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Fofino [41]
3 years ago
13

A refrigerator operates on average for 10.0 hours an day. If the power rating is the refrigerator is 709 w how much electrical e

nergy does the fridge use in one day
Engineering
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

61257.6 kW per day

Explanation:

The rating of the refrigerator is given as : 709 w

This means the power consumption is 709 joules per second

The operating time is given as 10 hours.

Change hours to second as

1 hour = 3600 seconds

10 hours = 36000 seconds

Apply the rating as;

709 w = 1 s

?         = 36000 seconds

perform cross product to get;

709 * 36000= 25524000 w

25524000 = 25524kW

For a day, 24 hours will be;

25524 kW = 10 hours

?                 =24 hours

={24*25524}/10 = 61257.6 kW per day

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Answer:

0.008

Explanation:

From the question, the parameters given are:

Velocity V = 5 m/s

Pressure = 10 pa

But pressure = F/A

10 = F/A

F = 10A

Substitute all the parameters into the formula below

Coefficient of viscosity (η) = F × r /[AV]

Where

F = tangential force,

r = distance between layers,

A = Area, and

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The A will cancel out

(η) = 10 × 0.004 /[5]

(η) = 0.04 /5

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Answer:

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In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75°C as it moves at 0.2 m/s through a
stepan [7]

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density \rho = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

m = pAV

m = \rho \dfrac{\pi}{4}D^2V

m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2

m = 0.0253 \ kg/s

To estimate the amount of the heat by using the expression:

q = mc_p(T_{final}-T_{initial})

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

q" = \dfrac{q}{A_s}

q" = \dfrac{q}{\pi D L}

q" = \dfrac{5060}{\pi \times 0.0127 \times 10}

q" =  12682.268 W/m²

The required heat flux = 12682.268 W/m²

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