48 = 12 (c) 1 bhp-hr = 75 kgf m x

3600 = 632.5 kcal/hr 427

Brake thermal efficiency ηth =

ihp 632.5 = = 0.211 or 21.1% fuel consumption x CV 0.3 x 10000

(d) Indicated thermal efficiency ηt is given by η th = η t x η m ∴ ηt =

ηth 21.1 = = 0.264 or 26.4 % ηm 80

(e) Fuel consumption per hour = bfsc x bhp = 0.3 x 48 = 14.4 kg (f) Air consumption per hour = 14 x 14.4 = 202 kg. 15. SI Engine: ma; air and mixture volume A SI engine has a fuel-air ratio of 0.07:1. How many kilograms of air per hour are required for an output of 100 bhp at an overall efficiency of 20%? How many m3 of air are required per hour if the density of air is 1.2 kg/m3? If the fuel vapour has a density four times than of air, how many m3 per hour of the mixture is required? The caloric value of the fuel is 10500 kcal/kg. 14

η=

( bhp x 4500 x 60) /427 fuel consumption in kg/hr x C.V.

∴ Fuel consumption in kg/hr =

bhp x 4500 x 60 100 x 4500 x 60 = = 30.05 427 x η x C.V. 427 x 0.20 x 10500

(a) ∴ Air consumption/hr = 30.05 x (b).

Air volume/hr =

1 = 430 kg 0.07

Airconsumption 430 = = 358 m3 ρair 1.2

Fuel volume/hr =

30.05 x 4 = 6.26 m 3 1.2

(c.) ∴ Mixture volume = 358 + 6.26 = 364.3 m3 1.6. Diesel engine; fuel consumption A diesel engine develops 5 bhp. Its indicated thermal efficiency is 30% and mechanical efficiency 75%. Estimate the fuel consumption of engine in (a) kg/hr, (b) liters/hr, (c) indicated specific fuel consumption, and (d) brake specific fuel consumption. Assume the specific gravity of fuel oil as 0.87 and caloric value (CV) of the fuel 10000 kcal/kg. Mechanical efficiency η m = ihp = Indicated thermal efficiency = 0.30 =

bhp ihp bhp 5 = = 6.66 η m 0.75

ihp fuel consumption x caloric value

( 6.66x4500x60) / 427 fuel consumption (kg/hr) x 10000

(a). Fuel consumption in kg/hr =

6.66 x 4500 x 60 = 1.405 kg/hr 0.30 x 427 x 10000

(b). Fuel consumption in liters/hr =

1.405 = 1.615 liter/hr 0.87

(c). Indicated specific fuel consumption = (d). Brake specific fuel consumption =

V=

W

γ

fuel consumption kg/hr 1.405 kg = = 0.205 ihp 6.66 hp.hr

fuel consumption in (kg/hr) 1.405 kg = = 0.204 bhp 5 hp.hr

1.7. Ihp; ηm; air/hr; ηt; ηth A two-stroke CI engine delivers 5000 bhp while using 1000 hp to overcome friction losses. It consumes 1800 kg of fuel per hour at an air-fuel ratio of 20 to 1. The heating value 15

of fuel is 10000 kcal/kg. Find (a) ihp, (b) mechanical efficiency, (c) air consumption per hr, (d) indicated thermal efficiency, and (e) brake thermal efficiency. (a). ihp = bhp + fhp = 5000 + 1000 = 6000 (b). Mechanical efficiency η m =

bhp 5000 = = 0.83 or 83% ihp 6000

(c). Air consumption/hr = A/F x fuel consumption/hr = 20/1 x 1800 = 36000 kg (d). Indicated thermal efficiency ηt =

ihp 6000 x 632.5 = = 0.211 or 21.1% Fuel consumption x CV 1800 x 10000

(e). Brake thermal efficiency

η tb = ηt x ηm = 0.211 x 0.83 = 0.175 or 17.5 % 2.1. Carnot engine: hp; Theat source A Carnot engine which rejects heat to a cooling pond at 27o C has an efficiency of 30 per cent. If the cooling pond receives 200 kcal per minute, determine the horse power of the engine. Also find temperature of the heat source.

η=

Q1 − Q 2 T1 − T2 = Q1 T1

0.30 =

T1 − 300 T1

− 0.7T1 = −300

T1 =

300 = 428.6 0 K = 155.6 0 C 0.7

0.30 =

Q1 − 200 Q1

− 0.7Q1 = −200

Q1 =

200 = 286 0.7

∴ Work done/min = Q1 – Q2 = 286 – 200 = 86 kcal hp =

86 = 8.16 hp 10.54

2.2. Otto cycle: Air standard efficiency The bore and stroke of an engine working on the Otto cycle are 17 cm and 30 cm respectively. The clearance volume is 0.002025 m3. Calculate the air standard efficiency. swept volume =

π 2 π 2 d  = (17 ) x 30 = 6800 cc 4 4

Clearance volume = 0.002025 x 106 cc = 2025 cc Total cylinder volume = 6800 + 2025 = 8825 cc ∴ Compressionratio(r ) =

8825 = 4.35 2025 16

∴ Air standard efficiency = 1 −

1 r

γ −1

= 1−

1

( 4.35) 1.4−1

= 1 − 0.566 = 44.4%

2.3. Otto Cycle: p, v, T at salient points; ratio of heat supplied to heat rejected. In an ideal Otto cycle the compression ratio is 6. The initial pressure and temperature of the air are 1 kgf/cm2 and 1000C. The maximum pressure in the cycle is 35 kgf/cm2. For 1 kg of air flow, calculate the values of the pressure, volume, and temperature at the four salient point of the cycle. What is the ratio of heat supplied to the heat rejected? For air, R = 29.27 kgf/kg0 K; γ = 1.4 Solution: 3

35.0 kg/cm2 P2

Q1 2

P

4

1.0 kg/cm2

1 v

Point 1 p1 = 1.0 kgf/cm2 = 1x 104 kgf/m2

T1 = 100 0C= 373 0K

p1V1 = mRT ∴ V1 =

m R T 1 x 29.27 x 373 = = 1.092 m 3 p1 1x10 4

Point 2

p1 V1γ = p 2 V2γ V p 2 = p1  1  V2 V2 =

γ

  = 1 x 61.4 = 12.3 kgf/cm 2 

V1 1.092 = = 0.182 m 3 6 6

p1 V1 p 2 V2 = T1 T2 ∴ T2 =

p 2 V2 12.3 x 0.182 x 373 T1 = = 765 0 K or 492 0 C p1 V1 1 x 1.092

Point 3 V3 = V2 = 0.182 m3,

p3 = 35.0 kgf/cm2 17

Q2

p3 p2 = T3 T2 T3 =

p3 35 T2 = x 765 = 2075 0 K or 1802 0C p2 12.3

Point 4

p 3 V3γ = p 4 V4γ V ∴ p 4 = p 3  3  V4

γ

 1  = 35x   6 

1.4

=

35 = 2.84kgf / cm 2 12.3

V4 = V1 = 1.092 m3 p 4 p1 = T4 T1 ∴ T4 = T1 cv =

p4 2.84 = 373 x = 1007 0 K or 734 0 C p1 1

R 29.27 = = 0.178 kcal/ 0 K (J γ − 1) 427 (1.4 − 1)

Heat supplied = cv(T3 – T2) = 0.178(1802 – 734) = 0.178 x 1068 = 19 kcal Heat rejected = cv(T4 – T1) = 0.178(734-100) = 0.178 x 634 = 11.6 kcal ∴

heat supplied 19 = = 1.64 heat rejected 11.6

2.4. Otto cycle: Tmax; η air standard; heat rejected. In an ideal Otto cycle the air at the beginning of isentropic compression is at 1 kgf/cm 2 and 15 0C. The ratio of compression is 8. If the heat added during the constant volume process is 250 kcal/kg, determine (a) the maximum temperature in the cycle, (b) the air standard efficiency, (c) the work done per kg of air, and (d) the heat rejected. Take cv = 0.17 and γ = 1.4. P

3

(a) T1 = 15 + 273 = 288 K p1 = 1 kgf/cm2 p2 = p1 (V1/V2)γ = 1 x (8)1.4

2

= 18.45 kgf/cm2 T2 = T1 x (V1/V2)γ-1

4

= 188 x (8)1.4-1 = 288 x 23 = 663 K

1 v Heat supplied = cv (T3 – T2) 250 = 0.17(T3 – 663) 18

T3 = (250/0.17) + 663 = 1470 + 663 = 2133 0K or 1860 0C The maximum temperature in the cycle is 18600C (b). Air standard efficiency = 1 −

1 r

γ −1

= 1−

1

( 8)

0.4

=1−

1 = 1 − 0.435 = 0.565 = 56.5% 2.3

(c). Work done = Heat supplied x Efficiency = 250 x 56.5 = 141.2 kcal or 141.2 x 4.12 kgf-m = 582 kgf-m (d). T4 = T3 x (V3/V2)γ-1 = 2133x(1/8)1.4-1 = 2133/23 = 927 0K Heat rejected = cv(T4 – T1) = 0.17(927-288) = 0.17x639 =108.6 kcal/kg 2.5. Otto Cycle: ηair standard; γ; mep Discuss the working of an engine on Otto cycle. In an Otto cycle air at 150C and 1.05 kgf/cm2 is compressed adiabatically until the pressure rises to 35 kgf/cm2. Calculate the air standard efficiency, the compression ratio, and the mean effective pressure for the cycle. Take cv = 0.1715, R = 29.27 p1 V1γ = p 2 V2γ

∴

V1  p 2 = V2  p1

  

1 γ

 13  r=   1.05 

η = 1− η = 1−

,

1 = 0,714 γ

0.714

=6

1 r

γ −1

1 = 1 − 0.488 = 51.2% 6 0.4

p1 V1 p 2 V2 = T1 T2 ∴ T2 = Now

p 2 V2 13 1 x T1 = x x 288 = 594 0 K p1 V1 1.05 6

p 3 V3 p 2 V2 = T3 T2

∴ T3 =

p 3 xT1 35 = x 594 = 1600 0 K p2 13

Heat Supplied = c v ( T3 − T2 ) = 0.1715(1600 − 594 ) = 171.5kcal / kg ∴ Work done = η x Q1 = 0.512 x 171.5 = 88 kcal/kg To find swept volume 19

p1 V1 = mRT1 ∴ V1 =

mRT = V1

20