A four cylinder four stroke petrol engine is to develop 40 KW at 40 rev/second when designed for a volumetric compression ratio of 10:1. The ambient air conditions are 1 bar and 80 C and the calorific value of the fuel is 44 MJ/Kg.

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A four cylinder four stroke petrol engine is to develop 40 KW at 40 rev/second when designed for a volumetric compression ratio of 10:1. The ambient air conditions are 1 bar and 80 C and the calorific value of the fuel is 44 MJ/Kg.

a) Calculate the specific fuel consumption in Kg/MJ of brake work if the indicated overall efficiency is 50% of the corresponding air standard Otto cycle. And the mechanical efficiency is 90%. The specific heat capacity ration for air is 1.4.

b) The required gravimetric air/fuel ratio is 15.4 and the volumetric efficiency is 92%. Estimate the required total swept volume and the cylinder bore if the bore is be equal to the stroke. Calculate also the brake mean effective pressure.

EXPERT ANSWER

In the given question it is given that “indicated overall efficiency is 50%” . There is no such thing like indicated overal efficiency. Indicated efficiency and overall efficiency are different things. Indicated efficiency is ratio of Indicated power to the Total Heat Input. While Overall efficiency is ratio of Shaft Power to the total heat input.

Here in this question it must be Indicated efficiency.

a) Specific fuel consumption is to be calculated

Indicated Efficiency (\eta_{indicated}) is 50% of the air standard Otto (\eta_{otto}) cycle efficiency

Air Std Efficiency = 1- \frac{1}{r^{(\gamma -1)}}

Where r = 10 ( Compression ratio)

\gamma = 1.4 ( Heat capacity ratio).

On substituting

\eta_{otto} = 1- \frac{1}{r^{(\gamma -1)}} = 1- \frac{1}{10^{(1.4 -1)}} = 0.6021
\eta_{indicated} = .5\times \eta_{otto} = .5\times 0.6021= 0.3009

Mechanical Efficiency is 90% or

\eta _{m} = \frac{Break power}{Indicated Power} = 0.9

Brake Power (BP) is given as 40 kW

Indicated Power (IP ) = BP/0.9 = 40/0.9 = 44.44 kW.

Indicated Efficiency is defined as

\eta _{indicated} = \frac{IP}{m_f \times C_v}

Where mf – Mass flow rate of fuel and

Cv – Calorific Value of the fuel.

Substituting in above equation

\eta _{indicated} = \frac{IP}{m_f \times C_v}
0.3009 = \frac{44 kJ/s}{m_f \times 44000 kJ/kg}
m_f = \frac{44 kJ/s}{0.3009\times 44000 kJ/kg}= 0.0033kg/s

Brake Specific fuel consumption bsfc is given by

bsfc = \frac{Mass Flow Rate of Fuel}{Break Power}
bsfc = \frac{0.003 kg/S}{40 kJ/s} = 0.000075 kg /kJ

bsfc = 0.075 kg/MJ(Final Answer)

b)

Gravimetric air/fuel ratio is 15.4

\frac{Mass Flow rate of Air}{Mass Flow Rate of Fuel} = 15.4
{Mass Flow rate of Air} = {Mass Flow Rate of Fuel} \times 15.4
= 0.003 Kg/s \times 15.4 = 0.0462 Kg/s

Volumetric Efficiency is defined as the ratio of actual volume entering the cylinder in a cycle to the swept volume per cycle

\eta _{vol} = \frac{V_a}{V_D}
V_a =\frac{ m_a}{\rho_a}

{\rho_a} is the density of the air in given temperature and pressure

{\rho_a} = \frac{P}{RT}

Where P is the pressure = 1 Bar = 100000 Pa

R is the Gas constant = 287 J/kg K

T is the absolute temperature = 80+273K = 353 k

{\rho_a} = \frac{P}{RT} = \frac{100000Pa}{287KJ/kgK\times 353 K} = 0.987 kg/m^3
V_a =\frac{ m_a}{\rho_a} = \frac{0.0462Kg/s}{0.987 Kg/m^3} = 0.0468 m^3/s

Va is 0.0468 m3/s so Vd should also be in m3/s. There are 40 revoltions per second, which means 20 Suction stroke per second as it is a four stroke engine. So actual displacement voulme per second is VDx20

\eta _{vol} = \frac{V_a}{V_D}
0.92 = \frac{0.0468 m^3/s}{V_D \times 20}

 (Final Answer)

This is a four cylinder engine. So volume of one cylinder is VD/4 and it is a square engine means stroke is equal to its bore volume is given by

\frac{V_D}{4} = \frac{\pi }{4}\times d^2\times d
\frac{0.0025}{4} = \frac{\pi }{4}\times d^3

Stroke = Bore =  ( FInal Answer)

Brake Mean Effective Pressure

bmep = \frac{BP_{per Cycle}}{V_D }

 is the work done per cycle, which is 40kW /20 Cycles.

bmep= \frac {2kJ/s per cycle}{0.0025 m^3} = 800KPa(Final Answer)