It has to do with mechanical engineering
Answer:
i) SF:
ii) BM : ![= \frac{(w_0*x)^3}{6L}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B%28w_0%2Ax%29%5E3%7D%7B6L%7D%20)
Explanation:
Let's take,
Making y the subject of formula, we have :
![y = \frac{x}{L} * w_0](https://tex.z-dn.net/?f=%20y%20%3D%20%5Cfrac%7Bx%7D%7BL%7D%20%2A%20w_0%20)
For shear force (SF), we have:
This is the area of the diagram.
![v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0](https://tex.z-dn.net/?f=%20v%28x%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20y%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%5Cfrac%7Bx%7D%7BL%7D%20%2A%20w_0)
The shear force equation =
For bending moment (BM):
![BM = v(x) * \frac{x}{3}](https://tex.z-dn.net/?f=%20BM%20%3D%20v%28x%29%20%2A%20%5Cfrac%7Bx%7D%7B3%7D%20)
![= \frac{(w_0*x)^3}{6L}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B%28w_0%2Ax%29%5E3%7D%7B6L%7D%20)
The bending moment equation =
![= \frac{(w_0*x)^3}{6L}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B%28w_0%2Ax%29%5E3%7D%7B6L%7D%20)
extension lines,sketches,leader lines,dimensions describes all illustrations created by freehand.
Answer:
1. ![\dot Q=19600\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D19600%5C%20W)
2. ![\dot Q=120\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D120%5C%20W)
Explanation:
1.
Given:
- height of the window pane,
![h=2\ m](https://tex.z-dn.net/?f=h%3D2%5C%20m)
- width of the window pane,
![w=1\ m](https://tex.z-dn.net/?f=w%3D1%5C%20m)
- thickness of the pane,
![t=5\ mm= 0.005\ m](https://tex.z-dn.net/?f=t%3D5%5C%20mm%3D%200.005%5C%20m)
- thermal conductivity of the glass pane,
![k_g=1.4\ W.m^{-1}.K^{-1}](https://tex.z-dn.net/?f=k_g%3D1.4%5C%20W.m%5E%7B-1%7D.K%5E%7B-1%7D)
- temperature of the inner surface,
![T_i=15^{\circ}C](https://tex.z-dn.net/?f=T_i%3D15%5E%7B%5Ccirc%7DC)
- temperature of the outer surface,
![T_o=-20^{\circ}C](https://tex.z-dn.net/?f=T_o%3D-20%5E%7B%5Ccirc%7DC)
<u>According to the Fourier's law the rate of heat transfer is given as:</u>
![\dot Q=k_g.A.\frac{dT}{dx}](https://tex.z-dn.net/?f=%5Cdot%20Q%3Dk_g.A.%5Cfrac%7BdT%7D%7Bdx%7D)
here:
A = area through which the heat transfer occurs = ![2\times 1=2\ m^2](https://tex.z-dn.net/?f=2%5Ctimes%201%3D2%5C%20m%5E2)
dT = temperature difference across the thickness of the surface = ![35^{\circ}C](https://tex.z-dn.net/?f=35%5E%7B%5Ccirc%7DC)
dx = t = thickness normal to the surface = ![0.005\ m](https://tex.z-dn.net/?f=0.005%5C%20m)
![\dot Q=1.4\times 2\times \frac{35}{0.005}](https://tex.z-dn.net/?f=%5Cdot%20Q%3D1.4%5Ctimes%202%5Ctimes%20%5Cfrac%7B35%7D%7B0.005%7D)
![\dot Q=19600\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D19600%5C%20W)
2.
- air spacing between two glass panes,
![dx=0.01\ m](https://tex.z-dn.net/?f=dx%3D0.01%5C%20m)
- area of each glass pane,
![A=2\times 1=2\ m^2](https://tex.z-dn.net/?f=A%3D2%5Ctimes%201%3D2%5C%20m%5E2)
- thermal conductivity of air,
![k_a=0.024\ W.m^{-1}.K^{-1}](https://tex.z-dn.net/?f=k_a%3D0.024%5C%20W.m%5E%7B-1%7D.K%5E%7B-1%7D)
- temperature difference between the surfaces,
![dT=25^{\circ}C](https://tex.z-dn.net/?f=dT%3D25%5E%7B%5Ccirc%7DC)
<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>
![\dot Q=k_a.A.\frac{dT}{dx}](https://tex.z-dn.net/?f=%5Cdot%20Q%3Dk_a.A.%5Cfrac%7BdT%7D%7Bdx%7D)
![\dot Q=0.024\times 2\times \frac{25}{0.01}](https://tex.z-dn.net/?f=%5Cdot%20Q%3D0.024%5Ctimes%202%5Ctimes%20%5Cfrac%7B25%7D%7B0.01%7D)
![\dot Q=120\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D120%5C%20W)