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Svetach [21]
2 years ago
11

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m K), and the wire/sheath interface i

s characterized by a thermal contact resistance of R"t,c= 3x10-4 m2 K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m2 K, and the temperature of the ambient air is 20o C. If the temperature of the insulation may not exceed 50o C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Engineering
1 answer:
Semmy [17]2 years ago
3 0

Answer:

maximum allowable electrical power=4.51W/m

critical radius of the insulation=13mm

Explanation:

Hello!

To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)

Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and  conduction as shown in the attached image

to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient

r=\frac{k}{h} =\frac{0.13}{10}=0.013m=13mm

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74 cycles it’s what u need
7 0
2 years ago
Which of the following is true about modern hydraulic lifts?
kaheart [24]

The modern hydraulic lifts make use of biodegradable fluid to transmit hydraulic power

<em>Question: The options are left out in the question. The details and facts about the modern hydraulic lift are presented here</em>

<em />

Details about the modern hydraulic lifts include;

The development of the  modern hydraulic occurred in the Industrial Revolution to perform task done previously by steam powered elevators  

The power of the hydraulic lift come from the hydraulic cylinder known as the actuator, which in turn is powered by pressurized hydraulic fluid such as oil

The hydraulic fluid is pushed by a piston rod through which energy is capable of being transferred, such that the applied force is multiplied, to provide more power for lifting

<u>Facts about the modern hydraulic lifts include;</u>

  • The dry motor in the modern hydraulic lift is more efficient and consumes 20% less energy
  • It comprises of valves that are controlled electronically such that the response is much rapid and the energy consumption is reduced by a further 20%
  • The cars used in the modern lift are lighter, as well as the slings, which reduces the power usage by 20%
  • It makes use of chemicals which are environmentally friendly as hydraulic fluid
  • The flash point of the fluid used is higher, as well as it posses 50% lower compressibility as well elasticity

Learn more here:

brainly.com/question/16942803

6 0
1 year ago
Two standard spur gears have a diametrical pitch of 10, a center distance 3.5 inches and a velocity ratio of 2.5. How many teeth
lubasha [3.4K]

Answer:50 , 20

Explanation:

Given

Diametrical Pitch\left ( P_D\right )=\frac{T}{D}

where T= no of teeths

D=diameter

module(m) of gears must be same

m=\frac{D}{T}=\frac{1}{P_D}=0.1

Let T_1 & T_2be the gears on two gears

Therefore Center distance is given by

m\frac{\left ( T_1+T_2\right )}{2}=3.5

thus

0.1\frac{\left ( T_1+T_2\right )}{2}=3.5

T_1+T_2=70----1

and Velocity ratio is given by

VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear}

2.5=\frac{T_1}{T_2}----2

From 1 & 2 we get

T_1=50, T_2=20

6 0
3 years ago
When a starter motor begins to turn it produces a high…. *
Georgia [21]

Answer:

The motor produces a high Torque when it begins to start.

8 0
2 years ago
A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, deter
uysha [10]

Answer:

4.83m/s^{2}

Explanation:

For a particle moving in a circular path the resultant  acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by a_{radial}=w^{2}r

Applying values we get  a_{radial}=(2t)^{2}X0.3m

Thus a_{radial}=1.2t^{2}

At time = 2seconds a_{radial}= 4.8m/s^{2}

The tangential acceleration is given by a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}

a_{tangential}=\frac{d(2tr)}{dt}

a_{tangential}= 2r

a_{tangential}=0.6m/s^{2}

Thus the resultant acceleration is given by

a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}

a_{res} =\sqrt{4.8^{2}+0.6^{2}  } =4.83m/s^{2}

8 0
3 years ago
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