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3 years ago

The input power for a thermostat is wired to the

Engineering

1 answer:

masya89 [10]3 years ago

6 0

Answer:

A. R

Explanation:

There are basically two wires that supply input power to the thermostat namely the C wire which is the common wire and the R wire .The G wire simply completes the input power circuit from the R-leg of the power supply.On the another hand the Y wire also completes the circuit from the compressor fan contactor to the R-leg of the power supply . While the W wire completes the path to the heater contactor coil.

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4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is

stellarik [79]

Answer:

5.833

Explanation:

Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.

$COP=\frac {RE}{P}$ where RE is refrigeration effect and P is power input

Here, the power input is given as 30 kW

We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW

Now the $COP=\frac {175}{30}=5.833$

6 0

2 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago

Gtjffs

grandymaker [24]

the required documents is 3000

4 0

2 years ago

Given asphalt content test data:

VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535, standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

5 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