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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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Applications of fleming hand rule
Eva8 [605]

Answer:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Explanation:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Fleming hand rule is used in the case of electric motors and electric generators.

Fleming hand rule is used to determine the following:

1. Direction of torque

2. Angular velocity

3. Angular acceleration

4 0
3 years ago
Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is
professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.

3 0
3 years ago
Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of
aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid T=500\ K

Einstein Temperature T_E=300\ K

Heat Capacity in the Einstein model is given by

C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}

e^{\frac{3}{5}}=1.822

Substitute the values

C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})

C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}

C_v=0.97\times (3R)            

6 0
3 years ago
A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of
viktelen [127]

Answer:

second-law efficiency  = 62.42 %

Explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C  =  293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same  temp reservoir

so here

efficiency ( reversible ) η1 = 1 - \frac{T2}{T1}      ............1

efficiency ( reversible ) η1  = 1 - \frac{293}{1473}  

so efficiency ( reversible ) η1  = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = \frac{thernal\ efficiency}{0.801}

second-law efficiency = \frac{50}{0.801}  

second-law efficiency  = 62.42 %

3 0
3 years ago
Ma poate ajuta cineva?
kari74 [83]
Da, sigur. cu ce ai nevoie de ajutor?
5 0
3 years ago
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