Tài liệu Handbook of Mechanical Engineering Calculations P6 pdf

6.1
SECTION 6
INTERNAL-COMBUSTION
ENGINES
Determining the Economics of
Reciprocating I-C Engine Cogeneration
6.1
Diesel Generating Unit Efficiency
6.7
Engine Displacement, Mean Effective
Pressure, and Efficiency
6.8
Engine Mean Effective Pressure and
Horsepower
6.9
Selection of an Industrial Internal-
Combustion Engine
6.10
Engine Output at High Temperatures
and High Altitudes
6.11
Indicator Use on Internal-Combustion
Engines
6.12
Engine Piston Speed, Torque,
Displacement, and Compression Ratio
6.13
Internal-Combustion Engine Cooling-
Water Requirements
6.14
Design of a Vent System for an Engine

Room
6.18
Design of a Bypass Cooling System for
an Engine
6.21
Hot-Water Heat-Recovery System
Analysis
6.26
Diesel Fuel Storage Capacity and Cost
6.27
Power Input to Cooling-Water and Lube-
Oil Pumps
6.29
Lube-Oil Cooler Selection and Oil
Consumption
6.31
Quantity of Solids Entering an Internal-
Combustion Engine
6.31
Internal-Combustion Engine
Performance Factors
6.32
Volumetric Efficiency of Diesel Engines
6.34
Selecting Air-Cooled Engines for
Industrial Applications
6.37
DETERMINING THE ECONOMICS OF
RECIPROCATING I-C ENGINE COGENERATION
Determine if an internal-combustion (I-C) engine cogeneration facility will be ec-

onomically attractive if the required electrical power and steam services can be
served by a cycle such as that in Fig. 1 and the specific load requirements are those
shown in Fig. 2. Frequent startups and shutdowns are anticipated for this system.
Calculation Procedure:
1. Determine the sources of waste heat available in the typical I-C engine
There are three primary sources of waste heat available in the usual I-C engine.
These are: (1) the exhaust gases from the engine cylinders; (2) the jacket cooling
water; (3) the lubricating oil. Of these three sources, the quantity of heat available
is, in descending order: exhaust gases; jacket cooling water; lube oil.
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
6.2
POWER GENERATION
FIGURE 1 Reciprocating-engine cogeneration system waste heat from the exhaust, and
jacket a oil cooling, are recovered. (Indeck Energy Services, Inc.)
FIGURE 2 Low-speed Diesel-engine cogeneration. (Indeck Energy Services, Inc.)
2. Show how to compute the heat recoverable from each source
For the exhaust gases, use the relation, H
A
ϭ
W(
⌬
t)(c
g
), where W
A
ϭ
rate of gas

flow from the engine, lb /h (kg/h);
⌬
t
ϭ
temperature drop of the gas between the
heat exchanger inlet and outlet,
Њ
F(
Њ
C); c
g
ϭ
specific heat of the gas, Btu/ lb
Њ
F
(J/kg
Њ
C). For example, if an I-C engine exhausts 100,000 lb / h (45,400 kg/h) at
700
Њ
F (371
Њ
C) to a HRSG (heat-recovery steam generator), leaving the HRSG at
330
Њ
F (166
Њ
C), and the specific heat of the gas is 0.24 Btu / lb
Њ
F (1.0 kJ / kg

Њ
C),
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.3
the heat recoverable, neglecting losses in the HRSG and connecting piping, is
H
A
ϭ
100,000(700
Ϫ
330)(0.24)
ϭ
8,880,000 Btu/ h (2602 MW).
With an average heat of vaporization of 1000 Btu / lb (2330 kJ / kg) of steam,
this exhaust gas flow could generate 8,880,000 /1000
ϭ
8880 lb/h (4032 kg/h) of
steam. If oil with a heating value of 145,000 Btu/gal (40,455 kJ/L) were used to
generate this steam, the quantity required would be 8,880,000 /145,000
ϭ
61.2
gal/h (232 L/ h). At a cost of 90 cents per gallon, the saving would be $0.90(61.2)
ϭ
$55.08/h. Assuming 5000 hours of operation per year, or 57 percent load, the
saving in fuel cost would be 5000($55.08)
ϭ

$275,400. This is a significant saving
in any plant. And even if heat losses in the ductwork and heat-recovery boiler cut
the savings in half, the new would still exceed one hundred thousand dollars a year.
And as the operating time increases, so too do the savings.
3. Compute the savings potential in jacket-water and lube-oil heat recovery
A similar relation can be used to compute jacket-water and lube-oil heat recovery.
The flow rate can be expressed in either pounds (kg) per hour or gallons (L) per
minute, depending on the designer’s choice.
Since water has a specific heat of unity, the heat-recovery potential of the jacket
water is H
W
ϭ
w(
⌬
t
w
), where w
ϭ
weight of water flow, lb per h (kg/h);
⌬
t
w
ϭ
change in temperature of the jacket water when flowing through the heat exchanger,
Њ
F(
Њ
C). Thus, if the jacket-water flow is 25,000 lb / h (11,350 kg / h) and the tem-
perature change during flow of the jacket water through and external heat exchanger
is 190 to 70

Њ
F (88 to 21
Њ
C), the heat given up by the jacket water, neglecting losses
is H
w
ϭ
25,000(190
Ϫ
70)
ϭ
3,000,000 Btu /h (879 MW). During 25 h the heat
recovery will be 24(3,000,000)
ϭ
72,000,000 Btu (75,960 MJ). This is a significant
amount of heat which can be used in process or space heating, or to drive an air-
conditioning unit.
If the jacket-water flow rate is expressed in gallons per minute instead of pounds
per hour (L/min instead of kg/ h), the heat-recovery potential, H
wg
ϭ
gpm(
⌬
t)(8.33)
where 8.33
ϭ
lb/gal of water. With a water flow rate of 50 gpm and the same
temperature range as above, H
wg
ϭ

50(120)(8.33)
ϭ
49,980 Btu/min (52,279 kJ /
min).
4. Find the amount of heat recoverable from the lube oil
During I-C engine operation, lube-oil temperature can reach high levels—in the
300 to 400
Њ
F (149 to 201
Њ
C) range. And with oil having a typical specific heat of
0.5 Btu/ lb
Њ
F (2.1 kJ / kg
Њ
C), the heat-recovery potential for the lube oil is
ϭ
H
w
o
w
o
(
⌬
t)(c
o
), where w
o
ϭ
oil flow in lb / h (kg / h);

⌬
t
ϭ
temperature change of the oil
during flow through the heat-recovery heat exchanger
ϭ
oil inlet temperature
Ϫ
oil
outlet temperature,
Њ
For
Њ
C; c
o
ϭ
specific heat of oil
ϭ
0.5 Btu/lb
Њ
F (kJ/kg
Њ
C).
With an oil flow of 2000 lb/h (908 kg/h), a temperature change of 140
Њ
F (77.7
Њ
C),
H
o

ϭ
2000(140)(0.50)
ϭ
140,000 Btu / h (41 kW). Thus, as mentioned earlier, the
heat recoverable from the lube oil is usually the lowest of the three sources.
With the heat flow rates computed here, an I-C engine cogeneration facility can
be easily justified, especially where frequent startups and shutdowns are anticipated.
Reciprocating Diesel engines are preferred over gas and steam turbines where fre-
quent startups and shutdowns are required. Just the fuel savings anticipated for
recovery of heat in the exhaust gases of this engine could pay for it in a relatively
short time.
Related Calculations. Cogeneration, in which I-C engines are finding greater
use throughout the world every year, is defined by Michael P. Polsky, President,
Indeck Energy Services, Inc., as ‘‘the simultaneous production of useful thermal
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INTERNAL-COMBUSTION ENGINES
6.4
POWER GENERATION
energy and electric power from a fuel source or some variant thereof. It is more
efficient to produce electric power and steam or hot water together than electric
power alone, as utilities do, or thermal energy alone, which is common in industrial,
commercial, and institutional plants.’’ Figures 1 and 2 in this procedure are from
the firm of which Mr. Polsky is president.
With the increased emphasis on reducing environmental pollution, conserving
fuel use, and operating at lower overall cost, cogeneration—especially with Diesel
engines—is finding wider acceptance throughout the world. Design engineers
should consider cogeneration whenever there is a concurrent demand for electricity
and heat. Such demand is probably most common in industry but is also met in

commercial (hotels, apartment houses, stores) and institutional (hospital, prison,
nursing-home) installations. Often, the economic decision is not over whether co-
generation should be used, but what type of prime mover should be chosen.
Three types of prime movers are usually considered for cogeneration—steam
turbines, gas turbines, or internal-combustion engines. Steam and/or gas turbines
are usually chosen for large-scale utility and industrial plants. For smaller plants
the Diesel engine is probably the most popular choice today. Where natural gas is
available, reciprocating internal-combustion engines are a favorite choice, especially
with frequent startups and shutdowns.
Recently, vertical modular steam engines have been introduced for use in co-
generation. Modules can be grouped to increase the desired power output. These
high-efficiency units promise to compete with I-C engines in the growing cogen-
eration market.
Guidelines used in estimating heat recovery from I-C engines, after all heat loses,
include these: (1) Exhaust-gas heat recovery
ϭ
28 percent of heat in fuel; (2) Jacket-
water heat recovery
ϭ
27 percent of heat in fuel; (3) Lube-oil heat recovery
ϭ
9
percent of the heat in the fuel. The Diesel Engine Manufacturers Association
(DEMA) gives these values for heat disposition in a Diesel engine at three-quarters
to full load: (1) Fuel consumption
ϭ
7366 Btu / bhp
⅐
h (2.89 kW/kW); (2) Useful
work

ϭ
2544 Btu / bhp
⅐
h (0.999 kW / kW); (3) Loss in radiation, etc.
ϭ
370 Btu/
bhp
⅐
h (0.145 kW/ kW); (4) To cooling water
ϭ
2195 Btu/bhp
⅐
h (0.862 kW/ kW);
(5) To exhaust
ϭ
2258 Btu/bhp
⅐
h (0.887 kW/ kW). The sum of the losses is 1
Btu/bhp
⅐
h greater than the fuel consumption because of rounding of the values.
Figure 3 shows a proposed cogeneration, desiccant-cooling, and thermal-storage
integrated system for office buildings in the southern California area. While directed
at the micro-climates in that area, similar advantages for other micro-climates and
building types should be apparent. The data presented here for this system were
prepared by The Meckler Group and are based on a thorough engineering and
economic evaluation for the Southern California Gas Co. of the desiccant-
cooling/thermal-energy-storage/cogeneration system, a proprietary design devel-
oped for pre- and post-Title-24 mid-rise office buildings. Title 24 is a section of
the State of California Administrative Code that deals with energy-conservation

standards for construction applicable to office buildings. A summary of the study
was presented in Power magazine by Milton Meckler.
In certain climates, office buildings are inviting targets for saving energy via
evaporative chilling. When waste heat is plentiful, desiccant cooling and cogener-
ation become attractive. In coupling the continuously available heat-rejection
capacity of packaged cogeneration units, Fig. 4, with continuously operating re-
generator demands, the use of integrated components for desiccant cooling, thermal-
energy storage, and cogeneration increases. The combination also ensures a rea-
sonable constant, cost-effective supply of essentially free electric power for general
building use.
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.5
FIGURE 3 Integrated system is a proposed off-peak desiccant/evaporative-cooling configu-
ration with cogeneration capability. (Power and The Meckler Group.)
Recoverable internal-combustion engine heat should at least match the heat re-
quirement of the regenerator, Fig. 3. The selected engine size (see a later procedure
in this section), however, should not cause the cogeneration system’s Purpa (Public
Utility Regulatory & Policies Act) efficiency to drop below 42.5 percent. (Purpa
efficiency decreases as engine size increases.) An engine size is selected to give
the most economical performance and still have a Purpa efficiency of greater than
42.5 percent.
The utility study indicated a favorable payout period and internal rate of return
both for retrofits of pre-Title-24 office buildings and for new buildings in compli-
ance with current Title-24 requirements (nominal 200 to 500 cooling tons). Al-
though the study was limited to office-building occupancies, it is likely that other
building types with high ventilation and electrical requirements would also offer

attractive investment opportunities.
Based on study findings, fuel savings ranged from 3300 to 7900 therms per year.
Cost savings ranged from $322,000 to $370,000 for the five-story-building case
studies and from $545,000 to $656,000 for 12-story-building case studies where
the synchronously powered, packaged cogeneration unit was not used for emer-
gency power.
Where the cogeneration unit was also used for emergency power, the initial cost
decreased from $257,000 to $243,000, representing a 31 percent drop in average
cost for the five-story-building cases; and from $513,000 to $432,000, a 22 percent
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INTERNAL-COMBUSTION ENGINES
6.6
POWER GENERATION
FIGURE 4 Packaged cogeneration I-C engine
unit supplies waste heat to desiccant regenerator.
(Power and The Meckler Group.)
dip in average cost for the 12-story-building cases. The average cost decrease shifts
the discounted payback period an average of 5.6 and 5.9 years for the five- and 12-
story-building cases, respectively.
Study findings were conservatively reported, since no credit was taken for po-
tential income resulting from Purpa sales to the serving utility at off-peak hours,
when actual building operating requirements fall below rated cogenerator output.
This study is another example of the importance of the internal-combustion engine
in cogeneration around the world today.
Worldwide there is a movement toward making internal-combustion engines, and
particularly diesel engines, cleaner-running. In general, this means reducing partic-
ulate emissions from diesel-engine exhaust gases. For cities with large numbers of
diesel-powered buses, exhaust emissions can be particularly unpleasant. And some

medical personnel say that diesel exhaust gases can be harmful to the health of
people breathing them.
The approach to making diesel engines cleaner takes two tacts: (1) improving
the design of the engine so that fewer particulates are emitted and (2) using cleaner
fuel to reduce the particulate emissions. Manufacturers are using both approaches
to comply with the demands of federal and state agencies regulating emissions.
Today’s engineers will find that ‘‘cleaning up’’ diesel engines is a challenging and
expensive procedure. However, cleaner-operating diesels are being introduced every
year.
*Elliott, Standard Handbook of Power Plant Engineering, McGraw-Hill, 1989.
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.7
DIESEL GENERATING UNIT EFFICIENCY
A 3000-kW diesel generating unit performs thus: fuel rate, 1.5 bbl (238.5 L) of
25
Њ
API fuel for a 900-kWh output; mechanical efficiency, 82.0 percent; generator
efficiency, 92.0 percent. Compute engine fuel rate, engine-generator fuel rate, in-
dicated thermal efficiency, overall thermal efficiency, brake thermal efficiency.
Calculation Procedure:
1. Compute the engine fuel rate
The fuel rate of an engine driving a generator is the weight of fuel, lb, used to
generate 1 kWh at the generator input shaft. Since this engine burns 1.5 bbl (238.5
L) of fuel for 900 kW at the generator terminals, the total fuel consumption is (1.5
bbl)(42 gal/bbl)
ϭ

63 gal (238.5 L), at a generator efficiency of 92.0 percent.
To determine the weight of this oil, compute its specific gravity s from s
ϭ
141.5/(131.5
ϩ Њ
API), where
Њ
API
ϭ
API gravity of the fuel. Hence, s
ϭ
141.5(131.5
ϩ
25)
ϭ
0.904. Since 1 gal (3.8 L) of water weighs 8.33 lb (3.8 kg)
at 60
Њ
F (15.6
Њ
C), 1 gal (3.8 L) of this oil weighs (0.904)(8.33)
ϭ
7.529 lb (3.39
kg). The total weight of fuel used when burning 63 gal is (63 gal)(7.529 lb/ gal)
ϭ
474.5 lb (213.5 kg).
The generator is 92 percent efficient. Hence, the engine actually delivers enough
power to generate 900 / 0.92
ϭ
977 kWh at the generator terminals. Thus, the engine

fuel rate
ϭ
474.5 lb fuel / 977 kWh
ϭ
0.485 lb/ kWh (0.218 kg/kWh).
2. Compute the engine-generator fuel rate
The engine-generator fuel rate takes these two units into consideration and is the
weight of fuel required to generate 1 kWh at the generator terminals. Using the
fuel-consumption data from step 1 and the given output of 900 kW, we see that
engine-generator fuel rate
ϭ
474.5 lb fuel/900 kWh output
ϭ
0.527 lb/ kWh (0.237
kg/kWh).
3. Compute the indicated thermal efficiency
Indicated thermal efficiency is the thermal efficiency based on the indicated horse-
power of the engine. This is the horsepower developed in the engine cylinder. The
engine fuel rate, computed in step 1, is the fuel consumed to produce the brake or
shaft horsepower output, after friction losses are deducted. Since the mechanical
efficiency of the engine is 82 percent, the fuel required to produce the indicated
horsepower is 82 percent of that required for the brake horsepower, or (0.82)(0.485)
ϭ
0.398 lb/ kWh (0.179 kg/kWh).
The indicated thermal efficiency of an internal-combustion engine driving a gen-
erator is e
i
ϭ
3413/ƒ
i

(HHV), where e
i
ϭ
indicated thermal efficiency, expressed as
a decimal; ƒ
i
ϭ
indicated fuel consumption, lb/kWh; HHV
ϭ
higher heating value
of the fuel, Btu / lb.
Compute the HHV for a diesel fuel from HHV
ϭ
17,680
ϩ
60
ϫ Њ
API. For this
fuel, HHV
ϭ
17,680
ϩ
60(25)
ϭ
19,180 Btu/ lb (44,612.7 kJ/kg).
With the HHV known, compute the indicated thermal efficiency from e
i
ϭ
3,413/[(0.398)(19,180)]
ϭ

0.447 or 44.7 percent.
4. Compute the overall thermal efficiency
The overall thermal efficiency e
o
is computed from e
o
ϭ
3413/ƒ
o
(HHV), where
ƒ
o
ϭ
overall fuel consumption, Btu /kWh; other symbols as before. Using the
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INTERNAL-COMBUSTION ENGINES
6.8
POWER GENERATION
engine-generator fuel rate from step 2, which represents the overall fuel consump-
tion e
o
ϭ
3413/[(0.527)(19,180)]
ϭ
0.347, or 34.7 percent.
5. Compute the brake thermal efficiency
The engine fuel rate, step 1, corresponds to the brake fuel rate ƒ
b

. Compute the
brake thermal efficiency from e
b
ϭ
3413/ƒ
b
(HHV), where ƒ
b
ϭ
brake fuel rate,
Btu/kWh; other symbols as before. For this engine-generator set, e
b
ϭ
3413/
[(0.485)(19,180)]
ϭ
0.367, or 36.7 percent.
Related Calculations. Where the fuel consumption is given or computed in
terms of lb / (hp
⅐
h), substitute the value of 2545 Btu/(hp
⅐
h) (1.0 kW/kWh) in
place of the value 3413 Btu / kWh (3600.7 kJ/kWh) in the numerator of the e
i
, e
o
,
and e
b

equations. Compute the indicated, overall, and brake thermal efficiencies as
before. Use the same procedure for gas and gasoline engines, except that the higher
heating value of the gas or gasoline should be obtained from the supplier or by
test.
ENGINE DISPLACEMENT, MEAN EFFECTIVE
PRESSURE, AND EFFICIENCY
A12
ϫ
18 in (30.5
ϫ
44.8 cm) four-cylinder four-stroke single-acting diesel engine
is rated at 200 bhp (149.2 kW) at 260 r/ min. Fuel consumption at rated load is
0.42 lb / (bhp
⅐
h) (0.25 kg/kWh). The higher heating value of the fuel is 18,920
Btu/lb (44,008 kJ/kg). What are the brake mean effective pressure, engine dis-
placement in ft
3
/(min
⅐
bhp), and brake thermal efficiency?
Calculation Procedure:
1. Compute the brake mean effective pressure
Compute the brake mean effective pressure (bmep) for an internal-combustion en-
gine from bmep
ϭ
33,000 bhp
n
/LAn, where bmep
ϭ

brake mean effective pressure,
lb/in
2
; bhp
n
ϭ
brake horsepower output delivered per cylinder, hp; L
ϭ
piston
stroke length, ft; a
ϭ
piston area, in
2
; n
ϭ
cycles per minute per cylinder
ϭ
crank-
shaft rpm for a two-stroke cycle engine, and 0.5 the crankshaft rpm for a four-
stroke cycle engine.
For this engine at its rated hbp, the output per cylinder is 200 bhp / 4 cylinders
ϭ
50 bhp (37.3 kW). Then bmep
ϭ
33,000(50)/[(18/ 12)(12)
2
(
␲
/4)(260/2)]
ϭ

74.8
lb/in
2
(516.1 kPa). (The factor 12 in the denominator converts the stroke length
from inches to feet.)
2. Compute the engine displacement
The total engine displacement V
d
ft
3
is given by V
d
ϭ
LAnN, where A
ϭ
piston
area, ft
2
; N
ϭ
number of cylinders in the engine; other symbols as before. For this
engine, V
d
ϭ
(18/12)(12/ 12)
2
(
␲
/4)(260/2)(4)
ϭ

614 ft
3
/min (17.4 m
3
/min). The
displacement is in cubic feet per minute because the crankshaft speed is in r/min.
The factor of 12 in the denominators converts the stroke and area to ft and ft
2
,
respectively. The displacement per bhp
ϭ
(total displacement, ft
3
/min)/bhp output
of engine
ϭ
614/200
ϭ
3.07 ft
3
/(min
⅐
bhp) (0.12 m
3
/kW).
3. Compute the brake thermal efficiency
The brake thermal efficiency e
b
of an internal-combustion engine is given by e
b

ϭ
2545/(sfc)(HHV), where sfc
ϭ
specific fuel consumption, lb /(bhp
⅐
h); HHV
ϭ
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.9
higher heating value of fuel, Btu/ lb. For this engine, e
b
ϭ
2545/[(0.42)(18,920)]
ϭ
0.32, or 32.0 percent.
Related Calculations. Use the same procedure for gas and gasoline engines.
Obtain the higher heating value of the fuel from the supplier, a tabulation of fuel
properties, or by test.
ENGINE MEAN EFFECTIVE PRESSURE
AND HORSEPOWER
A 500-hp (373-kW) internal-combustion engine has a brake mean effective pressure
of 80 lb / in
2
(551.5 kPa) at full load. What are the indicated mean effective pressure
and friction mean effective pressure if the mechanical efficiency of the engine is
85 percent? What are the indicated horsepower and friction horsepower of the

engine?
Calculation Procedure:
1. Determine the indicated mean effective pressure
Indicated mean effective pressure imep lb/in
2
for an internal-combustion engine is
found from imep
ϭ
bmep/e
m
, where bmep
ϭ
brake mean effective pressure, lb/
in
2
; e
m
ϭ
mechanical efficiency, percent, expressed as a decimal. For this engine,
imep
ϭ
80/0.85
ϭ
94.1 lb/ in
2
(659.3 kPa).
2. Compute the friction mean effective pressure
For an internal-combustion engine, the friction mean effective pressure ƒmep lb /
in
2

is found from ƒmep
ϭ
imep
Ϫ
bmep,orƒmep
ϭ
94.1
Ϫ
80
ϭ
14.1 lb/in
2
(97.3
kPa).
3. Compute the indicated horsepower of the engine
For an internal-combustion engine, the mechanical efficiency e
m
ϭ
bhp/ihp, where
ihp
ϭ
indicated horsepower. Thus, ihp
ϭ
bhp/e
m
,orihp
ϭ
500/0.85
ϭ
588 ihp

(438.6 kW).
4. Compute the friction hp of the engine
For an internal-combustion engine, the friction horsepower is ƒhp
ϭ
ihp
Ϫ
bhp.In
this engine, ƒhp
ϭ
588
Ϫ
500
ϭ
88 fhp (65.6 kW).
Related Calculations. Use a similar procedure to determine the indicated en-
gine efficiency e
ei
ϭ
e
i
/e, where e
ϭ
ideal cycle efficiency; brake engine efficiency,
e
eb
ϭ
e
b
e; combined engine efficiency or overall engine thermal efficiency e
eo

ϭ
e
o
ϭ
e
o
e. Note that each of these three efficiencies is an engine efficiency and cor-
responds to an actual thermal efficiency, e
i
, e
b
, and e
o
.
Engine efficiency e
e
ϭ
e
t
/e, where e
t
ϭ
actual engine thermal efficiency. Where
desired, the respective actual indicated brake, or overall, output can be substituted
for e
i
, e
b
, and e
o

in the numerator of the above equations if the ideal output is
substituted in the denominator. The result will be the respective engine efficiency.
Output can be expressed in Btu per unit time, or horsepower. Also, e
e
ϭ
actual
mep/ideal mep, and e
ei
ϭ
imep/ideal mep; e
eb
ϭ
bmep/ideal mep; e
eo
ϭ
overall
mep/ideal mep. Further, e
b
ϭ
e
m
e
i
, and bmep
ϭ
e
m
(imep). Where the actual heat
supplied by the fuel, HHV Btu/ lb, is known, compute e
i

e
b
and e
o
by the method
given in the previous calculation procedure. The above relations apply to any re-
ciprocating internal-combustion engine using any fuel.
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INTERNAL-COMBUSTION ENGINES
6.10
POWER GENERATION
TABLE 1
Internal-Combustion Engine Rating Table
SELECTION OF AN INDUSTRIAL
INTERNAL-COMBUSTION ENGINE
Select an internal-combustion engine to drive a centrifugal pump handling 2000
gal/min (126.2 L/s) of water at a total head of 350 ft (106.7 m). The pump speed
will be 1750 r / min, and it will run continuously. The engine and pump are located
at sea level.
Calculation Procedure:
1. Compute the power input to the pump
The power required to pump water is hp
ϭ
8.33GH/33,000e, where G
ϭ
water
flow, gal /min; H
ϭ

total head on the pump, ft of water; e
ϭ
pump efficiency,
expressed as a decimal. Typical centrifugal pumps have operating efficiencies rang-
ing from 50 to 80 percent, depending on the pump design and condition and liquid
handled. Assume that this pump has an efficiency of 70 percent. Then hp
ϭ
8.33(2000)/(350)/ [(33,000)(0.70)]
ϭ
252 hp (187.9 kW). Thus, the internal-
combustion engine must develop at least 252 hp (187.9 kW) to drive this pump.
2. Select the internal-combustion engine
Since the engine will run continuously, extreme care must be used in its selection.
Refer to a tabulation of engine ratings, such as Table 1. This table shows that a
diesel engine that delivers 275 continuous brake horsepower (205.2 kW) (the near-
est tabulated rating equal to or greater than the required input) will be rated at 483
bhp (360.3 kW) at 1750 r/ min.
The gasoline-engine rating data in Table 1 show that for continuous full load at
a given speed, 80 percent of the tabulated power can be used. Thus, at 1750 r/ min,
the engine must be rated at 252/0.80
ϭ
315 bhp (234.9 kW). A 450-hp (335.7-
kW) unit is the only one shown in Table 1 that would meet the needs. This is too
large; refer to another builder’s rating table to find an engine rated at 315 to 325
bhp (234.9 to 242.5 kW) at 1750 r/min.
The unsuitable capacity range in the gasoline-engine section of Table 1 is a
typical situation met in selecting equipment. More time is often spent in finding a
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.11
TABLE 2
Correction Factors for Altitude and Temperature
suitable unit at an acceptable price than is spent computing the required power
output.
Related Calculations. Use this procedure to select any type of reciprocating
internal-combustion engine using oil, gasoline, liquified-petroleum gas, or natural
gas for fuel.
ENGINE OUTPUT AT HIGH TEMPERATURES AND
HIGH ALTITUDES
An 800-hp (596.8-kW) diesel engine is operated 10,000 ft (3048 m) above sea
level. What is its output at this elevation if the intake air is at 80
Њ
F (26.7
Њ
C)? What
will the output at 10,000-ft (3048-m) altitude be if the intake air is at 110
Њ
F
(43.4
Њ
C)? What would the output be if this engine were equipped with an exhaust
turbine-driven blower?
Calculation Procedure:
1. Compute the engine output at altitude
Diesel engines are rated at sea level at atmospheric temperatures of not more than
90
Њ

F (32.3
Њ
C). The sea-level rating applies at altitudes up to 1500 ft (457.2 m). At
higher altitudes, a correction factor for elevation must be applied. If the atmospheric
temperature is higher than 90
Њ
F (32.2
Њ
C), a temperature correction must be applied.
Table 2 lists both altitude and temperature correction factors. For an 800-hp
(596.8-kW) engine at 10,000 ft (3048 m) above sea level and 80
Њ
F (26.7
Њ
C) intake
air, hp output
ϭ
(sea-level hp) (altitude correction factor), or output
ϭ
(800)(0.68)
ϭ
544 hp (405.8 kW).
2. Compute the engine output at the elevated temperature
When the intake air is at a temperature greater than 90
Њ
F (32.3
Њ
C), a temperature
correction factor must be applied. Then output
ϭ

(sea-level hp)(altitude correction
factor)(intake-air-temperature correction factor), or output
ϭ
(800)(0.68)(0.95)
ϭ
516 hp (384.9 kW), with 110
Њ
F (43.3
Њ
C) intake air.
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INTERNAL-COMBUSTION ENGINES
6.12
POWER GENERATION
TABLE 3
Atmospheric Pressure at Various
Altitudes
3. Compute the output of a supercharged engine
A different altitude correction is used for a supercharged engine, but the same
temperature correction factor is applied. Table 2 lists the altitude correction factors
for supercharged diesel engines. Thus, for this supercharged engine at 10,000-ft
(3048-m) altitude with 80
Њ
F (26.7
Њ
C) intake air, output
ϭ
(sea-level hp)(altitude

correction factor)
ϭ
(800)(0.74)
ϭ
592 hp (441.6 kW).
At 10,000-ft (3048-m) altitude with 110
Њ
F (43.3
Њ
C) inlet air, output
ϭ
(sea-level
hp)(altitude correction factor)(temperature correction factor)
ϭ
(800)(0.74)(0.95)
ϭ
563 hp (420.1 kW).
Related Calculations. Use the same procedure for gasoline, gas, oil, and liq-
uefied-petroleum gas engines. Where altitude correction factors are not available
for the type of engine being used, other than a diesel, multiply the engine sea-level
brake horsepower by the ratio of the altitude-level atmospheric pressure to the
atmospheric pressure at sea level. Table 3 lists the atmospheric pressure at various
altitudes.
An engine located below sea level can theoretically develop more power than
at sea level because the intake air is denser. However, the greater potential output
is generally ignored in engine-selection calculations.
INDICATOR USE ON
INTERNAL-COMBUSTION ENGINES
An indicator card taken on an internal-combustion engine cylinder has an area of
5.3 in

2
(34.2 cm
2
) and a length of 4.95 in (12.7 cm). What is the indicated mean
effective pressure in this cylinder? What is the indicated horsepower of this four-
cycle engine if it has eight 6-in (15.6-cm) diameter cylinders, an 18-in (45.7-cm)
stroke, and operates at 300 r/ min? The indicator spring scale is 100 lb/in (1.77
kg/mm).
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.13
Calculation Procedure:
1. Compute the indicated mean effective pressure
For any indicator card, imep
ϭ
(card area, in
2
) (indicator spring scale, lb)/(length
of indicator card, in) where imep
ϭ
indicated mean effective pressure, lb / in
2
. Thus,
for this engine, imep
ϭ
(5.3)(100)/4.95
ϭ

107 lb/ in
2
(737.7 kPa).
2. Compute the indicated horsepower
For any reciprocating internal-combustion engine, ihp
ϭ
(imep)LAn/33,000, where
ihp
ϭ
indicated horsepower per cylinder; L
ϭ
piston stroke length, ft; A
ϭ
piston
area, in
2
, n
ϭ
number of cycles/min. Thus, for this four-cycle engine where n
ϭ
0.5 r/ min, ihp
ϭ
(107)(18/12)(6)
2
(
␲
/4)(300/2) / 33,000
ϭ
20.6 ihp (15.4 kW) per
cylinder. Since the engine has eight cylinders, total ihp

ϭ
(8 cylinders)(20.6 ihp
per cylinder)
ϭ
164.8 ihp (122.9 kW).
Related Calculations. Use this procedure for any reciprocating internal-
combustion engine using diesel oil, gasoline, kerosene, natural gas, liquefied-
petroleum gas, or similar fuel.
ENGINE PISTON SPEED, TORQUE,
DISPLACEMENT, AND COMPRESSION RATIO
What is the piston speed of an 18-in (45.7-cm) stroke 300
ϭ
r/min engine? How
much torque will this engine deliver when its output is 800 hp (596.8 kW)? What
are the displacement per cylinder and the total displacement if the engine has eight
12-in (30.5-cm) diameter cylinders? Determine the engine compression ratio if the
volume of the combustion chamber is 9 percent of the piston displacement.
Calculation Procedure:
1. Compute the engine piston speed
For any reciprocating internal-combustion engine, piston speed
ϭ
ƒpm
ϭ
2L(rpm),
where L
ϭ
piston stroke length, ft; rpm
ϭ
crankshaft rotative speed, r/ min. Thus,
for this engine, piston speed

ϭ
2(18/12)(300)
ϭ
9000 ft/ min (2743.2 m / min).
2. Determine the engine torque
For any reciprocating internal-combustion engine, T
ϭ
63,000(bhp)/rpm, where
T
ϭ
torque developed, in
⅐
lb; bhp
ϭ
engine brake horsepower output; rpm
ϭ
crankshaft rotative speed, r/ min. Or T
ϭ
63,000(800)/300
ϭ
168,000 in
⅐
lb
(18.981 N
⅐
m).
Where a prony brake is used to measure engine torque, apply this relation: T
ϭ
(F
b

Ϫ
F
o
)r, where F
b
ϭ
brake scale force, lb, with engine operating; F
o
ϭ
brake
scale force with engine stopped and brake loose on flywheel; r
ϭ
brake arm, in
ϭ
distance from flywheel center to brake knife edge.
3. Compute the displacement
The displacement per cylinder d
c
in
3
of any reciprocating internal-combustion en-
gine is d
c
ϭ
L
i
A
i
where L
i

ϭ
piston stroke, in; A
ϭ
piston head area, in
2
. For this
engine, d
c
ϭ
(18)(12)
2
(
␲
/4)
ϭ
2035 in
3
(33,348 cm
3
) per cylinder.
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INTERNAL-COMBUSTION ENGINES
6.14
POWER GENERATION
The total displacement of this eight-cylinder engine is therefore (8 cylin-
ders)(2035 in
3
per cylinder)

ϭ
16,280 in
3
(266,781 cm
3
).
4. Compute the compression ratio
For a reciprocating internal-combustion engine, the compression ratio r
c
ϭ
V
b
/V
a
,
where V
b
ϭ
cylinder volume at the start of the compression stroke, in
3
or ft
3
; V
a
ϭ
combustion-space volume at the end of the compression stroke, in
3
or ft
3
. When

this relation is used, both volumes must be expressed in the same units.
In this engine, V
b
ϭ
2035 in
3
(33,348 cm
3
); V
a
ϭ
(0.09)(2035)
ϭ
183.15 in
3
.
Then r
c
ϭ
2035/183.15
ϭ
11.1:1.
Related Calculations. Use these procedures for any reciprocating internal-
combustion engine, regardless of the fuel burned.
INTERNAL-COMBUSTION ENGINE
COOLING-WATER REQUIREMENTS
A 1000-hbp (746-kW) diesel engine has a specific fuel consumption of 0.360 lb /
(bhp
⅐
h) (0.22 kg/kWh). Determine the cooling-water flow required if the higher

heating value of the fuel is 10,350 Btu/lb (24,074 kJ / kg). The net heat rejection
rates of various parts of the engine are, in percent: jacket water, 11.5; turbo-
charger, 2.0; lube oil. 3.8; aftercooling, 4.0; exhaust, 34.7; radiation, 7.5. How much
30 lb / in
2
(abs) (206.8 kPa) steam can be generated by the exhaust gas if this is a
four-cycle engine? The engine operates at sea level.
Calculation Procedure:
1. Compute the engine heat balance
Determine the amount of heat used to generate 1 bhp
⅐
h (0.75 kWh) from: heat
rate, Btu/ bhp
⅐
h)
ϭ
(sfc)(HHV), where sfc
ϭ
specific fuel consumption, lb / (bhp
⅐
h); HHV
ϭ
higher heating value of fuel, Btu / lb. Or, heat rate
ϭ
(0.36)(19.350)
ϭ
6967 Btu/ (bhp
⅐
h) (2737.3 W/kWh).
Compute the heat balance of the engine by taking the product of the respective

heat rejection percentages and the heat rate as follows:
Then the power output
ϭ
6967
Ϫ
4422
ϭ
2545 Btu/(bhp
⅐
h) (999.9 W /kWh),
or 2545/6967
ϭ
0.365, or 36.5 percent. Note that the sum of the heat losses and
power generated, expressed in percent, is 100.0.
2. Compute the jacket cooling-water flow rate
The jacket water cools the jackets and the turbocharger. Hence, the heat that must
be absorbed by the jacket water is 800
ϩ
139
ϭ
939 Btu/(bhp
⅐
h) (369 W/kWh),
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.15
using the heat rejection quantities computed in step 1. When the engine is devel-

oping its full rated output of 1000 bhp (746 kW), the jacket water must absorb
[939 Btu/ (bhp
⅐
h)(1000 bhp)
ϭ
939,000 Btu/ h (275,221 W).
Apply a safety factor to allow for scaling of the heat-transfer surfaces and other
unforeseen difficulties. Most designers use a 10 percent safety factor. Applying this
value of the safety factor for this engine, we see the total jacket-water heat load
ϭ
939,000
ϩ
(0.10)(939,000)
ϭ
1,032,900 Btu/ h (302.5 kW).
Find the required jacket-water flow from G
ϭ
H/500
⌬
t, where G
ϭ
jacket-water
flow, gal/min; H
ϭ
heat absorbed by jacket water, Btu/ h;
⌬
t
ϭ
temperature rise
of the water during passage through the jackets,

Њ
F. The usual temperature rise of
the jacket water during passage through a diesel engine is 10 to 20
Њ
F (5.6 to 11.1
Њ
C).
Using 10
Њ
F for this engine we find G
ϭ
1,032,900/[(500)(10)]
ϭ
206.58 gal / min
(13.03 L/ s), say 207 gal/min (13.06 L / s).
3. Determine the water quantity for radiator cooling
In the usual radiator cooling system for large engines, a portion of the cooling
water is passed through a horizontal or vertical radiator. The remaining water is
recirculated, after being tempered by the cooled water. Thus, the radiator must
dissipate the jacket, turbocharger, and lube-oil cooler heat, Fig. 5.
The lube oil gives off 264 Btu / (bhp
⅐
h) (103.8 W /kWh). With a 10 percent
safety factor, the total heat flow is 264
ϩ
(0.10)(264)
ϭ
290.4 Btu/ (bhp
⅐
h) (114.1

W/kWh). At the rated output of 1000 bhp (746 kW), the lube-oil heat load
ϭ
[290.4 Btu /(bhp
⅐
h)](1000 bhp)
ϭ
290,400 Btu / h (85.1 kW). Hence, the total heat
load on the radiator
ϭ
jacket
ϩ
lube-oil heat load
ϭ
1,032,900
ϩ
290,400
ϭ
1,323,300 Btu/ h (387.8 kW)
Radiators (also called fan coolers) serving large internal-combustion engines are
usually rated for a 35
Њ
F (19.4
Њ
C) temperature reduction of the water. To remove
1,323,300 Btu/h (387.8 kW) with a 35
Њ
F (19.4
Њ
C) temperature decrease will
require a flow of G

ϭ
H/(500
⌬
t)
ϭ
1,323,300/[(500)(35)]
ϭ
76.1 gal/min
(4.8 L/ s).
4. Determine the aftercooler cooling-water quantity
The aftercooler must dissipate 278 Btu/(bhp
⅐
h) (109.2 W / kWh). At an output of
1000 bhp (746 kW), the heat load
ϭ
[278 Btu/(bhp
⅐
h)](1000 bhp)
ϭ
278,000
Btu/h (81.5 kW). In general, designers do not use a factor of safety for the after-
cooler because there is less chance of fouling or other difficulties.
Witha5
Њ
F (2.8
Њ
C) temperature rise of the cooling water during passage through
the after-cooler, the quantity of water required G
ϭ
H/(500

⌬
t)
ϭ
278,000/
[(500)(5)]
ϭ
111 gal/ min (7.0 L /s).
5. Compute the quantity of steam generated by the exhaust
Find the heat available in the exhaust by using H
e
ϭ
Wc
⌬
t
e
, where H
e
ϭ
heat
available in the exhaust, Btu/h; W
ϭ
exhaust-gas flow, lb/h; c
ϭ
specific heat of
the exhaust gas
ϭ
0.252 Btu/(lb
⅐ Њ
F) (2.5 kJ/kg);
⌬

t
e
ϭ
exhaust-gas temperature
at the boiler inlet,
Њ
F
Ϫ
exhaust-gas temperature at the boiler outlet,
Њ
F.
The exhaust-gas flow from a four-cycle turbocharged diesel is about 12.5 lb/
(bhp
⅐
h) (7.5 kg /kWh). At full load this engine will exhaust [12.5 lb/(bhp
⅐
h)](1000
bhp)
ϭ
12,500 lb/ h (5625 kg/h).
The temperature of the exhaust gas will be about 750
Њ
F (399
Њ
C) at the boiler
inlet, whereas the temperature at the boiler outlet is generally held at 75
Њ
F (41.7
Њ
C)

higher than the steam temperature to prevent condensation of the exhaust gas. Steam
at 30 lb / in
2
(abs) (206.8 kPa) has a temperature of 250.33
Њ
F (121.3
Њ
C). Thus, the
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INTERNAL-COMBUSTION ENGINES
6.16
POWER GENERATION
FIGURE 5 Internal-combustion engine cooling systems: (a) radiator type;
(b) evaporating cooling tower; (c) cooling tower. (Power.)
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INTERNAL-COMBUSTION ENGINES
INTERNAL-COMBUSTION ENGINES
6.17
FIGURE 6 Slant diagrams for internal-combustion engine heat exchangers. (Power.)
exhaust-gas outlet temperature from the boiler will be 250.33
ϩ
75
ϭ
325.33
Њ
F

(162.9
Њ
C), say 325
Њ
F (162.8
Њ
C). Then H
e
ϭ
(12,500)(0.252)(750
Ϫ
325)
ϭ
1,375,000 Btu/ h (403.0 kW).
At 30 lb/in
2
(abs) (206.8 kPa), the enthalpy of vaporization of steam is 945.3
Btu/lb (2198.9 kJ /kg), found in the steam tables. Thus, the exhaust heat can gen-
erate 1,375,000 / 945.3
ϭ
1415 lb/h (636.8 kg/h) if the boiler is 100 percent effi-
cient. With a boiler efficiency of 85 percent, the steam generated
ϭ
(1415 lb/
h)(0.85)
ϭ
1220 lb/h (549.0 kg/h), or (1200 lb/h) / 1000 bhp
ϭ
1.22 lb/ (bhp
⅐

h)
(0.74 kg/ kWh).
Related Calculations. Use this procedure for any reciprocating internal-
combustion engine burning gasoline, kerosene, natural gas, liquified-petroleum gas,
or similar fuel. Figure 1 shows typical arrangements for a number of internal-
combustion engine cooling systems.
When ethylene glycol or another antifreeze solution is used in the cooling sys-
tem, alter the denominator of the flow equation to reflect the change in specific
gravity and specific heat of the antifreeze solution, a s compared with water. Thus,
with a mixture of 50 percent glycol and 50 percent water, the flow equation in step
2 becomes G
ϭ
H/(436
⌬
t). With other solutions, the numerical factor in the de-
nominator will change. This factor
ϭ
(weight of liquid lb / gal)(60 min/h), and the
factor converts a flow rate of lb /h to gal/min when divided into the lb/h flow rate.
Slant diagrams, Fig 6, are often useful for heat-exchanger analysis.
Two-cycle engines may have a larger exhaust-gas flow than four-cycle engines
because of the scavenging air. However, the exhaust temperature will usually be 50
to 100
Њ
F (27.7 to 55.6
Њ
C) lower, reducing the quantity of steam generated.
Where a dry exhaust manifold is used on an engine, the heat rejection to the
cooling system is reduced by about 7.5 percent. Heat rejected to the aftercooler
cooling water is about 3.5 percent of the total heat input to the engine. About 2.5

percent of the total heat input to the engine is rejected by the turbocharger jacket.
The jacket cooling water absorbs 11 to 14 percent of the total heat supplied.
From 3 to 6 percent of the total heat supplied to the engine is rejected in the oil
cooler.
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