Answer:
13.6mm
Explanation:
We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.
See attachment for the step by step solution of the problem
Answer:
38 kJ
Explanation:
The solution is obtained using the energy balance:
ΔE=E_in-E_out
U_2-U_1=Q_in+W_in-Q_out
U_2=U_1+Q_in+W_in-Q_out
=38 kJ
Answer:
Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.
Explanation:
Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that
represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:
- Wall Thickness. More thickness, more thermal resistance.
- Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
- Cross-section Area. More cross-section area, less thermal resistance.
A expression to define the thermal resistance for the wall is as follows:
, where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.
Answer:
|W|=169.28 KJ/kg
ΔS = -0.544 KJ/Kg.K
Explanation:
Given that
T= 100°F
We know that
1 °F = 255.92 K
100°F = 310 .92 K
![P _1= 15 psia](https://tex.z-dn.net/?f=P%20_1%3D%2015%20psia)
We know that work for isothermal process
![W=mRT\ln \dfrac{P_1}{P_2}](https://tex.z-dn.net/?f=W%3DmRT%5Cln%20%5Cdfrac%7BP_1%7D%7BP_2%7D)
Lets take mass is 1 kg.
So work per unit mass
![W=RT\ln \dfrac{P_1}{P_2}](https://tex.z-dn.net/?f=W%3DRT%5Cln%20%5Cdfrac%7BP_1%7D%7BP_2%7D)
We know that for air R=0.287KJ/kg.K
![W=RT\ln \dfrac{P_1}{P_2}](https://tex.z-dn.net/?f=W%3DRT%5Cln%20%5Cdfrac%7BP_1%7D%7BP_2%7D)
![W=0.287\times 310.92\ln \dfrac{15}{100}](https://tex.z-dn.net/?f=W%3D0.287%5Ctimes%20310.92%5Cln%20%5Cdfrac%7B15%7D%7B100%7D)
W= - 169.28 KJ/kg
Negative sign indicates compression
|W|=169.28 KJ/kg
We know that change in entropy at constant volume
![\Delta S=-R\ln \dfrac{P_2}{P_1}](https://tex.z-dn.net/?f=%5CDelta%20S%3D-R%5Cln%20%5Cdfrac%7BP_2%7D%7BP_1%7D)
![\Delta S=-0.287\ln \dfrac{100}{15}](https://tex.z-dn.net/?f=%5CDelta%20S%3D-0.287%5Cln%20%5Cdfrac%7B100%7D%7B15%7D)
ΔS = -0.544 KJ/Kg.K