1. Assignment of Power Plant-I Designed by Sir Engr. Masood Khan
SOLVED PROBLEMS OF CHAPTER # 13
TITLE: RECPROCTING INTERNAL COMBUSTION ENGINES
2. Reciprocating Internal Combustion Engines Designed by Sir Engr. Masood Khan
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.
Full load, break load = W = B = 433 N
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:
3. Reciprocating Internal Combustion Engines Designed by Sir Engr. Masood Khan
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 obtained at
5000 rev/min, and at the peak speed of 6750 rev/min
the load was 547 N. The specific gravity of the fuel
being 0.735, and Qnet,v=44200 kJ/kg.
Calculate the maximum bmep,the maximum bp,the
minimum specific fuel consumption, and the
maximum brake thermal efficiency at maximum
torque and compare this latter answer with the air
standard efficiency.
GIVEN DATA:
Four cylinder racing engine
n = no of cylinder = 4
Capacity of cylinder = V = 2.495 liters = 0.002495 m3
Dia of the bore = b = 94 mm= 0.94 m
Compression ratio= 12/1 = rv
Tested by dynamometer gives = k = 0.461 m
Wmax = 622N
4. Reciprocating Internal Combustion Engines Designed by Sir Engr. Masood Khan
N = 5000 rev/min = 83.3 rev/sec
Nmax = 6750/ 60 = 112. rev /sec
W = 547 N
Mini. Fuel consumption = 17.2 ml/s at 5000 rev/min
Specific gravity of the fuel = 0.735
Qnet,p = 44200 KJ/Kg
REQUIRED:
Max. bmep =? , (Pb) max. =?
Min. specific fuel consumption, (spc.)min. =?
(ηBT)max.= ? (At max. torque)
Compare this with the Air Standard Efficiency
SOLUTION:
As Tmax. = Wmax. * R = 622 * 0.461 = 286.74 Nm
At Tmax bp = 2πTmax. * N
(Sfc)min = mf / bp=0.012642 /150.13*103
=8.42*10-5
kg/kg = 8.42*10-5
*3600 kg/kg = 0.303
(BT)max = bp / mf*Qnet
=150.13*103
/0.012642*44200*103
=. 2686 = 26.86 %
For Air Standard Efficiency, we have a relation,
= 1-1/(rq)γ-1
= 1-1/120.4
= 0.63 = 63 %
PROBLEM # 13.6:
13.1 A four cylinder, four stroke diesel engine has a bore
of 212mm and a stroke of 292mm. at full load at 720rpm
the bmep is 5.93 bar ande the specific fule consumption is
0.226kg/kWh. The air-fuel ratio as determined by exhausty
gas analysis is 25/1. Calculate the brake thermal efficiency
and the volumetric efficiency of the engine. Atmospheric
conditions are 1.01bar and 15C and Qnet,v for the fuel may
be taken as 44200kj/kg.
GIVEN DATA:
Four cylinders, four stroke diesel engine
Dia of bore, d = 212 mm = 0.212 m
Length of stroke, L = 292 mm = 0.292 m
N = 720 rev/min = 720/60 rev/sec n = 4
Bmep = Pb = 5.93*105
N/m2
Sp = fuel consumption = (sfc) = 0.226 kg/kwh
= 0.226 (kg/h)/kw = 0.226 (kg/h)/(kj/h)
= 0.266/3600 kg/kj = 6.277*10-5
kg/kj = 6.277*10-8
kg/j
Air fuel ratio = 25 / 1
Atmospheric conditions are
P = 1.01 bar = 1.01*105
N/m2
, T = 15 C = 288
k
Qnet,v = 44200 kj/kg = 44200*103
j/kg
REQUIRED:
Brake Thermal Efficiency, BT =?
Volumetric Efficiency,v =?
SOLUTION:
As BT = bp / mf*Qnet,v-----------------------------(1)
bp
=(Pb.A.L.N.n)/2=(5.93*105
*⊼*(0.212)2
*(0.292)*12)/(
4*2) bp = 1.467*105
watt = 1.467 *102
kwatt
mf =? As
(Sfc) = mf / bp => mf = 9.208*10-3
kg/s
Now equation (1) becomes
BT = (1.467*105
)/(9.208*10-3
*44200*103
) = 0.36 =
36 %
v = V / Vs -------------------------------------------------
(2)
Vs = (ALnN)/(2*60) m3
/s =(d2
LnN)/(4*2*60) m3
/s
5. Reciprocating Internal Combustion Engines Designed by Sir Engr. Masood Khan
= (*(0.212) 2
*0.292*4*12)/(4*5*60) = 0.2473
As air mass flow rate = air fuel ratio*9.208*10-3
= (25/1)*9.208*10-3
= 0.2302 kg/s
Now as PV = mRT => V = mRT/P
V = 0.2362*287*288/1.01*105
= 0.1884 m3
/s
Now equation becomes
v = V / Vs = 0.1884/0.2473 = 0.7616 = 76.16 %
PROBLEM # 13.10:
Four cylinders petrol-engine has an output of 52KW at
2000rpm. A Morse Test is carried out and the brake torque
readings are 177, 170, 168 and 174Nm respectively. For
normal running at this speed the specific fuel consumption is
0.364kg/kWh. The Qnet,v of the fuel is 44200kJ/kg. Calculate
the mechanical and brake thermal efficiencies of the engine.
GIVEN DATA:
Four cylinder petrol engine, n = 4
bp = 52*103
watt
N = 2000 rev/min = 33.33 rev/sec
B1 = 177 N-m, B2 = 170 N-m,
B3 = 168 N-m, B4 = 174 N-m
Specific Fuel Consumption = (sfc) = 0.364 kg/kwh
=1.011*10-4
(kg/h)/s*1000 watt
=1.011*10-7
(kg/s)/(j/s) = 1.011*10-7
kg/j
Qnet,v = 44200 kj/kg = 44200*103
j/kg
REQUIRED:
M =? BT =?
SOLUTION:
As M = bp/ iP = (Pb.A.L.N.n)/(Pi*A*L*N*n/2) =bp/Pi
M = bT / iT-----------------------------------------------(1)
bp = 2 T N
T = bp / 2n = (52*103
)/(2*33.33) = 248.3 N-m
B = 248.3 N-m
As I1 = B-B1 = 248.3-177 = 71.3 N-m
I2 = B-B2 = 248.3-170 = 78.3 N-m
I3 = B-B3 = 248.3-168 = 80.3 N-m
I4 = B-B4 = 248.3-174 = 74.3 N-m
Now from equation (1)
M = B / I = 248.3/304.2 = 0.816 = 81.6 %
As BT = bp / (mf*Qnet,v)----------------------------(2)
To find mf, As
(Sfc) = mf / bp
mf = (Sfc)*bp = 1.011*10-7
*32*103
= 5.257*10-3
kg/s
Now equation becomes
BT = 52*103
/(5.257*10-3
*44200*103
)
BT = 0.2238 = 22.38 %
