Answer:
Taking responsibility for your own learning makes it easier to identify your strengths and weaknesses. Once these have been identified you can work on a learning plan that focuses on the areas that you need most help with, increasing the speed of your learning, and build the skills you have been trying to perfect.
Explanation:
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:
b)False
Explanation:
Those material have high thermal it is very difficult to weld because due to high thermal conductivity it transmit the heat in to the surrounding and can not reach a particular temperature required to melt the material.And when material does not melt then there is no possibility to weld the material.
So from above we can say that it is very difficult to weld the copper material due to high thermal conductivity.Generally welding of copper done by usiong gas welding technique.
So our option b is right.
Answer:
The answer is "583.042533 MPa".
Explanation:
Solve the following for the real state strain 1:
![\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}](https://tex.z-dn.net/?f=%5Cvarepsilon_%7BT%7D%3D%5CIn%20%5Cfrac%7BI_%7Bil%7D%7D%7BI_%7B01%7D%7D)
Solve the following for the real stress and pressure for the stable.![\sigma_{r1}=K(\varepsilon_{r1})^{n}](https://tex.z-dn.net/?f=%5Csigma_%7Br1%7D%3DK%28%5Cvarepsilon_%7Br1%7D%29%5E%7Bn%7D)
![K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5Csigma_%7Br1%7D%7D%7B%5B%5CIn%20%5Cfrac%7BI_%7Bil%7D%7D%7BI_%7B01%7D%7D%5D%5En%7D)
Solve the following for the true state stress and stress2.
![\sigma_{r2}=K(\varepsilon_{r2})^n](https://tex.z-dn.net/?f=%5Csigma_%7Br2%7D%3DK%28%5Cvarepsilon_%7Br2%7D%29%5En)
![=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Csigma_%7Br1%7D%7D%7B%5B%5CIn%20%5Cfrac%7BI_%7Bil%7D%7D%7BI_%7B01%7D%7D%5D%5En%7D%20%5Ctimes%20%5B%5CIn%20%5Cfrac%7BI_%7Bi2%7D%7D%7BI_%7B02%7D%7D%5D%5En%5C%5C%5C%5C%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B%5BIn%20%5Cfrac%7B54.4%7D%7B47.7%7D%5D%5E%7B0.2%7D%7D%20%5Ctimes%20%5BIn%20%5Cfrac%7B57.8%7D%7B47.7%7D%5D%5E%7B0.2%7D%5C%5C%5C%5C%20%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B%5B%20In%20%281.14046122%29%5D%5E%7B0.2%7D%7D%20%5Ctimes%20%5BIn%20%281.21174004%29%5D%5E%7B0.2%7D%5C%5C%5C%5C%20%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B%5B%20In%20%281.02663509%29%5D%7D%20%5Ctimes%20%5BIn%201.03915873%5D%5C%5C%5C%5C%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B0.0114161042%7D%20%5Ctimes%200.0166818905%5C%5C%5C%5C%3D%20399%20%5C%20MPa%20%5Ctimes%201.46125948%5C%5C%5C%5C%3D583.042533%5C%20%5C%20MPa)