A Brief Theory on IC Eng
Engine Testing and Performance Important Performance Parameters of I.C.Engines:The important performance parameters of I.C. engines are as follows: (a) Power and Mechanical Efficiency. (b) Mean Effective Pressure and Torque. (c) Specific Output. (d) Volumetric Efficiency. (e) Fuel-air Ratio. (f) Specific Fuel Consumption. (g) Thermal Efficiency and Heat Balance. (h) Exhaust Smoke and Other Emissions. (i) Specific Weight.
Engine Testing and Performance Indicated Power (IP) = Brake Power (BP) - Friction Power (FP)
INDICATE POWER DEVELOPED INSIDE THE ENGINE: IP
POWER AVAILABLE AT THE END OF CRANK SHAFT: BP
FRICTION POWER: FP
Indicated Power Power obtained at the cylinder. Obtained from the indicator diagram
where
Pi is the indicated mean effective pressure, in N/m
L is the stroke length, in m A is the area of cross section of the piston, m 2, N’ is the engine speed in rev/min, [=N’=N/2 for 4_S engine and N’=N for 2-S engine] n is the number of cylinders and
For 4-stroke engine-one cycle will be completed in two revolutions N’=N/2 For 2-stroke engine-one cycle will be completed in one revolutions N’=N
Mean Effective Pressure The mean effective pressure is a quantity related to the operation of an reciprocating engine and is a valuable measure of an engine's capacity to do work that is independent of engine displacement Indicated Mean Effective Pressure or imep (pi) - it may be defined as the average pressure over a cycle in the combustion chamber of the engine. Pi = (Net work of cycle)/Swept Volume in N/m2
= P mep = Pm = Pi
Let,
W = work per cycle in joule P = power output in watt pmep = mean effective pressure in pascal Vd = displacement volume in cubic metre nc = number of revolutions per cycle (for a 4-stroke engine nc = 2) N = number of revolutions per second T = torque in newton-metre
also
W = pmep * Vd
Since P = TN2π
so
Mean Effective Pressure is also obtained by engine indicator diagram as
Pm= Pmep = (s*a)/l = N/m2 Where: a = actual Indicator diagram cm2 l = base width of the indicator diagram, cm s = spring value or spring constant used for in indicator diagram,( N/m 2 )/cm
Brake mean effective pressure (BMEP) - Mean effective pressure calculated from brake torque
Brake Power MEASUREMENT OF B.P
1. Mechanical Dynamometer I. Prony Brake II. Rope Brake
BP = 2..N.T
/ 60
Watts
T = Torque = (W-S) De/2 W = Load on the Brake Drum in N S = Spring Balance Reading in N De = Effective Brake Drum Diameter = Drum Diameter(Db) + (2*Thickness of Rope) BP can also be written as
BP = (W-S) (Db +d).N Π (Db+d) is circumference of the brake drum
/ 60
in Watts
2. Hydraulic Dynamometer B.P=WN/K Watts W =Weight measured on the dynamometer, N K=Dynamometer constant (60*1000/2*pi*R) and N=RPM of the engine.
3. Electric Dynamometer I.Eddy current Type Dynamometer II.Swinging Field Dynamometer
Friction Power Friction power includes the frictional losses and the pumping losses. During suction and exhaust strokes the piston must move against a gaseous pressure and power required to do this is called the “pumping losses”. The friction loss is made up of the energy loss due to friction between the piston and cylinder walls, piston rings and cylinder walls, and between the crank shaft and camshaft and their bearings, as well as by the loss incurred by driving the essential accessories, such as water pump, ignition unit etc.
Following methods are used in the laboratory to measure friction
This method is also known as fuel rate extrapolation method. In this method a graph of fuel consumption (vertical axis) versus brake power (horizontal axis) is drawn and it is extrapolated on the negative axis of brake power (see Fig). The intercept of the negative axis is taken as the friction power of the engine at that speed. As shown in the figure, in most of the power range the relation between the fuel consumption and brake power is linear when speed of the engine is held constant and this permits extrapolation. Hence the extrapolated negative intercept of the horizontal axis will be the work representing the combined losses due to friction, pumping and as a whole is termed as the frictional loss of the engine. Further when the engine does not develop power, i.e. brake power = 0, it consumes a certain amount of fuel. This energy in the fuel would have been spent in overcoming the friction. This method of measuring friction power will hold good only for a particular speed and is applicable mainly for compression ignition engines.
The main draw back of this method is the long distance to be extrapolated from data between 5 and 40 % load towards the zero line of the fuel input. The directional margin of error is rather wide because the graph is not exactly linear.
From the Measurement of Indicated Power and Brake Power:- This is an ideal method by which friction
power is obtained by computing the difference between the indicated power and brake power. The indicated power is obtained from an indicator diagram and brake power is obtained by a brake dynamometer. This method requires elaborate equipment to obtain accurate indicator diagrams at high speeds.
This method can be used only for multi – cylinder IC engines
The test consists of making, in turn, each cylinder of the engine inoperative and noting the reduction in brake power developed.
In a petrol engine (gasoline engine), each cylinder is rendered inoperative by “shorting” the spark plug of the cylinder to be made inoperative. In a Diesel engine, a particular cylinder is made inoperative by cutting off the supply of fuel
It is assumed that pumping and friction are the same when the cylinder is inoperative as well as during firing.
In this test, the engine is first run at the required speed and the brake power is measured. Next, one cylinder is cut off by short circuiting the spark plug if it is a petrol engine or by cutting of the fuel supply if it is a diesel engine. Since one of the cylinders is cut of from producing power, the speed of the engine will change. The engine speed is brought to its original value by reducing the load on the engine. This will ensure that the frictional power is the same.
Total indicated power when all the cylinders are working n = ip1 + ip2 + ip3 + …………...+ ipn = ∑ ip j j=1
n We can write ∑ IPj = (BP)t + (FP)t ………………………………………..(1) j= 1 Where IPj is the indicated power produced by j th cylinder, n is the number of cylinders, (BP)t is the total brake power when all the cylinders are producing power and (FP)t is the total frictional power for the entire engine.
If the first cylinder is cut – off, then it will not produce any power, but it will have frictional losses. Then n we can write ∑ IPj = (BP)1 - (FP)t………………………………………..(2) j=2
where (BP)1 = total brake power when cylinder 1 is cut - off and (FP)t = Total frictional power.
Subtracting Eq. (2) from Eq. (1) we have the indicated power of the cut off cylinder. Thus
(IP)1 = (BP)t – (BP)1 ………………………………………..(3). Similarly we can find the indicated power of all the cylinders, viz., ip2, ip3, …..ipk. Then the total indicated power is calculated as
k (IP)total = ∑ IPj ……………………………………….(4) j=1 The frictional power of the engine is therefore given by
(FP)t = (IP)total – (BP)t
MORSE TEST Morse Test is applicable to multi-cylinder engines. The engine is run at desired speed and output is noted. Then one of the cylinders is cut out by short circuiting spark plug. Under this condition other cylinders “motor” this cut cylinder. The output is measured by keeping speed constant to original value. The difference in output is measure of the indicated power of cut-out cylinder. Thus for each cylinder indicated power is obtained to find out total indicated power Let, BP = Brake Power when all cylinders are in working condition. BP1 = Brake Power when first cylinder cut-off. BP2 = Brake Power when second cylinder cut-off. BP3 = Brake Power when third cylinder cut-off. IP = Indicated Power of Engine IP1 = Indicated Power of first cylinder IP2 = Indicated Power of second cylinder IP3 = Indicated Power of third cylinder FP1, FP2, FP3 = Friction power of each cylinder
When, All cylinders in working condition, IP = (IP1 + IP2 + IP3) …………………………………………………………………(i) BP = (IP1 + IP2 + IP3) – ( FP1+ FP2 + FP3 ) ……………………………………….….(ii) First Cylinder Cut-off, BP1 = (IP2 + IP3) – ( FP1+ FP2 + FP3 ) ………………………………………………. (iii) Where, ( FP1+ FP2 + FP3 ) in above both eqs.(ii)&(iii) remains almost constant at constant speed. Subtracting Eq.(iii) from Eq.(ii), We get, Indicated Power of first cylinder,
IP1 = (BP - BP1) ……………………………………………………………………. (iv) Similarly, Indicated Power of second cylinder IP2 = (BP - BP2) …………………………………………………………………….. (v) Indicated Power of third cylinder IP3 = (BP - BP3) ……………………………………………………………………. (vi) Putting the values of IP1, IP2, IP3¬ in eq.(i),we get, IP = (BP - BP1) + (BP - BP2) + (BP - BP3) …………………………………. ……(vii) Frictional Power, FP = ( IP – BP ) ……………………………………………………………………(viii)
Efficiencies
Efficiencies
Indicates air capacity of a 4 stroke engine.
ήvol
= Actual Air Admitted at intake condition / Theoretical Volume Available(Vs)
Volumetric efficiency of an engine is an indication of the measure of the degree to which the engine fills its swept volume. It is defined as the ratio of the mass of air inducted into the engine cylinder during the suction stroke to the mass of the air corresponding to the swept volume of the engine at atmospheric pressure and temperature. Alternatively, it can be defined as the ratio of the actual volume inhaled during suction stroke measured at intake conditions to the swept volume of the piston.
v
m N Vs i 2
2m iVs N
m-- is the mass flow rate of fresh mixture. N-- is the engine speed in rev/unit time. Vs --is the piston displacement (swept
Also Vs = ApL = s = 2LN L is the piston stroke and s is the linear piston speed (m/s). N
v
2m s Ap L a 2L
Specific Fuel Consumption
4m a Ap s
s 2L
The power out put of an IC engine is measured by a rope brake dynamometer. The diameter of brake pulley is 700 mm and rope diameter is 25 mm. The load on the tight side of the rope is 50 kg and spring balance read 50N. The engine is running at 900 rpm consumes fuel of calorific value of 44000 kJ/kg, at a rate of 4 kg/hr. Calculate i. Brake specific fuel consumption, ii. Brake thermal Efficiency bsfc = mf(kg/hr)/BP(kW) BP=(2πNT) /60*1000= 2*π*(W-S) (Db+dr)/2 = (50*9.81-50)*π*(0.025+0.7)*900/ 60*1000 =15.05 kW So, bsfc = 4 / 15.05 = 0.266 kg/kW hr
Brake thermal efficiency (ɳbt) = BP kW/ mf (kg/sec)*CV (kJ/kg) = 15.05/(4/3600)*44000 = 0.3878 = 38.78%
A 4 cylinder 4 stroke SI engine has a compression ratio of 8 and bore of 100mm, with stroke equals to bore. The volumetric efficiency of each cylinder is 75%. The engine speed is 4800 rpm with an air fuel ratio of 15. CV of fuel is 42NJ/kg, mean effective pressure in the cylinder=10 bar and mechanical efficiency of the engine = 80% determine Indicated thermal efficiency nd Brake Power IP= pm L A N’/60000 Pm=mep= 10*105 N/m2 So IP = [10*105 * (π/4) 0.12* 0.1*(4800/2)*4]/60000 = 125.66kW To find fuel consumption in kg/sec Volumetric Efficiency = Actual Air consumed/ Theoretical Air consumed
Theoretical air = Vs* N’ = (π/4) 0.12* 0.1*4800/2 = 1.884m3/min = 0.0314 m3/sec = 0.1256 m3/sec for 4 cylinders
So, air consumed = 0.1256*0.75 = 0.094252 m3/sec
Air consumed in kg/s = 0.094252 m3/sec* 1.12 kg/sec = 0.1056 kg/s
Now to find mass of fuel consumed, use air fuel ratio as A/F= Air used/Fuel used
Therefore mass of fuel used = Air used/A:F = 0.1056/15 = 0.00704 kg/s
So Indicated Thermal Efficiency = IP/mf*CV = 125.66/0,00704*42*106 = 42.5%
Mechanical Efficiency= BP/IP
So, BP = IP*
ɳ
m
= 125.66*0.8 = 100.53 Kw
A 4 cylinder 2 stroke petrol engine develop 30 kW at 2500 rpm. The mean effective pressure on each piston is 8 bar and mechanical efficiency is 80%. Calculate the diameter and stroke of each cylinder if stroke to bore ratio is 1.5. Also evaluate the fuel consumption of the engine, if brake thermal efficiency is 28%. The Calorific Value of the fuel is 43900 kJ/kg. ɳm = BP/IP, so IP = 30/0.8 = 37.5kW IP = pm l A N’/ 60000 37.7 = [(π/4)*D2 (1.5D) 2500 * 8*105*5]/60000 D3 = 0.0002387
D = 0.062 m
L = 0.62*1.5 = 93 mm
Fuel Consumption Brake Thermal Efficiency (ɳbt) = BP/mf*CV 0.28 = 30/mf*43900 so mf = 0.00244 kg/sec
A six cylinder 4-S, SI engine having a piston displacement of 700 cm3 per cylinder developed 78kW at 3200 rpm and consumed 27 kg of petrol per hour. The calorific value of Petrol is 44 MJ/kg. Estimate i. The volumetric efficiency of the engine if the A:F is 12 and intake air is at 0.9 bar, 320C. ii. Find Brake Thermal Efficiency and iii. The Brake Torque.
i.Volumetric Efficiency = Actual Volume of Intake Air/Theoretical Air (Vs) Mass of air = A:F * Mass of Fuel = 12* 27 (kg/hr)=324 kg/hr This should be converted to m3/hr based on inlet condition, use PV=mRT i.e, Va = m R T/P = 324*287*305/0.9*105 = 315.126 m3/h Now swept volume per hour = Piston displacement/cylinder*No. cylinder*N/2*60
Vs=700*10-6 * 6*3200/2*60 = 403.2m3/h
Now volumetric efficiency = Volume of Intake Air/ Swept Volume Va/Vs= 325.126/403.2 = 0.781 = 78.1%
Brake Thermal Efficiency = BP/mf*CV = 78 kW/ (27/3600)*44*103 = 23.64%
The Brake Torque = T
BP = 2πNT/60000 so T = BP*60000/2πN =
0.2328kN
The following particulars are obtained in a trial on a 4-S gas engine Duration of trial = 1 hour Revolutions = 14000 Number of missed cycles = 500 Net Brake Load = 1470 N MEP= 7.5 bar Gas Consumption = 20000 liters LCV of fuel at admit conditions = 21kJ/liters Cylinder Diameters = 250 mm Stroke = 400mm Effective Brake Circumference = 4m CR=6.5:1
Calculate i.IP, ii. BP, iii. ɳm .iv. ɳit, v. Relative efficiency Indicated Power = Pm*L*A*N’/60000 Here N=14000 rev/1 hour = 14000/60 = 233.33 rpm and for 4-S engine, it is 233.33/2= 116.665 rpm
But there are 500 misfire/hr=500/60=8.33 misfire per minute which should not be considered for calculating Indicated Power So, Number of working cycles= 116.665-8.33=108.33 working cycles/min IP=[ 7.5*105 * (π/4)*0.252*0.4*108.33]/60000 = 26.59 kW Brake Power (BP) = 2πN T/60000 where T= (W-S)* R effective = (W-S)* (D+d)/2 Here circumference = π(D+d) = 4m i.e. BP = πN (W-S)* (D+d )/60000 = 3.14*1470*(14000/60)(4)/60000 = 22.86kW Mechanical Efficiency = BP/IP = 22.86/26.59 = 85.9% Indicated Thermal Efficiency = IP/mf*CV
Here mf = 20000/3600=5.55 lit/sec
So ɳth =[26.59/5.55 lit/sec*21kJ/lit ] = 23% Relative Efficiency = Thermal efficiency/Air Standard Efficiency
Air Standard Efficiency – 1 – [1/r γ-1] = 1- 1/ 6.5 1.4-1 = 52.7%
Relative Efficiency = 0.23/0.527 = 0.436 or 43.6%
During the test of 40 minutes on a single cylinder gas engine of 200 mm cylinder bore and 400 mm stroke, working on the fourstroke cycle and governed by hit and miss method of governing, the following readings were taken: Total number of revolutions = 9400 Total number of explosions = 4200 Area of indicator diagram = 550 mm2 Length of indicator diagram = 72 mm Spring number = 0.8 bar/mm Brake load = 540 N Brake wheel diameter = 1.6 m Brake rope diameter = 2 cm Gas used = 8.5 m3 Calorific value of gas = 15900 kJ/m3 Calculate : (i) Indicated power, and brake thermal efficiency
(ii) Brake power, and (iii) Indicated
Indicated power (I.P) Indicated mean effective pressure Pm = (Area of indicator diagram x spring number)/Length of the diagram = (550 x 0.8)/72 = 6.11 bar For IP calculations , Number of firing or explosions are considered, For 9400 rev, there must have been 9400/2= 4700 firings but due to misfires there are only 4400 explosions in 40 min, that is 105 firings/min
I.P. = ( Pm. L A N’ x 10)/ 6 = (6.11 x105 x 0.4 x π/4 x 0.22 x 105)/ 6oooo Brake power B.P.
B.P.=(W-S) π(Db + d)N/(60 x 1000)
= 13.4 kW
=540 x π(1.6 + 0.02) x (9400/40)/(60 x 1000) = 10.76kW Indicated thermal efficiency ηth.(I) = I.P./(Vg x C) = 13.4 x (0.00354 x 15900) = 0.238 or 23.8%
Brake thermal efficiency
ηth.(B) = B.P./(Vg x C) = 10.76 x (0.00354 x 15900) = 0.191 or 19.1%
The following observations were recorded during the test on a 6cylinder, 4-stroke Diesel engine : Bore = 125 mm Stroke = 125 mm Engine speed = 2400 r.p.m. Load on a dynamometer = 490 N Dynamometer constant = 16100 Air orifice diameter = 55 mm Co-efficient of discharge = 0.66 Head causing flow through orifice = 310 mm of water Barometer reading = 760 mm Hg Ambient temperature = 25◦ C Fuel consumption = 22.1 kg/h Calorific value of fuel = 45100 kJ/kg Per cent carbon in the fuel = 85% Per cent hydrogen in the fuel = 15% Pressure of air at the end of suction stroke = 1.013 bar Temperature at the end of stroke = 25◦ C Calculate : •Brake mean effective pressure, (ii) Specific fuel consumption, (iii) Brake thermal efficiency, (iv) Volumetric efficiency, Percentage of excess air supplied.
Brake mean effective pressure, Pmb: B.P. = (W x N)/CD = (490 x 2400)/16100 = 73 kW Also B.P. = (n pmLAN’ x 10)/6 73 = (6 x pmb x 0.125 x π/4 x 0.1252 x 2400 x ½ x 10)/6 Pm = (76 x 6 x 4 x 2)/(6 x 0.125 x π x 0.1252 x 2400 x 10) = 3.96 bar
Brake thermal efficiency, ηth.(B) : ηth.(B) = B.P./(mf x C) = 73/(0.00614 x 45100) =0.2636 or 26.36%
Specific fuel consumption, b.s.f.c : b.s.f.c = 22.1/73 = 0.3027 kg/kWh
Volumetric efficiency, ηvol : Stroke volume of cylinder = π/4 x D2 x L = π/4 x 0.1252 x 0.125 = 0.00153 m3
The volume of air passing through the orifice of the air box per minute is given by, hw x ρw=ha x ρa hw
ha = hw x ρw/ρa ρw = density of water=1000kg/m3 hw=manometer reading
Now velocity Head of air through orifice ha =
½g va2 or va = √ 2gha [ m/s2 x m=√m2/s2 = m/s]
But ha = hw x ρw/ρa so Velocity of air va = √ 2gha = √ 2x9.81x hw x 1000/ρa m/sec The volume flow rate of air (Qa) = cd x va
x
a0
The volume of air passing through the orifice of the air box per minute is given by,
Va = √ 2x9.81x hw x 1000/ρa
Cd
Ao = Area of cross –section of orifice, = π/4 do 2 = π/4 x (0.055)2 =0.00237 m2, hw = Head causing flow through orifice in cm of water, = 310/10 = 31 cm or 0.31m , and ῥa = Density of air at 1.013 bar and 25oC = P/RT = (1.013 x 105)/(287 x (25 + 273)) = 1.18 kg/m3 Volume of air, Qa = 0.00237 x 0.66√(2x9.81x0.31x1000/1.18) = 6.73 m3/sec (x60) =6.73 m3/min Actual volume of air per cylinder = 6.73/n = 6.73/6 = 1.12 m3/min Air supplied per stroke per cylinder= 1.12/(2400/2) = 0.000933 m3 Ηvol= Volume of air actually supplied / Volume of air theoretically required = 0.000933/0.00153 =0.609 or 60.9%
(v)Percentage of excess air supplied : Quantity of air required per kg of fuel for complete combustion = 100/23[ C x 8/3 + H2 x 8/1 ] Where C is the fraction of carbon and H 2 is the fraction of hydrogen present in the fuel respectively. =100/23[ 0.85 x 8/3 + 0.15 x 8/1 ] = 15.07 kg/kg of fuel Actual quantity of air supplied per kg of fuel = (Va x ῥa x 60)/22.1 = (6.73 x 1.18 x 60)/22.1 = 21.56 kg Percentage excess air = [(21.56 – 15.07)/15.07] x 100 = 43.06
%
Find the air-fuel ratio of a 4-stroke, 1 cylinder, air cooled engine with fuel consumption time for 10 cc as 20.0 sec. and air consumption time for 0.1 m3 as 16.3 sec. The load is 16 kg at speed of 3000 rpm. Also find brake specific fuel consumption in g/kWh and brake thermal efficiency. Assume the density of air as 1.175 kg/m3 and specific gravity of fuel to be 0.7. The lower heating value of fuel is 44 MJ/kg and the dynamometer constant is 5000.
A two stroke two cylinder engine runs with speed of 3000 rpm and fuel consumption of 5 litres/hr. The fuel has specific gravity of 0.7 and air-fuel ratio is 19. The piston speed is 500 m/min and indicated mean effective pressure is 6 bar. The ambient conditions are 1.013 bar, 15ºC. The volumetric efficiency is 0.7 and mechanical efficiency is 0.8. Determine brake power output considering R for gas = 0.287 kJ/kg · K Take piston speed, m/min = 2 LN where L is stroke (m) and N is rpm) Let the bore of cylinder be ‘D’ meter Using piston speed, 500 = 2 × L × 3000 ⇒ L = 0.0833 m
During trial of four strokes single cylinder engine the load on dynamometer is found 20 kg at radius of 50 cm. The speed of rotation is 3000 rpm. The bore and stroke are 20 cm and 30 respectively. Fuel is supplied at the rate of 0.15 kg/min. The calorific value of fuel may be taken as 43 MJ/kg. After some time the fuel supply is cut and the engine is rotated with motor which required 5 kW to maintain the same speed of rotation of engine.
Determine
the
brake power, indicated power, mechanical efficiency, brake thermal efficiency, indicated thermal efficiency, brake mean effective pressure, indicated mean effective pressure.
After switching off fuel supply the capacity of motor required to run engine will be the friction power required at this speed of engine Friction power = 5 kW
A four stroke four cylinder diesel engine running at 600 rpm produces 250 kW of brake power. The cylinder dimensions are 30 cm bore and 25 cm stroke. Fuel consumption rate is 1 kg/min while air fuel ratio is 10. The average indicated mean effective pressure is 0.8 MPa. Determine indicated power, mechanical efficiency, brake thermal efficiency and volumetric efficiency of engine. The calorific value of fuel is 43 MJ/kg. The ambient conditions are 1.013 bar, 27ºC. Given, D = 0.3 m, L = 0.25 m, N = 300 rpm, mf = 1 kg/min, F/A = 20, Pimep = 0.8 MPa, Brake power = 250 kW
Problems on Heat Balance Sheet A heat balance sheet is an account of heat supplied and heat utilized in various ways in the system. Necessary information concerning the performance of the engine is obtained from the heat balance.
The heat balance is generally done on second basis or minute basis or hour basis.
The heat supplied to the engine is only in the form of fuel-heat and that is given by Qs = mf X CV
Problems on Heat Balance Sheet The various ways in which heat is used up in the system is given by a. Heat equivalent of BP = kW = kJ/sec. = x60 kJ/min. b. Heat carried away by cooling water = Cpw X mw (Two – Twi) kJ/min. c. Heat carried away by exhaust gases = mg Cpg (Tge – Ta) (kJ/min.) or (kJ/sec) d. A part of heat is lost by convection and radiation as well as due to the leakage of gases. Part of the power developed inside the engine is also used to run the accessories as lubricating pump, cam shaft and water circulating pump. These cannot be measured precisely and so this is known as unaccounted ‘losses’. This unaccounted heat energy is calculated by the different between heat supplied Qs and the sum of (a) + (b) (c). Heat input per minute Heat supplied by the combustion fuel
Total
kcal (kj)
%
Heat expenditure per minute
Qs
100%
(a) Heat in BP. (b) Heat carried by jacket cooling water (c) Heat Carried by exhaust gases (d) Heat unaccounted for = Qs – (a + b + c)
Qs
100%
kcal (kj) -----
% -----
100%
Problems on Heat Balance Sheet The following observations were recorded in a test of one hour duration on a single cylinder oil engine working on four stroke cycle. Bore =300 mm Stroke =450 mm Fuel used = 8.8 kg Calorific value of fuel = 41800 kJ/kg Average speed =200 rpm m.e.p = 5.8 bar brake friction load = 1860 N Quantity of cooling water = 650 kg Temperature rise = 22oC Diameter of brake wheel = 1.22 m Calculate : (i) Mechanical efficiency, (ii) Brake thermal efficiency. Draw the heat balance sheet. Mechanical efficiency, ɳmech : Indicated power I.P.= (n pmiLAN’)/60000 =( 1 x 5.8 x105 x 0.45 x π/4 x 0.32 x 200 x ½ x)/6 = 30.7 kW
Brake power, B.P. = ((W-S) π DN)/(60 x1000) = (1860 x π x 1.22 x 200)/(60 x 1000) =23.76 kW
ɳmech = B.P./I.P. = 23.76/30.7 = 0.773 or 77.3% (ii) Brake thermal efficiency, ɳthb : ɳth.(B) =B.P./(mf x C) = 23.76/((8.8/3600) x 41800) = 0.232 or 23.2 %
heat supplied =8.8 x 41800 =367840 kJ/h (i) Heat equivalent of I.P. =I.P. x 3600 kJ/h = 30.7 x 3600 = 110520 kJ/h (ii) Heat carried away by cooling water: = mw x cpw x (tw2 - tw1) = 650 x 4.18 x 22 = 59774 kJ/h
Heat balance sheet (hourly basis) Item
kJ
Percent
Heat supplied by fuel
367840
100
(i)Heat absorbed in I.P. (ii)Heat taken away by cooling water
110520 59774
30.05 16.25
(iii)Heat carried by exhaust gases, radiation etc. (by difference) Total
197546
53.70
367840
100
In a trial of single cylinder oil engine working on a dual cycle, the following observations were made : Compression ratio = 15 Oil consumption = 10.2 kg/h Calorific value of fuel = 43890 kJ/kg Air consumption = 3.8 kg/min Speed =1900 rpm Torque on the brake drum = 186 N-m Quantity of cooling water used = 15.5kg/min Temperature rise = 36oC Exhaust gas temperature = 410oC Room temperature = 20oC Cpfor exhaust gases =1.17 kJ/kg K
Calculate : (i) Brake power, (ii) Brake specific fuel consumption, (iii) Brake thermal efficiency. Draw heat balance sheet on minute basis.
(i)Brake Power,B.P. : B.P. = 2πNT/(60 x 1000) = (2π x 1900 x 186)/(60 x 1000) = (ii) Brake specific fuel consumption, b.s.f.c. : b.s.f.c. =10.2/37 =0.2756 kg/kWh (iii) Brake thermal efficiency ɳth.(B) =B.P./(mf x C) = 37/((10.2/3600) x 43890) = 0.2975 or 29.75 %
Heat supplied by fuel per minute = (10.2/60) x 43890 = 7461 kJ/min
37
(i)Heat equivalent of B.P. = B.P. x 60 = 37 x 60 = 2220 kJ/min (ii) Heat carried away by cooling water = mw x cpw x (tw2 x tw1) = 15.5 x 4.18 x 36 = 2332 kJ/min (iii) Heat carried away by exhaust gases = mg x cpg x (tg – tr) = ((10.2/60)+3.80) x 1.17 x (410-20) = 1811 kJ/min
Item Heat supplied by fuel (i)Heat absorbed in B.P. (ii)Heat taken away by cooling water (iii)Heat carried by exhaust gases (iv)Heat unaccounted for (by difference) Total
kJ 7461 2220 2332 1811 1098 7461
Percent 100 29.8 31.2 24.3 14.7 100
During an experiment on four stroke single cylinder engine the indicator diagram obtained has average height of 1 cm while indicator constant is 25 kN/m2 per mm. The engine run at 300 rpm and the swept volume is 1.5 × 104 cm3. The effective brake load upon dynamometer is 60 kg while the effective brake drum radius is 50 cm. The fuel consumption is 0.12 kg/min and the calorific value of fuel oil is 42 MJ/kg. The engine is cooled by circulating water around it at the rate of 6 kg/min. The cooling water enters at 35º C and leaves at 70ºC. Exhaust gases leaving have energy of 30 kJ/s with them. Take specific heat of water as 4.18 kJ/kg K. Determine indicated power output, brake power output and mechanical efficiency. Also draw the overall energy balance in kJ/s Indicated mean effective pressure = 10 × 25 = 250 kPa
During 15 minutes trial of an internal combustion engine of 2-stroke single cylinder type the total 4 kg fuel is consumed while the engine is run at 1500 rpm. Engine is cooled employing water being circulated at 15 kg/min with its inlet and exit temperatures as 27ºC and 50ºC. The total air consumed is 150 kg and the exhaust temperature is 400ºC. The atmospheric temperature is 27ºC. The mean specific heat of exhaust gases may be taken as 1.25 kJ/kg K. The mechanical efficiency is 0.9. Determine, the brake power, brake specific fuel consumption and indicated thermal efficiency. Also draw energy balance on per minute basis. Brake torque is 300 Nm and the fuel calorific value is 42 MJ/kg.
During trial of a four cylinder four stroke petrol engine running at full load it has speed of 1500 rpm and brake load of 250 N when all cylinders are working. After some time each cylinder is cut one by one and then again brought back to same speed of engine. The brake readings are measured as 175 N, 180 N, 182 N and 170 N. The brake drum radius is 50 cm. The fuel consumption rate is 0.189 kg/min with the fuel whose calorific value is 43 MJ/kg and A/F ratio of 12. Exhaust gas temperature is found to be 600ºC. The cooling water flows at 18 kg/min and enters at 27ºC and leaves at 50ºC. The atmospheric air temperature is 27ºC. Take specific heat of exhaust gas as 1.02 kJ/kg K. Determine the brake power output of engine, its indicated power and mechanical efficiency. Also draw a heat balance on per minute basis
During the trial of a single acting oil engine, cylinder diameter is 20 cm, stroke 28 cm, working on two stroke cycle and firing every cycle, the following observations were made: •Duration of trial :1 hour •Total fuel used :4.22 kg •Calorific value :44670 kJ/kg •Proportion of hydrogen in fuel : 15% •Total number of revolutions : 21000 •Mean effective pressure : 2.74 bar •Net brake load applied to a drum of 100 cm diameter : 600 N •Total mass of cooling water circulated : 495 kg •Total mass of cooling water : inlet 13ºC, outlet 38ºC •Air used : 135 kg •Temperature of air in test room : 20ºC •Temperature of exhaust gases : 370ºC •Assume Cp, gases = 1.005 kJ/kg K, •Cp, steam at atmospheric pressure = 2.093 kJ/kg K •Calculate thermal efficiency and draw up the heat balance.
