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azamat
4 years ago
9

Evaluate the performance of the proposed heat pump for three locations Using R134a. Discuss the effect of outdoor temperature on

the performance of the heat pump. What happens to the COP if the heat exchangers are only 80% effective, Philadelphia only. Discuss improvements to the design for Philadelphia that will increase the COP closer to the theoretical max (you must show supporting numbers). Your report should include tables that list the information at each state, the evaluated work, heat transfer, and calculated COP values. Sample calculations of your work should be included in an appendix.
Engineering
1 answer:
Phoenix [80]4 years ago
4 0

Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).

1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42

.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43

-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82

A paired t-test is the standard procedure for testing this null hypothesis.

We use a paired t-test because each worker was measured twice, once for Paired t-test for

each workplace, so the observations on the two workplaces are dependent. paired data

Fast workers are probably fast for both workplaces, and slow workers are

slow for both. Thus what we do is compute the difference (standard − er-

gonomic) for each worker, and test the null hypothesis that the average of

these differences is zero using a one sample t-test on the differences.

Table 2.2 gives the differences between standard and ergonomic times.

Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-

ences in the sample. We assume that these differences are independent sam-

ples from a normal distribution with mean µ and variance σ

2

, both unknown.

Our null hypothesis is that the mean µ equals prespecified value µ0 = 0

(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the

workers to be faster in the ergonomic workplace.

The formula for a one sample t-test is

t =

¯d − µ0

s/√

n

,

where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test

sample size, and s is the sample standard deviation (of the differences)

s =

vuut

1

n − 1

Xn

i=1

(di − ¯d )

2 .

If our null hypothesis is correct and our assumptions are true, then the t-

statistic follows a t-distribution with n − 1 degrees of freedom.

The p-value for a test is the probability, assuming that the null hypothesis

is true, of observing a test statistic as extreme or more extreme than the one The p-value

we did observe. “Extreme” means away from the the null hypothesis towards

the alternative hypothesis. Our alternative here is that the true average is

larger than the null hypothesis value, so larger values of the test statistic are

extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of

freedom from the observed t-value to the right. (If the alternative had been

µ < µ0, then the p-value is the area under the curve to the left of our test

Explanation: The curve represents the sum total of the evaluation

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Q5
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The C++ code that would draw all the iterations in the selection sort process on the array is given below:

<h3>C++ Code</h3>

#include <stdio.h>

#include <stdlib.h>

int main() {

   int i, temp1, temp2;

   int string2[16] = { 0, 4, 2, 5, 1, 5, 6, 2, 6, 89, 21, 32, 31, 5, 32, 12 };

   _Bool check = 1;

   while (check) {

       temp1 = string2[i];

       temp2 = string2[i + 1];

       if (temp1 < temp2) {

           string2[i + 1] = temp1;

           string2[i] = temp2;

           i = 0;

       } else {

           i++;

           if (i = 15) {

               check = !check;

           }

       }

   }

   

   return 0;

}

Read more about C++ programming here:

brainly.com/question/20339175

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