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velikii [3]
3 years ago
6

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarterof th

e time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is(a) $3.56 (b) 55.18 (a 53.54 {0159.26 [6) $20.74
Engineering
1 answer:
ryzh [129]3 years ago
4 0

Answer:

$5.184

Explanation:

The cost can be calculated using the formula: Cost = Load \ factor \times Number \ of \ hours \ \\M_{month} = M_{units} \times W\\

Before using this, we require the following conversions:

<em>320 W → kW:</em>

\frac {320}{1000} = 0.32

<em>30 Days → Hours:</em>

30 \times 24 = 720

Using the above stated formula:

M_{month} = 0.09 \times 0.32 \times \frac{1}{4} \times 720 = 5.184

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The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin
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Answer:

Ф = 0.02838 ft

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2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

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Ф = ∝  × ΔT  × L  ... where Ф= Gap Delta in ft

                                          ΔT= Temperature rise in F

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Ф =  28,380 × 10^(-6) ft

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Stress induced in rails is given by,

   σ     =  ∝  × ΔT  × E

          =  6.45 ×10^(-6)   × 110  × 200

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Now, let's find axial force in rails,

Here,we have to consider  ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

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