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trapecia [35]
3 years ago
14

6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

rature inside the refrigerator is 38°F. Room temperature is 72°F. [answer in BTUs]
Engineering
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}

Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t

Therefore, we have;

Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

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3 years ago
A motor is mounted on a platform that is observed to vibrate excessively at an operating speed of 6000 rpm producing a 250-N for
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Answer:

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Explanation:

given data

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modulus of elasticity E = 29.700 kpsi

we get here

\sigma a = \sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

\sigma a =  \sqrt{(12 \times 1)^2+3\times (0\times 1)^2}  

\sigma a = 12 kpsi

and

\sigma m = \sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}     ...........2

put here value and we get

\sigma m = \sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}  

\sigma m = 17.34 kpsi

now we apply here goodman line equation here that is

\frac{\sigma m}{Sut} +  \frac{\sigma a}{Se} = \frac{1}{FOS}     ...................3

here Se = 0.5 × Sut

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If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement b, no actuator movement c.
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If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement.

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