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Pavlova-9 [17]
2 years ago
11

A composite plane wall consists of a 5-in.-thick layer of insulation (ks = 0.029 Btu/h*ft*°R) and a 0.75-in.-thick layer of sidi

ng (ks = 0.058 Btu/h*ft*°R). The inner temperature of the insulation is 67°F. The outer temperature of the siding is -8°F.
Determine at steady state:


(a) the temperature at the interface of the two layers, in °F, and (b) the rate of heat transfer through the wall in Btu/h*ft^2 of surface area.
Engineering
1 answer:
Leno4ka [110]2 years ago
3 0

We have that  the <em>temperature </em>at the <em>interface </em>of the two <em>layers</em>, in °F, and (b) the rate of heat transfer through the wall in Btu/h*ft^2 of surface area is given as

a)    T_i-12.11^oF

b) H_r=3.80 btu/hft^2\\\\

From the question we are told

A composite plane wall cons<em>i</em>sts of a 5-in.-thick layer of insulation (ks = 0.029 Btu/h*ft*°R) and a 0.75-in.-thick layer of siding (ks = 0.058 Btu/h*ft*°R).

The inner temperature of the insulation is 67°F. The outer temperature of the siding is -8°F.

<h3>Interface Temperature</h3>

Generally the equation for the Terminal resistance   is mathematically given as

R_1=\frac{t_1}{k_1A}\\\\R_1=14.3667hft^2R/btu\\\\Also\\\\R_2=\frac{t_2}{k_2A}\\\\R_2=1.077hft^2R/btu\\\\

b)

Heat transfer rate\\\\H_r=\frac{67-8}{15.444}\\\\H_r=3.80 btu/hft^2\\\\

a)Interface Temp(T_i)

T_i=\frac{67-t_2}{R1}=3.820

For more information on Temperature visit

brainly.com/question/15267055

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A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig
jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

             Section: 9-ft section of pipe.

                     Sum of forces perpendicular to member AC = 0

                     F_d - 0.8*W*L = 0

                     F_d = 0.8*10*9 = 72 lb

                     Sum of forces perpendicular to member BC = 0

                     F_e - 0.6*W*L = 0

                     F_e = 0.6*10*9 = 54 lb

              F_d = 72 lb ,  F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

              Section: Entire Frame.

                    Sum of moments about point B = 0

                    -A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

                    -A_y*(1.5625) + 15 + 39.375 = 0

                    A_y = 34.8 lb  

                   Sum of forces in vertical direction = 0

                     A_y + B_y - 0.8*F_d - 0.6*F_e = 0

                     B_y = 0.8*(72) + 0.6*(54) - 34.8

                     B_y = 55.2 lb  

                   Sum of forces in horizontal direction = 0

                     A_x + B_x - 0.6*F_d + 0.8*F_e = 0

                     A_x + B_x = 0

               Section: Member AC

                    Sum of moments about point C = 0

                     F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

                     72*2.5 - 34.8*12 - 9*A_x = 0

                     A_x = -237.6 / 9 = - 26.4 lb

                     B_x = - A_x = 26.4 lb

                     A_x = -26.4 lb  ,  B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

               Member AC:

                    From point A till just before point D

                     -0.6*A_x*x - A_y*0.8*x + M = 0

                     15.84*x - 27.84*x + M = 0

                      M = 12*x   ..... max value at D, x = 12.25 in

                      M_max = 12*12.25/12 = 12.25 lb-ft

               Member BC:

                    From point B till just before point E

                     -0.8*B_x*x + B_y*0.6*x + M = 0

                     -21.12*x + 33.12*x + M = 0

                      M = -12*x   ..... max value at E, x = 11.25 - 2.5 = 8.75 in

                      M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

                      AC: at D , M_max = 12.25 lb-ft

                      BC: at E , M_max = 8.75 lb-ft

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