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LenKa [72]
3 years ago
13

2 Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as

The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow
Engineering
1 answer:
lutik1710 [3]3 years ago
6 0

Answer:

Heat flux on the plate surface is 1.45 x 10⁴ W/m², Heat transfer coefficient is 72.6 W/m². K

Explanation:

<u>The temperature profile of the airflow is given as</u>

<u>T(y) = T∞ – (T∞ - Ts)exp[(-V/</u>α<u> fluid)y]</u>

<u>Calculate the mean film temperature of air</u>

T(f) = T(s) + T∞/2, where T(s) is the surface temperature of the plate and T∞ is the temperature of free stream

Substitute Ts = 220 C and T∞ = 20 C

T(f) = 220 +20/2 = 120 C

Obtain the properties of air from Table A- 15 (Properties of air at 1 atm pressure) in the text book of from the internet at T(f) = 120 C, which would give us the quantities of thermal diffusivity (α fluid) and the thermal conductivity (k) of air

α fluid = 3.565 x 10⁻⁵ m²/s

k = 0.03235 W/m.k

<u>Calculate the temperature gradient for the given profile</u>

T(y) = T∞ – (T∞ - Ts)exp({-V/α fluid}y)

Where, T∞ is the ambient temperate, V is the free stream velocity of air and y is the temperature displacement thickness

<u>Differentiate the expression with respect to y</u>

dT/dY = - (T∞ - Ts){-V/α fluid}e[(-V/α fluid) x 0]

dT/dy = -(T∞ - Ts){-V/α fluid}

<u>Substitute T∞ = 20 C, Ts = 220 C, V = 0.08 m/s and </u>α<u> = 3.565 x 10⁻⁵ m²/s</u>

dT/dy = - (20 - 220)(-0.08/3.565 x 10⁻⁵)

dT/dy = -448807.8541

<u>Calculate the heat flux on the plate surface</u>

Q = -kdT/dy

K = 0.03235 W/m.k, dT/dy = -448807.8541

Q = -(0.03235)(-448807.8541)

Q = 1.45 x 10⁴ W/m²

Therefore, the heat flux on the plate surface is 1.45 x 10⁴ W/m²

<u>Calculate the heat transfer coefficient</u>

Q = h(Ts - T∞), where h is the transfer coefficient

1.45 x 10⁴ = h(220 - 20)

H = 72.6 W/m2.k

Therefore, the heat transfer coefficient is 72.6 W/m². K

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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)

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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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3 years ago
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