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Problems Ice

The document provides data from a trial of a 4 cylinder gas engine including duration, revolutions, missed cycles, net load, mean effective pressure, gas consumption, LCV of gas supply, cylinder diameter, stroke, effective brake circumference, and compression ratio. It then shows the calculations to determine the indicated power, brake power, mechanical efficiency, and indicated thermal efficiency.

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Sriramulu Jaichandar
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40 views2 pages

Problems Ice

The document provides data from a trial of a 4 cylinder gas engine including duration, revolutions, missed cycles, net load, mean effective pressure, gas consumption, LCV of gas supply, cylinder diameter, stroke, effective brake circumference, and compression ratio. It then shows the calculations to determine the indicated power, brake power, mechanical efficiency, and indicated thermal efficiency.

Uploaded by

Sriramulu Jaichandar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as ODT, PDF, TXT or read online on Scribd
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19 b ANS

The following data obtained in trial of 4 cylinder gas engine

Duration of trial = 1 hr

Revolution = 14000

No. of missed cycle = 500

Net load = 1470N

Mean eff. Pressure = 7.5 bar

Gas consumption = 20000 lit

LCV of gas supply condition = 21 KJ/liters

Cylinder Diameter = 250mm

Stroke = 450mm

Effective break circumference = 4m

Compression ratio = 6.5

Determine

1. Indicated power
2. brake power
3. mechanical effeciency
4. indicated thermal effeciency
Solution:-
NW = N = 1400 = 700 rpm
2 2
Missed cycle = 500
NW=700-500=200 rpm

to find out I.P ¿ p m   L A N W = 100 * 7.5 * 0.45 * 3.14 * 0.25 * 200 =220.78 kw
2

60000 60

to find out B.P ¿ ( W −S ) π D N = 1470 * π * 4 * 400 =123.154kw


60000 60000

To find mechanical effeciency ηm ¿ B P =55.78%


IP

Indicated Thermal Efficiency ηith ¿ IP = 220.78 = 18.94%


E ne r g y  i n  f u el   per   s ec ond 0.55* 21000
17 b ANS

Given,

Mass of the air = 6 kg/min = 0.1 kg/sec


Mass of the fuel = 0.4 kg/min
Density of the fuel = 770 kg/m3
Initial Pressure P1= 1 bar
Initial Temperature T1 =290 K
Velocity of air at the throat C2 = 100 m/sec
Cda = 0.84
Cdf = 0.65
Ratio of specific heat = 1.4
Pressure drop across the fuel metering orifice is 0.85 of the pressure at the throat
We know that,

√
r−1
P2
C2 = ( 1− )
P1
2C P T 1
r

√
P 2 1.4 −1
100 = (1− )
1
2* 1005* 290
1.4
P2 = 0.941232 bar
mf actual = Cdf Af
√ 2ρ f (P1−P2)
0.4 = 0.65* Af
60 √ 2* 770* ( 1−0.941232 ) * 105
Af = 3.40 *10-6 m2
π df2 = 3.40 *10-6
4
df = 2.08 mm

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