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3 years ago

5

A rigid tank of 1 in3 contains nitrogen gas at 600 kPa, 400 K. By mistake someone lets 0.5 kg flow out. If the final temperature

is 375 K what is then the final pressure?

1 answer:

tigry1 [53]3 years ago

6 0

Answer:

Final Pressure in the tank after letting 0.5 kg out of the tank is 504.6 kPa

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

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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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3 years ago