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lara [203]
2 years ago
14

A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1

.3, find the final temperature. charge of air at 1 bar & 20°C and a) T2= 1093 K b) T2=151.8 K c) T2=983.6 K d) T2 =710.9 K e) T2= 424.8 K
Engineering
1 answer:
Gelneren [198K]2 years ago
5 0

Answer:

final temperature is 424.8 K

so correct option is e 424.8 K

Explanation:

given data

pressure p1 = 1 bar

pressure p2 = 5 bar

index k = 1.3

temperature t1 = 20°C = 293 k

to find out

final temperature  t2

solution

we have given compression is reversible and has an index k

so we can say temperature is

\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }  ...........1

put here all these value and we get t2

\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }

t2 = 424.8

final temperature is 424.8 K

so correct option is e

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2 years ago
A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
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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

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And since the new total time would be given by 250-14=236 s

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 (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

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For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

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

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

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 (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

And we can quantify the decrease using the relative change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% 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.

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2 years ago
A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?
stiks02 [169]

Answer:

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Explanation:

Given

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