Answer:
i). Compression ratio = 3.678
ii). fuel consumption = 0.4947 kg/hr
Explanation:
Given :
![PV^{1.3}=C](https://tex.z-dn.net/?f=PV%5E%7B1.3%7D%3DC)
Fuel calorific value = 45 MJ/kg
We know, engine efficiency is given by,
![\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}](https://tex.z-dn.net/?f=%5Ceta%20%3D%201-%5Cleft%20%28%20%5Cfrac%7B1%7D%7Br_%7Bc%7D%7D%20%5Cright%20%29%5E%7B1.3-1%7D)
where
is compression ratio = ![\frac{v_{c}+v_{s}}{v_{c}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bc%7D%2Bv_%7Bs%7D%7D%7Bv_%7Bc%7D%7D)
= ![1+\frac{v_{s}}{v_{c}}](https://tex.z-dn.net/?f=1%2B%5Cfrac%7Bv_%7Bs%7D%7D%7Bv_%7Bc%7D%7D)
where
is compression volume
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}](https://tex.z-dn.net/?f=v_%7B30%7D%3Dv_%7Bc%7D%2B0.7v_%7Bs%7D)
and at 70% compression, 30% of the swept volume remains
∴
We know,
![\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%3D%5Cleft%20%28%20%5Cfrac%7BV_%7B1%7D%7D%7BV_%7B2%7D%7D%20%5Cright%20%29%5E%7Bn%7D)
![\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}](https://tex.z-dn.net/?f=%5Cfrac%7B2.75%7D%7B1.5%7D%3D%5Cleft%20%28%20%5Cfrac%7Bv_%7Bc%7D%2B0.7%5Ctimes%20v_%7Bs%7D%7D%7Bv_%7Bc%7D%2B0.3%5Ctimes%20v_%7Bs%7D%7D%20%5Cright%20%29%5E%7B1.3%7D)
![\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\](https://tex.z-dn.net/?f=%5Cleft%20%28%201.833%20%5Cright%20%29%5E%7B%5Cfrac%7B1%7D%7B1.3%7D%7D%3D%5Cfrac%7Bv_%7Bc%7D%2B0.7v_%7Bs%7D%7D%7Bv_%7Bc%7D%2B0.3v_%7Bs%7D%7D%5C%5C)
![1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}](https://tex.z-dn.net/?f=1.594%3D%5Cfrac%7Bv_%7Bc%7D%2B0.7v_%7Bs%7D%7D%7Bv_%7Bc%7D%2B0.3v_%7Bs%7D%7D)
![v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}](https://tex.z-dn.net/?f=v_%7Bc%7D%2B0.7v_%7Bs%7D%3D1.594v_%7Bc%7D%2B0.4782v_%7Bs%7D)
![0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}](https://tex.z-dn.net/?f=0.7v_%7Bs%7D-0.4782v_%7Bs%7D%3D1.594v_%7Bc%7D-v_%7Bc%7D)
![0.2218v_{s} = 0.594v_{c}](https://tex.z-dn.net/?f=0.2218v_%7Bs%7D%20%3D%200.594v_%7Bc%7D)
![v_{c}=0.3734 v_{s}](https://tex.z-dn.net/?f=v_%7Bc%7D%3D0.3734%20v_%7Bs%7D)
∴ ![r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}](https://tex.z-dn.net/?f=r_%7Bc%7D%3D%201%2B%5Cfrac%7Bv_%7Bs%7D%7D%7B0.3734v_%7Bs%7D%7D)
Therefore, compression ratio is
= 3.678
Now efficiency, ![\eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}](https://tex.z-dn.net/?f=%5Ceta%20%3D%5Cleft%20%28%201-%5Cfrac%7B1%7D%7Br_%7Bc%7D%7D%20%5Cright%20%29%5E%7B0.3%7D)
![\eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}](https://tex.z-dn.net/?f=%5Ceta%20%3D%5Cleft%20%28%201-%5Cfrac%7B1%7D%7B3.678%7D%20%5Cright%20%29%5E%7B0.3%7D)
, this is the ideal efficiency
Therefore actual efficiency, ![\eta_{act} =0.5\times \eta _{ideal}](https://tex.z-dn.net/?f=%5Ceta_%7Bact%7D%20%3D0.5%5Ctimes%20%5Ceta%20_%7Bideal%7D)
![\eta_{act} =0.5\times 0.32342](https://tex.z-dn.net/?f=%5Ceta_%7Bact%7D%20%3D0.5%5Ctimes%200.32342)
![\eta_{act} =0.1617](https://tex.z-dn.net/?f=%5Ceta_%7Bact%7D%20%3D0.1617)
Therefore total power required = 1 kW x 3600 J
= 3600 kJ
∴ we know efficiency, ![\eta=\frac{W_{net}}{Q_{supply}}](https://tex.z-dn.net/?f=%5Ceta%3D%5Cfrac%7BW_%7Bnet%7D%7D%7BQ_%7Bsupply%7D%7D)
![Q_{supply}=\frac{W_{net}}{\eta _{act}}](https://tex.z-dn.net/?f=Q_%7Bsupply%7D%3D%5Cfrac%7BW_%7Bnet%7D%7D%7B%5Ceta%20_%7Bact%7D%7D)
![Q_{supply}=\frac{3600}{0.1617}](https://tex.z-dn.net/?f=Q_%7Bsupply%7D%3D%5Cfrac%7B3600%7D%7B0.1617%7D)
![Q_{supply}=22261.78 kJ](https://tex.z-dn.net/?f=Q_%7Bsupply%7D%3D22261.78%20kJ)
Therefore fuel required = ![\frac{22261.78}{45000}](https://tex.z-dn.net/?f=%5Cfrac%7B22261.78%7D%7B45000%7D)
= 0.4947 kg/hr