# Internal Combustion Engine and Specific Fuel Consumption

Topics: Internal combustion engine, Compression ratio, Diesel engine Pages: 7 (1620 words) Published: September 15, 2013
SOLVED PROBLEMS OF CHAPTER # 13

TITLE:RECPROCTING INTERNAL COMBUSTION ENGINES
PROBLEM # 13.1:

A quantity governed four-stroke, single-cylinder gas engine has a bore of 146mm and a stroke of 280mm.At 475 rev/min and full load the net load on the friction brake is 433 N, and the torque arm is 0.45 m. The indicator diagram gives a net area of 578 mm² and a length of 70 mm with a spring rating of 0.815 bar per mm. calculate the ip,bp, and mechanical efficiency.

GIVEN DATA:
Four stroke single cylinder gas engine
n = no. of cylinders = 1
Dia. of cylinder bore = 146 mm = 0.146 m
Length of stroke = L = 280 mm = 0.28 m
N = 475 revolution/min = 475/ 60 rev/sec. = 7.5416 rev/sec.
Torque arm = R = 0.45 m
Net area of indicator diagram = 578 mm2 = 578 * 10-6 m2
Length of indicating diagram = 70 mm = 70 * 10 -3 m
Spring rating constt. = 0.815 bar/mm
= 0.815 / 10-3 bar /m = 815 bar/m

REQUIRED:
Indicated power = ip =?Brake power = bp =?
ηmech = ?

DIAGRAM:

SOLUTION:
Indicated power is given by
ip = (Pi * ALnN)/2 _________________ (1)
This formulae is for 4 stroke engine, But
Pi = indicated pressure =?
Pi = (net area of diagram)* constt. / Length of the diagram = 578 * 10-6 * 815/ (70 * 10-3) = 6.73 bar
A = Area of the bore = (π/4)* d2 = 0.01674 m2
Now equation (1) becomes
ip = 6.73 * 105 * 0.01674 * 0.28 * 7.916 * 1/ 2
= 12485.4 watts = 12.4854 Kwatts

Now T = W * R = 194.85 Nm

bp = 2 π T N = 9.691 K watts
Now as
ηmech = bp / ip = 9.691 / 12.485 = 0.776 = 77.6 %

PROBLEM # 13.2:
A two-cylinder, four-stroke gas engine has a bore of 380 mm and a stroke of 585 mm.At 240 rev/min the torque developed is 11.86 kNm. Calculate (i) the bp;(ii) the mean piston speed in m/s;(iii) the bmep.

GIVEN DATA:
Two cylinders, four-stroke gas engine.
No. of cylinder = n = 2,
Dia. Of bore = d = 380mm = 0.38 m
Radius = r = 0.38/ 2 = 0.19 m
Length of the stroke = L = 585 mm = 0.85m
Engine speed = N = 240 rev/min = 4 rev/sec
Torque developed, T =11.86 KN.m = 11.86 * 103 N-m

REQUIRED:
1. Braking power = bp=?
2. Mean piston speed in m/s= v =?
3. brake mean effective pressure = bmep = Pb

DIAGRAM:

SOLUTION:
1) As braking power = bp = 2π TN
= 2π* 11.86* 103 * 4 = 298KW
2) As bp= F*d/ t = F* v
Where F = T/moment arm = T/r
bp= (T/r) * v
⇨ v = bp * r/T = 298 * 0.19/ 11.86 = 4.77m/s
3) Pb = 4πT/(ALn) = 11.23 bar

PROBLEM # 13.3:
The engine of problem 13.2 is supplied with a mixture of gas and air with a proportion of 1 to 7 by volume. The estimated volumetric efficiency is 85% and the Qnet,p of the gas is 38.6 MJ/m³, calculate the brake thermal efficiency of the engine.

GIVEN DATA:
(Data from problem 13.2)
Mixture of gas and air in propulsion by volume = 1 gas
and 7 air.
Volumetric efficiency = ηv = 85% = 0.85
Net calorific value of the gaseous fuel = Qnet,p = 38.6 MJ/m3

REQUIRED:
Brake thermal efficiency, ηBT =?

SOLUTION:
For gas engine,
ηBT = bp/ (Vf * Qnet,p) __________________________(1)
As bp = 2πTN = 298.07 KW
As swept volume of the engine = Vs = (ALNm/ 120) m3/sec
Vs = πd2* N * n * L / 480 = 0.26538 m3/sec
ηv= V/ Vs where V = volume flow rate of a mixture. V = ηv * Vs = 0.2257 m/s
Volume of the mixture = vol. of the air + vol. of the gas = 7 volume of air + 1 volume of gas
Dividing by time to get flow rate
V = 7 Vf + 1 Vf = 8 Vf
==> Vf = V/ 8 = 0.2257/ 8 = 0.28196 m3/s
Now equation (1) becomes
ηBT = 0.27387 = 27.4%

PROBLEM # 13.4:
A four-cylinder racing engine of capacity 2.495 liters has a bore of 94 mm and a compression ratio of 12/1, when tested against a dynamometer with a torque arm of 0.461 m a maximum load of 622 N was...