ICE Lecture 2A

8/14/2019 ICE Lecture 2A

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ILUSTRATIVE EXAMPLES-1

1.1. Otto Cycle Engine V s; V c

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.

Swept volume of one cylinder = 2d4

= 7.95x8.255x4 2

= 425.5 cm 3 (or cc)

Cubic capacity of the engine = total swept volume of all cylinders= 425.5 x 6 = 2553 cm 3

Compression ratio,volumeclearence

meswept voluvolumeclearancevolumeclearance

volumetotalr

+==

Or volumeclearance

meswept volu17.8 +=

8.618.71r volumeclearance

meswept volu ===

Thus, clearance volume of each cylinder

3cm62.586.8

425.56.8

meswept voluolumeclearencev ===

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 ( m) = ihp bhp

bhp = m 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 80%or 0.80125100

25100100

fhp bhp bhp

m ==+=+=

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(a) At half load

Mechanical efficiency 66.7%or 0.6677550

255050

fhp bhp bhp

m ==+=+=

(b) At quarter load

Mechanical efficiency %50or 0.55025

255225

fhp bhp bhp m ==+=+=

1.3. Petrol Engine: ihp; fhp; th ; t; 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,ihp

bhpm =

608.0

48 bhpihp

m

===

(b) fhp = ihp bhp = 60 48 = 12

(c) 1 bhp-hr = 75 kgf m x427

3600= 632.5 kcal/hr

Brake thermal efficiency

21.1%or 0.21110000x0.3

632.5CVn xconsumptiofuel

ihp th ===

(d) Indicated thermal efficiency t is given by

mtth x =

%26.4or 0.26480

1.21

m

tht ===

(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: m a; 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 m 3 of air are

required per hour if the density of air is 1.2 kg/m 3? 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 is10500 kcal/kg.

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( )C.V.kg/hr xinnconsumptiofuel

/42760x4500x bhp =

Fuel consumption in kg/hr = 30.0510500x0.20x427

60x4500x100C.V.xx427

60x4500x bhp ==

(a) Air consumption/hr = kg4300.071x30.05 =

(b). Air volume/hr =3

air

m3581.2430tionAirconsump ==

Fuel volume/hr 3m6.264x1.2

30.05 ==

(c.) Mixture volume = 358 + 6.26 = 364.3 m 3

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 efficiencyihp

bhpm =

66.675.05 bhpihp

m

===

Indicated thermal efficiencyvaluecaloricn xconsumptiofuel

ihp=

( )10000x(kg/hr)nconsumptiofuel

427/606.66x4500x30.0 =

(a). Fuel consumption in kg/hr kg/hr 1.40510000x427x0.30

60x4500x6.66 ==

(b). Fuel consumption in liters/hr liter/hr 1.6150.87

1.405 ==

(c). Indicated specific fuel consumptionhp.hr

kg0.205

6.661.405

ihpkg/hr nconsumptiofuel ===

(d). Brake specific fuel consumptionhp.hr

kg0.204

51.405

bhp(kg/hr)innconsumptiofuel ===

1.7. Ihp; m; air/hr; t; th

A two-stroke CI engine delivers 5000 bhp while using 1000 hp to overcome frictionlosses. It consumes 1800 kg of fuel per hour at an air-fuel ratio of 20 to 1. The heating value

15

WV =


4/8

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 83%or 0.83

6000

5000

ihp

bhp m ===

(c). Air consumption/hr = A/F x fuel consumption/hr

= 20/1 x 1800 = 36000 kg

(d). Indicated thermal efficiency

21.1%or 0.21110000x1800632.5x6000

CVn xconsumptioFuelihp

t ===

(e). Brake thermal efficiency

%17.5or 0.1750.83x0.211x mttb===

2.1. Carnot engine: hp; T heat source

A Carnot engine which rejects heat to a cooling pond at 27 o 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.

1

21

1

21

TTT

QQQ ==

1

1

T300T

30.0= 300T7.0 1 = C6.155K 6.4287.0

300T 001 ===

1

1

Q200Q

30.0= 200Q7.0 1 = 2867.0

200Q1 ==

Work done/min = Q 1 Q 2 = 286 200 = 86 kcal

hp8.1610.54

86hp ==

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 m 3. Calculate the air standard efficiency.

( ) cc680030x174

d4

meswept volu 22 ===

Clearance volume = 0.002025 x 10 6 cc = 2025 cc

Total cylinder volume = 6800 + 2025 = 8825 cc

35.420258825)r (nratioCompressio ==

16


5/8

( )44.4%0.5661

4.35

11

r 1

1efficiencystandardAir 11.41 ====

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 1000

C. The maximum pressure in the cycle is 35 kgf/cm2

. For 1kg 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/kg 0 K; = 1.4

Solution:

Point 1

p1 = 1.0 kgf/cm 2 = 1x 10 4 kgf/m 2 T1 = 100 0C= 373 0K

p1V1 = mRT

34

11 m1.0921x10

373x29.27x1 p

TR mV ===

Point 2

2211 V pV p =

21.4

2

112 kgf/cm12.36x1V

V p p ==

=

312 m0.1826

1.0926

VV ===

2

22

1

11

TV p

TV p =

C492or K 7651.092x1

373x0.182x12.3T

V pV p

T 00111

222 ===

Point 3

V3 = V 2 = 0.182 m 3, p 3 = 35.0 kgf/cm 2

17

1.0 kg/cm 2

35.0 kg/cm 2

P2

Q1

Q2


6/8

3

2

2

3

3

T p

T p =

C1802or K 2075765x12.335

T p p

T 0022

33 ===

Point 4 4433 V pV p =

24.1

4

334 cm/kgf 84.23.12

3561

x35VV

p p ==

=

=

V4 = V 1 = 1.092 m 3

1

1

4

4

T p

T p =

C734or K 10071

2.84x373

p p

TT 00

1

414 ===

( ) ( )K kcal/0.178

11.442729.27

1JR

c 0v ===

Heat supplied = c v(T3 T 2) = 0.178(1802 734) = 0.178 x 1068 = 19 kcal

Heat rejected = c v(T 4 T 1) = 0.178(734-100) = 0.178 x 634 = 11.6 kcal

1.6411.6

19

rejectedheat

suppliedheat ==

2.4. Otto cycle: T max ; 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 c v = 0.17 and =

1.4.

(a) T 1 = 15 + 273 = 288 K p1 = 1 kgf/cm 2

p2 = p 1 (V 1/V 2) = 1 x (8) 1.4

= 18.45 kgf/cm 2

T2 = T 1 x (V 1/V 2)-1

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

= 663 K

Heat supplied = c v (T 3 T 2)

250 = 0.17(T 3 663)

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7/8

T3 = (250/0.17) + 663 = 1470 + 663 = 2133 0K or 1860 0C

The maximum temperature in the cycle is 1860 0C

(b). Air standard efficiency ( )%5.56565.0435.01

3.21

18

11

r 1

1 4.01 ======

(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). T 4 = T 3 x (V 3/V 2)-1 = 2133x(1/8) 1.4-1 = 2133/23 = 927 0K

Heat rejected = c v(T 4 T 1) = 0.17(927-288) = 0.17x639 =108.6 kcal/kg

2.5. Otto Cycle: air standard ; ; mep

Discuss the working of an engine on Otto cycle.

In an Otto cycle air at 150

C and 1.05 kgf/cm2

is compressed adiabatically until the pressure rises to 35 kgf/cm 2. Calculate the air standard efficiency, the compression ratio, and

the mean effective pressure for the cycle. Take c v = 0.1715, R = 29.27 2211 V pV p =

1

1

2

2

1

p p

VV

= , 714,0

1 =

605.113

r

714.0

= =

1r 1

1 =

%2.51488.0161

1 4.0 ===

2

22

1

11

TV p

TV p =

K 594288x61x

1.0513Tx

V pV pT 01

11

222 ===

Now2

22

3

33

TV p

TV p =

K 1600594x1335

pxT p

T 0

2

133 ===

Heat Supplied ( ) ( ) kg/kcal5.17159416001715.0TTc 23v ===

kcal/kg88171.5x0.512QxdoneWork 1 ===To find swept volume

19


8/8

111 mRTV p =

==1

1 VmRT

V

20

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ICE Lecture 2A