I.C.E.
POWER PLANT Sample Diagram of spring scale (𝒔)
Engine Performance
Sources of Energy: Fuels (e.g. Diesel and
Gasoline)
where:
𝐄𝐂 = Energy Chargeable
TP = Theoretical Power
IP = Indicated Power
BP = Brake Power ➢ 𝑙 = length of card
EP = Electrical Power or Combined Power
• 𝑳 = length of stroke
Theoretical Power (TP) • 𝑨= piston face area
𝜋
𝑇𝑃 = 𝑊𝑛𝑒𝑡 𝐴 = (𝐷)2
4
CYLINDER DIMENSION:𝐷 𝑥 𝐿 (Bore x
Indicated Power (IP) Stroke)
𝐼𝑃 = 𝑃𝐼 𝐿𝐴𝑁𝑛 𝑥 2 ∗
Note: *for double acting cylinder and neglecting the effect
of piston • 𝑵 = power stroke per cylinder
𝑐𝑦𝑐𝑙𝑒𝑠
rod diameter, multiply by 2 in 𝑡𝑖𝑚𝑒−𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
• 𝑷𝑰 = indicated mean effective
pressure ❖ For 2-stroke cycles:
(𝐴𝑐 )(𝑠) 𝑵 = 𝑠𝑝𝑒𝑒𝑑 in rpm
𝑃𝐼 = 𝑠𝑝𝑒𝑒𝑑
𝑟𝑒𝑣 𝑐𝑦𝑐𝑙𝑒𝑠
𝑙 𝑚𝑖𝑛 𝑚𝑖𝑛
𝑁 = 𝑟𝑒𝑣 = 𝑟𝑒𝑣 = =
1 𝑐𝑦𝑙
𝑐𝑦𝑐𝑙𝑒 𝑐𝑦𝑐𝑙𝑒
where: 𝑐𝑦𝑐𝑙𝑒𝑠
➢ 𝐴𝑐 = indicator card area 𝑚𝑖𝑛−𝑐𝑦𝑙
➢ 𝑠 = spring scale
❖ For 4-stroke cycles:
𝑠𝑝𝑒𝑒𝑑
𝑵= in rpm
2
𝑟𝑒𝑣 𝑐𝑦𝑐𝑙𝑒𝑠
𝑠𝑝𝑒𝑒𝑑 𝑚𝑖𝑛 𝑚𝑖𝑛
𝑁 = 2𝑟𝑒𝑣/𝑐𝑦𝑐𝑙𝑒 = 2𝑟𝑒𝑣 = =
1 𝑐𝑦𝑙
𝑐𝑦𝑙
𝑐𝑦𝑐𝑙𝑒𝑠
𝑚𝑖𝑛−𝑐𝑦𝑙
Note: 1 revolution = 2 strokes
• 𝒏 = no. of cylinders
1
UNIT ANALYSIS OF INDICATED POWER (𝐼𝑃) where:
• 𝑽𝑫 = volume or engine
from, displacement
𝐼𝑃 = 𝑃𝐼 𝐿𝐴𝑁𝑛 𝑥 2
UNIT ANALYSIS OF BRAKE POWER (𝐵𝑃)
𝐼𝑃
𝑘𝑁 𝑚3 𝑐𝑦𝑐𝑙𝑒 1𝑚𝑖𝑛 from,
= 2[ ][ ] [𝑐𝑦𝑙] 𝑥 [ ]
𝑚 𝑐𝑦𝑐𝑙𝑒 𝑚𝑖𝑛 − 𝑐𝑦𝑙 60 𝑠𝑒𝑐 𝐵𝑃 = 𝑃𝐵 𝐿𝐴𝑁𝑛 𝑥 2
𝑘𝑁 − 𝑚 𝑘𝐽 2𝜋 𝑟𝑒𝑣 𝑚𝑖𝑛
𝐼𝑃 = = = 𝑘𝑊 𝐵𝑃 = [ ] [(𝑘𝑁)(𝑚) [ ]𝑥 ]
𝑠 𝑠 𝑟𝑒𝑣 𝑚𝑖𝑛 60𝑠
𝑘𝑁 − 𝑚 𝑘𝐽
𝐵𝑃 = = = 𝑘𝑊
𝑠 𝑠
Brake Power (BP) Friction Power (FP)
𝐵𝑃 = 2𝜋𝑇(𝑠𝑝𝑒𝑒𝑑) 𝐹𝑃 = 𝐼𝑃 − 𝐵𝑃
where: 𝐹𝑃 = 2𝜋𝑇𝐹 (𝑠𝑝𝑒𝑒𝑑)
• 𝑻 = brake torque
where:
𝑇 = 𝐹𝑅 𝑻𝑭 = Friction Torque
Electrical Power (EP)
Single Phase
𝐸𝑃 = 𝐸𝐼 𝑥 𝑃𝐹
3 – Phase
𝐸𝑃 = (√3)𝐸𝐼 𝑥 𝑃𝐹
where:
• 𝑬 = Potential Voltage Difference
➢ 𝑭 = net dynamometer force (in V or kV)
• 𝑰 = Current (in Amperes)
𝐹 = 𝑔𝑟𝑜𝑠𝑠 𝑤𝑡. −𝑡𝑎𝑟𝑒 𝑤𝑡. • 𝑷𝑭 = Power Factor
UNIT ANALYSIS OF ELECTRICAL POWER (𝐸𝑃)
➢ 𝑹 = brake dynamometer arm 𝐸𝑃 = VA × PF (Watts)
𝐸𝑃 = kVA × PF (kW)
also, Piston Speed (𝝂𝒑 )
𝐵𝑃 = 𝑃𝐵 𝐿𝐴𝑁𝑛 𝑥 2*
𝜈𝑝 = 2𝐿 (𝑠𝑝𝑒𝑒𝑑)
𝐵𝑃
⇒ 𝑃𝐵 = UNIT ANALYSIS OF PISTON SPEED (𝜈𝑝 )
𝐿𝐴𝑁𝑛 𝑥 2 ∗
from,
where: 𝜈𝑝 = 2𝐿 (𝑠𝑝𝑒𝑒𝑑)
2 𝑠𝑡𝑟𝑜𝑘𝑒𝑠 𝑚 𝑟𝑒𝑣
• 𝑷𝑩 = brake mean effective pressure 𝜈𝑝 = ( ) (𝑠𝑡𝑟𝑜𝑘𝑒 ) (𝑚𝑖𝑛)
𝑟𝑒𝑣
and,
𝑚
𝑉𝐷 = 𝐿𝐴𝑁𝑛 𝑥 2* 𝜈𝑝 =
𝑚𝑖𝑛
2
Engine Efficiency (𝜼) If 𝐸𝐶 is fixed as Heat Input,
𝑃𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 Theoretical Thermal Efficiency (𝒆𝑻 )
𝜼= 𝑥 100%
𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡 𝑇𝑃
𝑒𝑇 = × 100%
𝐸𝐶
If TP is fixed as Power Input,
Indicated Thermal Efficiency (𝒆𝑰 )
Indicated Engine Efficiency (𝜼𝑰 )
𝐼𝑃
IP 𝑒𝐼 = × 100%
𝜂𝐼 = × 100% 𝐸𝐶
TP
Brake Thermal Efficiency (𝒆𝑩 )
Brake Engine Efficiency (𝜼𝑩 )
𝐵𝑃
BP 𝑒𝐵 = × 100%
𝜂𝐵 = × 100% 𝐸𝐶
TP
Combined Engine Efficiency (𝜼𝑲 ) Combined Thermal Efficiency (𝒆𝑲 )
EP 𝐸𝑃
𝜂𝐾 = × 100% 𝑒𝐾 = × 100%
TP 𝐸𝐶
for other𝜂;
Mechanical Efficiency (𝜼𝒎 )
BP Heat Rate (𝑯𝑹)
𝜂𝑚 = × 100%
IP
𝐻𝑒𝑎𝑡 𝑖𝑛𝑝𝑢𝑡 𝑘𝐽 𝐵𝑇𝑈
𝑯𝑹 = 𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
; 𝑘𝑊−ℎ𝑟
𝑜𝑟 𝐻𝑃−ℎ𝑟
Generator or Electrical Efficiency (𝜼𝒈 )
EP If 𝐸𝐶 is fixed as Heat Input,
𝜂𝑔 = × 100%
BP
Theoretical Heat Rate (𝑻𝑯𝑹)
Combined Mechanical and Electrical
Efficiency (𝜼𝒎𝒈 ) 𝐸𝐶
𝑇𝐻𝑅 =
𝑇𝑃
EP
𝜂𝑚𝑔 = × 100% Indicated Heat Rate (𝑰𝑯𝑹)
IP
𝐸𝐶
𝐼𝐻𝑅 =
𝐼𝑃
Thermal Efficiency (𝓮)
Brake Heat Rate (𝑩𝑯𝑹)
𝑃𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑝𝑢𝑡
𝓮 = 𝑥 100% 𝐵𝐻𝑅 =
𝐸𝐶
𝐻𝑒𝑎𝑡 𝐼𝑛𝑝𝑢𝑡
𝐵𝑃
Combined Heat Rate (𝑪𝑯𝑹)
𝐸𝐶
𝐶𝐻𝑅 =
𝐸𝑃
3
Specific Fuel Consumption I.C.E. Heat Balance
(𝑺𝑭𝑪)
or Fuel Rate For 5-item heat balance
𝑚𝑓 𝑘𝑔 𝑙𝑏
𝑺𝑭𝑪 = 𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
; 𝑘𝑊−ℎ𝑟
𝑜𝑟 𝐻𝑃−ℎ𝑟
Theoretical Specific Fuel Consumption
(𝑻𝑺𝑭𝑪)
𝑚𝑓
𝑇𝑆𝐹𝐶 =
𝑇𝑃
𝐸𝐶 = ΣQTotal
Indicated Specific Fuel Consumption
(𝑰𝑺𝑭𝑪)
𝐸𝐶 = 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 + 𝑄5
𝑚𝑓
𝐼𝑆𝐹𝐶 =
𝐼𝑃
where:
Brake Specific Fuel Consumption (𝑩𝑺𝑭𝑪)
𝑚𝑓 • 𝑸𝟏 => Useful Heat Output
𝐵𝑆𝐹𝐶 =
𝐵𝑃
𝑄1 = 𝐵𝑃
Combined Specific Fuel Consumption
(𝑪𝑺𝑭𝑪)
𝑚𝑓
𝐶𝑆𝐹𝐶 =
𝐸𝑃
• 𝑸𝟐 => Heat loss due to Friction
Some Conversion Factors (CF):
𝑙𝑏𝑓 −𝑓𝑡 𝑄2 = 𝐹𝑃
550 𝐻𝑝−𝑠𝑒𝑐
where:
2545
𝐵𝑇𝑈
➢ 𝐹𝑃 = 𝐼𝑃 − 𝐵𝑃
𝐻𝑝−ℎ𝑟
𝑘𝐽
3600
• 𝑸𝟑 => Heat loss due to Coolant
𝑘𝑊−ℎ𝑟
𝑘𝑔𝑓 −𝑚
4562.8 𝐻𝑝−𝑚𝑖𝑛
𝑄3 = 𝑚𝑐 𝑐𝑝𝑐 ∆𝑇𝑐
𝑙𝑏𝑓 −𝑓𝑡
33000 𝐻𝑝−𝑚𝑖𝑛
where:
➢ ∆𝑇𝑐 = 𝑇𝑐𝑜𝑢𝑡 − 𝑇𝑐𝑖𝑛
Note: If coolant is not specified, assume 𝐻2 𝑂
so,
➢ 𝑐𝑝𝑐 = 𝑐𝑝 𝐻 (𝑠𝑡𝑑. 𝑣𝑎𝑙𝑢𝑒)
2𝑂
4
• 𝑸𝟒 = Heat loss due to Exhaust Effect of Elevation
Gases
𝑄4 = 𝑚𝑔 𝑐𝑝𝑔 ∆𝑇𝑔
Note: If 𝑐𝑝𝑔 is not given nor can’t be calculated
then,
➢ 𝑐𝑝𝑔 = 𝑐𝑝𝑎𝑖𝑟 (𝑠𝑡𝑑. 𝑣𝑎𝑙𝑢𝑒)
➢ ∆𝑇𝑔 = 𝑇𝑔 − 𝑇𝑎𝑖𝑟
• 𝑸𝟓 = Radiation and Unaccounted
Losses Conditions @ sea level:
from, 𝑃𝑠 = 𝑃𝑎𝑡𝑚𝑠 = 101.325 𝑘𝑃𝑎 = 760 𝑚𝑚𝐻𝑔
𝐸𝐶 = 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 + 𝑄5 𝑇𝑠 = 𝑇𝑎𝑖𝑟𝑠𝑡𝑑 = 80℉ or 26.67℃
𝑄5 = 𝐸𝐶 − [ 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 ] Change in atmospheric conditions
𝑃𝑎𝑡𝑚𝑒 = 𝑃𝑎𝑡𝑚𝑠 − (𝐶𝐹 )𝐴
For 4-item heat balance
where:
▪ 𝑄2 and 𝑄5 is combined
• 𝑪𝑭 =Correction Factor
WHERE:
𝑄2&5 = Friction, Radiation, and 1𝑖𝑛𝐻𝑔 83.312𝑚𝑚𝐻𝑔
𝑪𝑭 = 𝑜𝑟
Unaccounted losses 1000𝑓𝑡 1000𝑚
so,
•𝑨= Altitude =𝑓𝑡 𝑜𝑟 𝑚
𝐸𝐶 = 𝑄1 + 𝑄2&5 + 𝑄3 + 𝑄4 Temperature at any elevation
In terms of %𝑄𝑖 6.5𝑜 𝐶
𝑇𝑒 = 𝑇𝑠 − (𝐴 )
𝑄𝑖 1000𝑚
%𝑄𝑖 = 𝑥100%
𝐸𝐶
where:
Note: ∑ %𝑄𝑖 = 100%
• 𝑨 = Altitude in m
• 𝑇𝑠 = 𝑇𝑎𝑡𝑚𝑠𝑡𝑑 = 90𝑜 𝐹 𝑜𝑟 26.67𝑜 𝐶
SAE Derating Method
from,
𝐵𝑃 ∝ 𝜌𝑎𝑖𝑟 ⇒ 𝐵𝑃 = (𝐶)(𝜌𝑎𝑖𝑟 )
𝐵𝑃
= 𝐶
𝜌𝑎𝑖𝑟
5
so, If both P and T changes;
𝐵𝑃𝑒 𝐵𝑃𝑠 then,
=
𝜌𝑎𝑖𝑟𝑒 𝜌𝑎𝑖𝑟𝑠 𝑃𝑒
𝐵𝑃𝑒 𝑅𝑎𝑖𝑟 𝑇𝑒
Note: =
𝐵𝑃𝑠 𝑃𝑠
For air, considering air as Ideal gas 𝑅𝑎𝑖𝑟 𝑇𝑠
𝑚 𝑃 𝐵𝑃𝑒 𝑃𝑒 𝑇𝑠
𝑃𝑉 = 𝑚𝑅𝑇 ⇒ =
𝑉 𝑅𝑇 = [ ][ ]
𝐵𝑃𝑠 𝑃𝑠 𝑇𝑒
but,
𝑚
𝜌= Engine Volumetric Efficiency (η𝑣 )
𝑉
𝑉𝑎𝑖𝑟
so, η𝑣 = 𝑥 100%
𝑉𝐷
𝑃
𝜌=
𝑅𝑇 For air, Volume is constant
so,
so; 𝑉𝑎𝑖𝑟 = 𝐶
For 𝝆𝒂𝒊𝒓 at any elevation
Also, engine displacement is constant
𝑃𝑒 so,
𝜌𝑎𝑖𝑟𝑒 = 𝑉𝐷 = 𝐶
𝑅𝑎𝑖𝑟 𝑇𝑒
For 𝝆𝒂𝒊𝒓 at sea level If 𝑉𝑎𝑖𝑟 and 𝑉𝐷 is constant
then,
𝑃𝑠 𝜼𝒗 = 𝐶
𝜌𝑎𝑖𝑟𝑠 =
𝑅𝑎𝑖𝑟 𝑇𝑠
Items not affected:
Say only 𝑃 changes: ▪ Mass of fuel
𝑇𝑒 = 𝑇𝑠 𝒎𝒇 = 𝐶
then, ▪ Friction Power
𝑃𝑒
𝐵𝑃𝑒 𝑅𝑎𝑖𝑟 𝑇𝑒 𝑭𝑷 = 𝐶
=
𝐵𝑃𝑠 𝑃𝑠
▪ Volume of air
𝑅𝑎𝑖𝑟 𝑇𝑠
𝑽𝒂𝒊𝒓 = 𝐶
𝐵𝑃𝑒 𝑃𝑒
=
𝐵𝑃𝑠 𝑃𝑠
6
SAMPLE PROBLEMS operation. The SG of diesel fuel is 0.8 and 1
bbl = 159 L.
(a) Determine the no. of spherical
1. A test on one cylinder Otto cycle yields the tanks needed with diameter of 7
following data: m by the plant to ensure
Torque = 950 N-m continuous operation for 30 days
HHV of fuel = 41860 kJ/kg without refilling.
Indicated mep = 758 kPa (b) If all the tanks in (a) are full and
Speed = 300 rpm 3
plant is to run continuously at 4
Fuel consumption = 0.003 kg/s
Bore x stroke = 28cm x 30.5 cm load, how many hrs will the fuel
Determine: last (without refilling)?
𝑘𝑔
(a) Engine thermal efficiency Engine sp. fuel consumption is 0.82 𝑘𝑊−ℎ𝑟
(b) Engine mechanical efficiency 3 𝑘𝑔
at load and 0.8 at full load.
(c) The fuel cost per hr if fuel costs 4 𝑘𝑊−ℎ𝑟
PhP 30 per liter.
5. A 40 cm x 56 cm, 235 rpm, 16 cylinder, 4-
SG of fuel is 0.82.
stroke cycle, stationary Diesel Engine is
connected to a 3125 kVA (80% Power Factor)
2. Data for a single-acting, four stroke Diesel generator. It also drives a 30 kW exciter.
engine under test are as follows: Other data are as follows:
Fuel consumption = 10.8 kg/hr of Generator Efficiency = 90%
24°API Mechanical Efficiency = 70%
fuel Calculate:
Prony brake test data: (a) Brake mep
Arm = 107 cm (b) Indicated mep
Speed = 420 rpm (c) Indicated sp. fuel consumption if
Gross wt. = 127 kg heating value of fuel is 40 500
Bore x stroke = 230 mm x 355 mm kJ/kg and 𝑒𝐼 = 38%
Tare wt. = 12 kg
Calculate:
6. An internal combustion engine is situated
(a) Brake thermal efficiency
at a height of 1600 m where the surrounding
(b) Piston speed
temperature is 18°𝐶. The capacity of the
(c) Engine displacement
engine is 150 kW with a mechanical
(d) Brake sp. fuel consumption
efficiency of 90%. The sp. fuel consumption
𝑘𝑔
of the engine is0.52 𝑘𝑊−ℎ𝑟 and the air-fuel
3. The following data are observed from an
ratio is 11 kg/kg.If the engine is to operate at
I.C.E.
sea level.
Power Plant:
Determine the new:
Brake Torque = 1000 N-m
(a) Brake power
Mechanical efficiency = 80%
(b) Sp. fuel consumption
Speed = 400 rpm
(c) Air-fuel ratio
Water jacket loss = 30%
(d) Mechanical efficiency
Indicated thermal efficiency = 25%
Determine:
(a) Brake heat rate
(b) Indicated heat rate
(c) Gallons per min. of cooling water
needed if temperature rise for
the coolant is 8°𝐶
4. A Diesel electric power plant consumes
130 bbl of diesel per 24 hrs continuous
