TITLE: Performance testing of IC engine (heat balance sheet)
Aim: To conduct test on four stroke diesel engine to prepare heat balance sheet at various
loading conditions
Outcomes:
The students should be able to:
1. Conduct performance test on engine at various loading conditions
2. Calculate heat supplied, brake power, heat lost to cooling water, heat carried away by
exhaust gases and miscellaneous losses at different loads
3. Prepare heat balance sheet for the engine and interpret the results
Theory:
In internal combustion engines, chemical energy of fuel is converted to thermal energy on
combustion. This thermal energy is not totally converted into mechanical power. A part of
this thermal energy is converted to useful power, a part is absorbed by the cooling system, a
part is carried away by exhaust gases and a part is miscellaneous losses which includes
radiation, incomplete combustion etc. These heat losses reduce the thermal efficiency of the
engine. Frictional losses are also there which consume about 10 % of the total heat supplied.
However, the energy lost in friction in converted to heat energy due to friction which in turn
is carried away by cooling media (air or water). Due to this reason, frictional losses are not
measured separately while preparing heat balance sheet.
Heat balance sheet is a representation of heat supplied to engine with fuel and heat
expenditure in tabular form. The heat expenditure typically includes heat equivalent to
brake power, heat lost to cooling water, heat carried away by miscellaneous losses etc. Table
1 represents typical values of components of heat expenditure in percentage for SI and CI
engine at full load.
Table 1: Components of Heat Balance Sheet in Percent at Full Load
Engine Heat Equivalent to Heat Lost with
Heat Carried away by Miscellaneous
Type
Brake Power (%)
Cooling Water (%)
Exhaust Gases (%)
Losses (%)
SI
21-28
12-27
30-55
3-55
CI
29-42
15-35
25-45
5-21
Procedure:
1. Check oil level and fuel supply.
2. Start cooling and calorimeter water supply.
3. Start the engine by cranking and allow it to run at constant temperature at to achieve
steady state conditions.
4. Load the engine by applying weight at the end of rope.
5. Note the spring balance and dead weight such that effective load on dynamometer
corresponds to given loading condition.
6. Measure speed of the engine using tachometer
7. Count time required for 20 or 30 ml of fuel consumption using stopwatch.
8. Note cooling and calorimeter water flow rate in liters per hour.
9. Record all temperatures from T1 to T7
10. Increase the load on engine by adding weight at the end of the rope corresponding to
next loading condition.
11. Repeat steps 5 to 9 for different loading conditions.
Specifications:
Density of fuel: 820 kg/m3
Calorific value of diesel: 44000 kJ/kg
Effective radius of dynamometer = 0.136 m
Rated power of engine (100 % load) = 5.9 kW at 1800 rpm
Observation Table:
Load
N
S
W
T
mjw
mcw
T1
T2
T3
T4
T5
T6
T7
(rpm)
(kg)
(kg)
(sec)
(LPH)
(LPH)
(0C)
(0C)
(0C)
(0C)
(0C)
(0C)
(0C)
25%
50%
75%
100%
Where,
N is crankshaft speed in rpm,
S is spring balance reading in kg,
W is Dead load at rope end in kg,
t is time required for ______ ml (generally 20 or 30 ml) fuel consumption,
mjw is flow rate of water through engine jacket in liters per hour (LPH),
mcw is flow rate of water through exhaust gas calorimeter in liters per hour (LPH),
T1 and T2 are inlet and outlet temperatures of jacket cooling water respectively (0C)
T3 and T4 are inlet and outlet temperatures of calorimeter cooling water respectively (0C)
T5 and T6 are the temperatures of exhaust gases before and after calorimeter respectively
(0C)
Sample Calculations:
Loading Conditions:
Rated power of engine is 5.9 kW at 1800 rpm.
B. P. =
2∗π∗N∗T
kW
60 ∗ 103
∴ 5.9 =
2 ∗ π ∗ 1800 ∗ T
60 ∗ 103
T = 31.3 N − m (At 100% load).
At ______ % load condition, torque is _____ * 31.3 = _____ N-m
But, T = (s-w)*g*R
Using above equation, calculate value of term (s-w) for different loading conditions.
Heat Input:
Mass flow rate of fuel (mf)
mf =
_______ ∗ 0.820
=
time (t) ∗ 1000
30 ∗ 0.820
=
∗ 1000
kg
sec
Heat supplied with fuel,
Input Energy = mf ∗ C. V. =
∗ 44000 =
kW
Heat Expenditure:
i.
Heat equivalent to B. P.
B. P. =
2∗π∗N∗T 2∗π∗
=
60 ∗ 103
∗
60 ∗ 103
=
kW
ii. Heat lost with cooling water,
Mass flow rate of cooling water jacket, mjw =
mjw (LPH)
=
3600
3600
=
kg
sec
Heat lost with cooling water = mjw ∗ Cpw ∗ (T2 − T1 )
∗ 4.186 ∗ (
Heat lost with cooling water =
−
iii. Heat carried away by exhaust gases,
In exhaust gas calorimeter,
Heat given by exhaust gases = Heat received by cooling water
mg ∗ Cg ∗ (T5 − T6 ) = mcw ∗ Cpw ∗ (T4 − T3 )
∴ mg ∗ Cg =
mcw ∗Cpw ∗(T4 −T3 )
(T5 −T6 )
=
)=
kW
Hence,
Heat carried away by exhaust gases = mg ∗ Cpg ∗ (T5 − T7 ) =
kW
iv. Miscellaneous losses
Miscellaneous losses = Heat supplied – (Brake power – Heat lost with cooling water – Heat
carried away by exhaust gases) =
kW
Repeat above calculations for different loading conditions.
Result Table:
Heat
kW
Input
25
50
75
100
%
%
%
%
Heat
Heat
kW
%
Expenditure
25
50
75
100
25
50
75
100
%
%
%
%
%
%
%
%
Brake power
supplied
Heat lost to
cooling water
Heat carried
away by
exhaust gases
Miscellaneous
losses
Total
Draw a plot of load vs heat supplied, load vs brake power, load vs heat lost to cooling water,
load vs heat carried away by exhaust gas and load vs miscellaneous loss.
Conclusion:
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