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enyata [817]
3 years ago
14

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons

tant. The pressure developed in the cylinder during 30% and 70% of the compression stroke is 1.5 bar and 2.75 bar, respectively. The relative efficiency of the engine with respect to the air standard efficiency is 50%. Determine the (i) compression ratio (ii) fuel consumption per hour per kW of power generated
Engineering
1 answer:
antoniya [11.8K]3 years ago
5 0

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given  :

PV^{1.3}=C

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}

where r_{c} is compression ratio = \frac{v_{c}+v_{s}}{v_{c}}

           r_{c} = 1+\frac{v_{s}}{v_{c}}

where v_{c} is compression volume

           v_{s} is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, v_{30}=v_{c}+0.7v_{s}

and at 70% compression, 30% of the swept volume remains

    ∴    v_{70}=v_{c}+0.3v_{s}  

We know,

\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}

\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}

\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\

1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}

v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}

0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}

0.2218v_{s} = 0.594v_{c}

v_{c}=0.3734 v_{s}

∴   r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}

Therefore, compression ratio is r_{c} = 3.678

Now efficiency, \eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}

 \eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}

 \eta =0.32342 , this is the ideal efficiency

Therefore actual efficiency, \eta_{act} =0.5\times \eta _{ideal}

           \eta_{act} =0.5\times 0.32342

           \eta_{act} =0.1617

Therefore total power required = 1 kW x 3600 J

                                                    = 3600 kJ

∴ we know efficiency, \eta=\frac{W_{net}}{Q_{supply}}

Q_{supply}=\frac{W_{net}}{\eta _{act}}

Q_{supply}=\frac{3600}{0.1617}

Q_{supply}=22261.78 kJ

Therefore fuel required = \frac{22261.78}{45000}

                                        = 0.4947 kg/hr      

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