A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th
e temperature is 2050 K.
Using a cold air-standard analysis, i.e., constant specific heats at 300 K, find the compression ratio, compression specific work, and the highest pressure in the cycle. Repeat your analysis assuming temperature-dependent specific heats and comment on your results.
Using a cold air-standard analysis, i.e., constant specific heats at 300 K, find the compression ratio, compression specific work, and the highest pressure in the cycle. Repeat your analysis assuming temperature-dependent specific heats and comment on your results.
1 answer:
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Answer:
a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
Explanation:
From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.
Part 1
For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
Part 2
For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
And we can quantify the decrease using the relative change:
of reduction
Part 3
A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.