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BaLLatris [955]
3 years ago
11

A building wall has dimensions of 3 m tall and 10 m wide. It is constructed of 2 cm. wallboard (k = 0.5 W/m-C) on the inside, 3

cm. of loose fill mineral wool (k = 0.052 W/m-C), and 10 cm. thick brick (0.7 W/m-C). The inside room temperature is 22 C and the outside air is -20 C. The inside heat transfer coefficient is 8 W/m2-C and the outside coefficient is 32 W/m2- C. a. Determine the wall heat loss (W). b. Draw the temperature profile from the inside room to the outside. Be qualitative in your drawing, no numbers necessary except for the inside and outside temperature.

Engineering
1 answer:
Art [367]3 years ago
4 0

Answer: heat loss through wall is 16.58034kW

Temperature of inside wall surface is 47°c

Temperature of outside wall surface is -2.7°c

Explanation:detailed calculation and explanation is shown in the image below.

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a) V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}

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Using the principle of conservation of momentum:

M_AU_A + M_BU_B = M_AV_A + M_BV_B............(1)

Coefficient of restitution, e = \frac{V_B - V_A}{U_A - U_B}

V_A = V_B - e(U_A - U_B)........................(2)

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M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B..............(3)

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V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

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Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

b)

M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

Rod A is said to be at rest after the impact, V_A = 0 m/s

Substitute these parameters into equation (7)

0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s

To calculate the final velocity, V_B, substitute the given parameters into (4):

V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s

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I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

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d) %\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%

%\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%

% \triangle KE = -47.85 \%

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