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azamat
3 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]3 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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Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the
Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0
3 years ago
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0
3 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
2 years ago
Visual perception is a mental process that is non selective<br> True<br> False
NARA [144]

The statement "Visual perception is a mental process that is non selective" is false, it is a psychic function that allows the organism to capture, elaborate and interpret selective information that comes from the environment.

<h2>What is visual perception?</h2>

Visual perception is that inner sensation of apparent knowledge, resulting from a specific stimulus or light impression recorded by the eyes.

<h3>Characteristics of visual perception</h3>

  • It incorporates the sensory stimuli received from objects, situations or events and converts them into a meaningful interpretation experience.

  • It is an active process of the brain through which an external reality is created by transforming the light information captured by the eye.

Therefore, we can conclude that visual perception is the interpretation made by the brain of the different organisms of the stimuli received through the senses.

Learn more about visual perception here: brainly.com/question/10259599

3 0
2 years ago
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