UNIVERSITY OF ENGINEERING AND
TECNOLOGY PESHAWAR
Power Plant Assignment
Name: Iftikhar Ahmad
Registration No: 22PWMEC5227
Section: D
Subject: Power Plant - I
Assignment No: 01
Date: 27-Mar-2025
Submitted To: Dr. Alam Zaib Khan
Question No 01:
A twin-cylinder 2-stroke engine has a swept volume of 150 cm3. The maximum power
output is 19 kW at 11,000 rpm. At this condition, the bsfc is .11kg/MJ and the
gravimetric air/fuel ratio is 12:1. If ambient test conditions are 10 C, 1.03 bar and the
fuel has a calorific value of 44 MJ/kg, calculate the bmep, the arbitrary overall efficiency
and the volumetric efficiency.
Given Data:
Swept Volume (Vs) = 150 cm3
Power Output = 19 kW
Speed (N) = 11000 rpm
BSFC = 11kg/MJ
Air/Fuel Ratio = 12:1
Tempreture = 10o = 283 K
Pressure = 1.03 bar
CV Of Fuel Is = 44 MJ/kg
Required Data:
(i) BMEP ?
(ii) Arbitrary Overall Efficiency ηoverall ?
(iii) ηvolumetric
Solution:
(i)
We know that,
BMEP = Brake power/ Vs × N
Putting values so,
19 × 103/(150 × 10-6)(11000/60) = 6.91 × 105 = 6.91 bar
BMEP = 6.91 bar Ans
(ii)
Arbitrary Overall Efficiency ηoverall will be
ηoverall = 1/(bsfc × CV)
Putting values so
ηoverall = 1/ (0.11×10-6)(44×106) = 0.207
ηoverall = 20.7% Ans
(iii)
As we know
Volumetric efficiency is equall to,
ηvolumetric = Va/(Vs×N) ………………………………... eq (A)
where Va = ma/ρa ………………………………… eq (B)
ma = AFR × mf = 12 × 0.00209 = 0.0251 kg/s,
ρa = p/RT = 1.03×105/(287×283) = 1.268 kg/m3
Put ma and ρa in eq B so
Va = ma/ρa = 0.0251/1.268 = 0.0198 m3/s
Now volumetric efficiency will be:
ηvolumetric = Va/(Vs×N) = 0.0198/(150 × 10-6)(11000/60) = 7.20
ηvolumetric = 72% Ans
……………………………………………………………………………………………………………………………………
Question No 02:
Use Lotus Engine Simulation Software for a Single Cylinder Model shows the BSFC (brake
specific fuel consumption), BMEP (brake mean effective pressure), Torque, and Power
produced by the engine given in the table at different RPM. Also, for the same engine
show Variation of Exhaust Constituents with AFR and RPM using different Fuel.
Graphical Result:
Lotus Model:
…………………………………………………………………………………………………………………………………….
Question No 03:
The following is the analysis of a supply of coal gas:
H2 = 49.4%, CO = 18%, CH4 = 20%, C4H8 = 2%, O2 = 0.4% CO2 = 4%, N2 = 6.2%
(i) Calculate the stoichiometric A/F ratio.
(ii) Find also the wet and dry analysis of the products of combustion if the actual
mixture of the products of combustion if the actual mixture is 20% weak.
Solution:
(i)
Kmol/kmol Combustion O2 kmo/ Products Products
fuel equation Kmol fuel CO2 H2O
H2 0.494 2H2+O2--->2H2O 0.247 ---- 0.494
CO 0.18 2CO+O2-2CO2 0.09 0.18 ----
CH4 0.2 CH4+2O2-2CO2+H2O 0.4 0.2 0.4
C4H8 0.02 C4H8+6O2-4CO2+4H2O 0.12 0.08 0.08
O2 0.004 ---- 0.004 ---- ----
CO2 0.04 ---- 0.04 ----
N2 0.062 ---- ---- ----
Total 0.853 0.50 0.974
Air contain 21% oxygen so
Stoichiometric A/F ratio = 4.062/0.21 = 4.062 by volume
Stoichiometric A/F ratio = 4.062 Ans
(ii)
Mixture which is 20% weak so
Actual A/F ratio will be = 4.062 + (0.2 × 4.062)
= 4.874 by volume
Excess oxygen = (0.21 × 4.874) – 0.853
= 0.1706 kmol/kmol gas
Associated nitrogen = 0.79 × 4.874 = 3.851 kmol/kmol gas
Now nitrogen in the product = 3.851 + 0.062 = 3.913 kmol/kmol gas
Solution for product
Product Kmol/ kmol fuel % by vol (dry) % by vol (wet)
CO2 0.50 10.90 9.0
H2O 0.974 ---- 17.5
O2 0.171 3.72 3.08
N2 3.912 85.4 70.4
Total Total wet= 5.557 100.02 99.98
Total dry= 4.583
THE END
