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klasskru [66]
3 years ago
7

The performance of a heat pump degrades (i.e., its COP decreases) as the temperature of the heat source decreases. This makes us

ing heat pumps at locations with severe weather conditions unattractive. Consider a house that is heated and maintained at 26°C by a heat pump during the winter. What is the maximum COP for the heat pump if heat is extracted the out door air at 10 C?
Engineering
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

COP_max = 18.69

Explanation:

We are given;

Heated space temperature; T_H = 26°C = 273K + 26 = 299K

Temperature at which heat is extracted; T_L = 10°C = 273 + 10 = 283K

Now the Coefficient of performance (COP) of a heat pump will be a maximum when the heat pump operates in a reversible manner. The COP of a reversible heat pump depends on the temperature limits in the cycle only and is determined by the formula;

COP_max = 1/(1 - (T_L/T_H))

Thus,

COP_max = 1/(1 - (283/299))

COP_max = 1/(1 - 0.9465)

COP_max = 1/0.0535 = 18.69

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Question Set 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.2.2 In
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Answer:

# Program is written in python

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3 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
Oxana [17]

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

(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

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}

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

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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