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2 years ago

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

Engineering

1 answer:

stellarik [79]2 years ago

6 0

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

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A fair die is thrown, What is the probability gained if you are told that 4 will

Semmy [17]

Answer:

1/6

Explanation:

A dice has 6 sides, the probability of 4 appearing is 1/6.

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2 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

State five applications of thermochromic materials

rusak2 [61]

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

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2 years ago

A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving thr

Eva8 [605]

Answer:

Explanation:

It wouldn't work because the wind energy she would be collecting would actually come from the car engine.

The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.

u' = u + v

While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.

Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.

A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.

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3 years ago

The in-situ dry density of a sand is 1.72Mg/m3. The maximum and minimum drydensities, determined by standard laboratory tests, a

Stells [14]

Answer:

Relative density = 0.7 or 70%

Explanation:

The following information was provided by this question

Pd = 1.72mg/mg³

Pd max = 1.81 mg/mg³

Pd min = 1.54 mg/mg³

We substitute into the formula. This formula is contained in the attachment.

[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]

= 0.649350 - 0.581395 / 0.649350 - 0.552486

= 0.067955/0.096864

= 0.7015

= 0.7

The relative density is Therefore 0.7 or 70% when converted to percentage

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2 years ago