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vova2212 [387]
3 years ago
5

Why does an object under forced convection reach a steady-state faster than an object subjected to free-convection?

Engineering
1 answer:
bonufazy [111]3 years ago
5 0

Answer:

Free convection:

   When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.

Force convection:

   When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.

Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.

We know that convective heat transfer given as

 q = h  A ΔT

h=Heat transfer coefficient

A= Surface area

ΔT = Temperature difference

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Explanation:

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In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Conductors include metal, steel, aluminum, copper, frying pan, pot, spoon etc.

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The waffle slab is: a) the two-way concrete joist framing system. b) a one-way floor and roof framing system. c) the one-way con
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The current in a 20 mH inductor is known to be: 푖푖=40푚푚푚푚푡푡≤0푖푖=푚푚1푒푒−10,000푡푡+푚푚2푒푒−40,000푡푡푚푚푡푡≥0The voltage across the induct
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Answer:

a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.

Explanation:

The question text is corrupted. I found the complete question on the web and it goes as follow:

The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.

a. Find the numerical expressions for i and v for t>0.

b. Specify the time intervals when the inductor is storing energy and is delivering energy.

A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:

i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))

40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)

40*10^(-3) = (A1)*1 + (A2)*1

40*10^(-3) = A1 + A2

A1 + A2 = 40*10^(-3)

Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:

V = L*di/dt

We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:

di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt

di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)

di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)

By applying t = 0 to this expression we have:

di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)

di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0

di/dt (at t = 0) = -10,000*A1- 40,000*A2

We can now use the voltage equation for the inductor at t=0, that is:

v = L di/dt (at t=0)

68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)

68 = -400*A1 -800*A2

-400*A1 - 800*A2 = 68

We now have a system with two equations and two variable, therefore we can solve it for both:

A1 + A2 = 40*10^(-3)

-400*A1 - 800*A2 = 68

Using the first equation we have:

A1 = 40*10^(-3) - A2

We can apply this to the second equation to solve for A2:

-400*[40*10^(-3) - A2] - 800*A2 = 68

-1.6 + 400*A2 - 800*A2 = 68

-1.6 -400*A2 = 68

-400*A2 = 68 + 1.6

A2 = 69.6/400 = 0.174

We use this value of A2 to calculate A1:

A1 = 40*10^(-3) - 0.174 = -0.134

Applying these values on the expression we have the equations for both the current and tension on the inductor:

i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]

v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]

v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.

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