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Snezhnost [94]
3 years ago
15

An inventor claims to have developed a food freezer that at steady state requires a power input of 0.25 kW to extract energy by

heat transfer at a rate of 3000 J/s from freezer contents at 270 K. Investigate this claim for an ambient temperature of 293 K.
Engineering
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

The claim is false. (COP_{real} > COP_{ideal}).

Explanation:

The real coefficient of performance of the food freezer is:

COP_{real} = \frac{\dot Q_{L}}{\dot W}

COP_{real} = \frac{3\,kW}{0.25\,kW}

COP_{real} = 12

The ideal coefficient of performance, that is, when freezer has a reversible process, is:

COP_{ideal} = \frac{T_{L}}{T_{H}-T_{L}}

COP_{ideal} = \frac{270\,K}{293\,K-270\,K}

COP_{ideal} = 11.739

A real freezer has a coefficient of performance lesser than or equal to ideal coefficient of performance. Since supposed real coefficient of performance is greater than ideal coefficient of performance. The claim is proved to be false.

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Verify if 83 is a term in the arithmetic sequence 11,15,19,23
EleoNora [17]

Answer:

yes, it is

Explanation:

The sequence: (+4)

23,27,31,35,39,43,47,51,55,59,63,67,71,75,79,83

Hope this helps! :)

3 0
3 years ago
Consider the following grooves, each of width W, that have been machined from a solid block of material. (a) For each case obtai
kogti [31]

Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

8 0
3 years ago
In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0
3 years ago
1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0
3 years ago
64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump eff
grin007 [14]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

5 0
3 years ago
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