15x -/c/ fb is the answer
Answer:
b. A view of a building seen from one side, a flat representation of one façade. This is the most common view used to describe the external appearance of a building.
Explanation:
An elevation is a three-dimensional, orthographic, architectural projection that reveals just a side of the building. It is represented with diagrams and shadows are used to create the effect of a three-dimensional image.
It reveals the position of the building from ground-depth and only the outer parts of the structure are illustrated. Elevations, building plans, and section drawings are always drawn together by the architects.
Answer:
,
, ![\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdx%7D%20%3D%20-v_%7Bin%7D%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%20%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%20%5Ccdot%20%5Cleft%5B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D%20-1%20%5Cright%29%20%5Ccdot%20x%20%5Cright%5D%5E%7B-2%7D)
Explanation:
Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:
![\dot m_{in} - \dot m_{out} = 0](https://tex.z-dn.net/?f=%5Cdot%20m_%7Bin%7D%20-%20%5Cdot%20m_%7Bout%7D%20%3D%200)
![\dot m_{in} = \dot m_{out}](https://tex.z-dn.net/?f=%5Cdot%20m_%7Bin%7D%20%3D%20%5Cdot%20m_%7Bout%7D)
![\dot V_{in} = \dot V_{out}](https://tex.z-dn.net/?f=%5Cdot%20V_%7Bin%7D%20%3D%20%5Cdot%20V_%7Bout%7D)
![v_{in} \cdot A_{in} = v_{out}\cdot A_{out}](https://tex.z-dn.net/?f=v_%7Bin%7D%20%5Ccdot%20A_%7Bin%7D%20%3D%20v_%7Bout%7D%5Ccdot%20A_%7Bout%7D)
The following relation are found:
![\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bout%7D%7D%7Bv_%7Bin%7D%7D%20%3D%20%5Cfrac%7BA_%7Bin%7D%7D%7BA_%7Bout%7D%7D)
The new relationship is determined by means of linear interpolation:
![A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x](https://tex.z-dn.net/?f=A%20%28x%29%20%3D%20A_%7Bin%7D%20%2B%5Cfrac%7BA_%7Bout%7D-A_%7Bin%7D%7D%7BL%7D%5Ccdot%20x)
![\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x](https://tex.z-dn.net/?f=%5Cfrac%7BA%28x%29%7D%7BA_%7Bin%7D%7D%20%3D%201%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%20%5Cfrac%7BA_%7Bout%7D%7D%7BA_%7Bin%7D%7D-1%5Cright%29%5Ccdot%20x)
After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:
![\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bin%7D%7D%7Bv%28x%29%7D%20%3D%201%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%5Cright%29%20%5Ccdot%20x)
![v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}](https://tex.z-dn.net/?f=v%28x%29%20%3D%20%5Cfrac%7Bv_%7Bin%7D%7D%7B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%5Ccdot%20x%7D)
![v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}](https://tex.z-dn.net/?f=v%20%28x%29%20%3D%20v_%7Bin%7D%5Ccdot%20%5Cleft%5B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%5Ccdot%20x%20%5Cright%5D%5E%7B-1%7D)
The acceleration can be calculated by using the following derivative:
![a = v\cdot \frac{dv}{dx}](https://tex.z-dn.net/?f=a%20%3D%20v%5Ccdot%20%5Cfrac%7Bdv%7D%7Bdx%7D)
The derivative of the velocity in terms of position is:
![\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdx%7D%20%3D%20-v_%7Bin%7D%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%20%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%20%5Ccdot%20%5Cleft%5B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D%20-1%20%5Cright%29%20%5Ccdot%20x%20%5Cright%5D%5E%7B-2%7D)
The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.
Answer:
The maximum load the bar can withstand = 35.43 KN
Explanation:
Ultimate tensile strength of the given aluminium bar
= 540 M pa
Cross section area of the bar =
= 65.61 ![mm^{2}](https://tex.z-dn.net/?f=mm%5E%7B2%7D)
We know that the ultimate strength of the bar is calculated from
![\sigma = \frac{P_{max} }{A}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7BP_%7Bmax%7D%20%7D%7BA%7D)
![540 = \frac{P_{max} }{65.61}](https://tex.z-dn.net/?f=540%20%3D%20%5Cfrac%7BP_%7Bmax%7D%20%7D%7B65.61%7D)
= 540 × 65.61
= 35.43 KN
Therefore the maximum load the bar can withstand = 35.43 KN
:nanjan lang din yan sa binasa mo ne........