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3 years ago

Which type of load generates a magnetic field?

Engineering

1 answer:

kozerog [31]3 years ago

5 0

Answer:

inductive loads

Explanation:

Inductive: When a changing current passes through an inductor, it induces a magnetic field around itself. Turning the inductor into a coil increases the magnetic field. Examples of inductive loads include transformers, fans, electric motors and coils.

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A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

Viefleur [7K]

Answer:

Both come from the sun

Both are reusable sources

and both don't cause pollution

Explanation:

3 0

2 years ago

Can someone pls give me the answer to this?

olganol [36]

I think option c 12 is currect

7 0

3 years ago

Which two pieces of evidence best support the statement: “Germs do not always cause disease”?

Ratling [72]

Answer:

1) "Most germs are too small to cause disease" http://ecufrog.weebly.com/uploads/8/6/7/7/8677638/850_meet_the_microbes.pdf

2) "Some bacteria that live in your body are helpful. For instance, Lactobacillus acidophilus... helps you digest food, destroys some disease-causing organisms and provides nutrients." https://www.mayoclinic.org/diseases-conditions/infectious-diseases/in-depth/germs/art-20045289

8 0

3 years ago

Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2

tresset_1 [31]

1.i am superman

2.175 175 175

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8 0

2 years ago