Respuesta :
Answer:
i). Compression ratio = 3.678
ii). fuel consumption = 0.4947 kg/hr
Explanation:
Given :
[tex]PV^{1.3}=C[/tex]
Fuel calorific value = 45 MJ/kg
We know, engine efficiency is given by,
[tex]\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}[/tex]
where [tex]r_{c}[/tex] is compression ratio = [tex]\frac{v_{c}+v_{s}}{v_{c}}[/tex]
[tex]r_{c}[/tex] = [tex]1+\frac{v_{s}}{v_{c}}[/tex]
where [tex]v_{c}[/tex] is compression volume
[tex]v_{s}[/tex] is swept volume
Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.
Then, [tex]v_{30}=v_{c}+0.7v_{s}[/tex]
and at 70% compression, 30% of the swept volume remains
∴ [tex]v_{70}=v_{c}+0.3v_{s}[/tex]
We know,
[tex]\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}[/tex]
[tex]\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}[/tex]
[tex]\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\[/tex]
[tex]1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}[/tex]
[tex]v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}[/tex]
[tex]0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}[/tex]
[tex]0.2218v_{s} = 0.594v_{c}[/tex]
[tex]v_{c}=0.3734 v_{s}[/tex]
∴ [tex]r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}[/tex]
Therefore, compression ratio is [tex]r_{c}[/tex] = 3.678
Now efficiency, [tex]\eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}[/tex]
[tex]\eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}[/tex]
[tex]\eta =0.32342[/tex] , this is the ideal efficiency
Therefore actual efficiency, [tex]\eta_{act} =0.5\times \eta _{ideal}[/tex]
[tex]\eta_{act} =0.5\times 0.32342[/tex]
[tex]\eta_{act} =0.1617[/tex]
Therefore total power required = 1 kW x 3600 J
= 3600 kJ
∴ we know efficiency, [tex]\eta=\frac{W_{net}}{Q_{supply}}[/tex]
[tex]Q_{supply}=\frac{W_{net}}{\eta _{act}}[/tex]
[tex]Q_{supply}=\frac{3600}{0.1617}[/tex]
[tex]Q_{supply}=22261.78 kJ[/tex]
Therefore fuel required = [tex]\frac{22261.78}{45000}[/tex]
= 0.4947 kg/hr
