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krok68 [10]
3 years ago
15

An aluminum electrical cable is 20 mm in diameter is covered by a plastic insulation (k = 1 W/m-k) of critical thickness. This w

ire is placed in a room with an air flow heat transfer coefficient of 50 W/m^2-K. Compared to the bare aluminum wire, the heat loss from this insulated wire will be a) LESS b) MORE c) SAME d) ZERO
Engineering
1 answer:
Anastaziya [24]3 years ago
5 0

Answer:

the heat loss from this insulated wire is less

Explanation:

Given data in question

diameter of cable (d)  =  20 mm

( K ) = 1 W/m-k

heat transfer coefficient (h) = 50 W/m²-K

To find out

the heat loss from this insulated wire

solution

we will find out thickness of wire

heat loss is depend on wire thickness also

we have given dia 20 mm

so radius will be d/2 = 20/ 2 = 10 mm

Now we find the critical thickness i.e.

critical thickness = K / heat transfer coefficient

critical thickness = 1 / 50 = 0.02 m i.e. 20 mm

now we can see that critical thickness is greater than radius 10 mm

so our rate of heat loss will be decreasing

so we can say our correct option is (a) less

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A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det
Vinil7 [7]

Answer:

a) T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C), b) \dot Q_{H} = 26.875\,kW

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

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

After some algebraic handling, the temperature of the cold reservoir is determined:

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

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)

T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)

b) The heating load provided by the heat pump is:

\dot Q_{H} = COP_{HP}\cdot \dot W

\dot Q_{H} = (12.5)\cdot (2.15\,kW)

\dot Q_{H} = 26.875\,kW

4 0
3 years ago
Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th
jek_recluse [69]

Answer:

False

Explanation:

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A 3-in-thick slab is 10 in wide and 15 ft long. The thickness of the slab is reduced by 20% and width increases by 3% in a hot-r
xxMikexx [17]

Answer: l = 2142.8575 ft

v = 193.99 ft/min.

Explanation:

Given data:

Thickness of the slab = 3in

Length of the slab = 15ft

Width of the slab = 10in

Speed of the slab = 40ft/min

Solution:

a. After three phase

three phase = (0.2)(0.2)(0.2)(3.0)

= 0.024in.

wf = (1.03)(1.03)(1.03)(10.0)

= 10.927 in.

Using constant volume formula

= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf

Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)

= 6750 /0.2625

= 25714.28in = 2142.8575 ft

b.

vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)

= (0.12)(424.36)/0.2625

= 50.9232/0.2625

= 193.99 ft/min.

4 0
3 years ago
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