Answer:
![Q_{cv}=-339.347kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%3D-339.347kJ)
Explanation:
First we calculate the mass of the aire inside the rigid tank in the initial and end moments.
(i could be 1 for initial and 2 for the end)
State1
![1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295](https://tex.z-dn.net/?f=1bar%2A%7C%5Cfrac%7B100kPa%7D%7B1%7D%7C%2A2%3Dm_1%2A0.287%2A295)
![m_1=232kg](https://tex.z-dn.net/?f=m_1%3D232kg)
State2
![8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350](https://tex.z-dn.net/?f=8bar%2A%7C%5Cfrac%7B100kPa%7D%7B1bar%7D%7C%2A2%3Dm_2%2A0.287%2A350)
![m_2=11.946](https://tex.z-dn.net/?f=m_2%3D11.946)
So, the total mass of the aire entered is
![m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg](https://tex.z-dn.net/?f=m_v%3Dm_2-m_1%5C%5Cm_v%3D11.946-2.362%5C%5Cm_v%3D9.584kg)
At this point we need to obtain the properties through the tables, so
For Specific Internal energy,
![u_1=210.49kJ/kg](https://tex.z-dn.net/?f=u_1%3D210.49kJ%2Fkg)
For Specific enthalpy
![h_1=295.17kJ/kg](https://tex.z-dn.net/?f=h_1%3D295.17kJ%2Fkg)
For the second state the Specific internal Energy (6bar, 350K)
![u_2=250.02kJ/kg](https://tex.z-dn.net/?f=u_2%3D250.02kJ%2Fkg)
At the end we make a Energy balance, so
![U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e](https://tex.z-dn.net/?f=U_%7Bcv%7D%28t%29-U_%7Bcv%7D%28t%29%3DQ_%7Bcv%7D-W%7Bcv%7D%2B%5Csum_i%20m_ih_i%20-%20%5Csum_e%20m_eh_e)
No work done there is here, so clearing the equation for Q
![Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20m_2u_2-m_1u_1-h_1%28m_v%29)
![Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20%2811.946%2A250.02%29-%282.362%2A210.49%29-%28295.17%2A9.584%29)
![Q_{cv}=-339.347kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%3D-339.347kJ)
The sign indicates that the tank transferred heat<em> to</em> the surroundings.
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius,
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax ![= \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4q%7D%7B3%5Cpi%20%7D%20%28%5Cfrac%7Br2%5E2%20%2B%20r2r1%20%2B%20r1%5E2%7D%7Br2%5E4%20-%20r1%5E4%7D%20%29)
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
Technician a is correct because he says that Many common rail injectors filters can be bypassed by dirt, which can lead to an injector sticking open and continuously fueling a cylinder.
Coalescence is used to separate the water and fuel. To the fuel injector cleaning kit, fasten your air compressor. Diesel engines run at compression ratios that are greater than those of gasoline engines. greater ratio compared to gasoline engines. increased thermal expansion as a result. more fuel energy that is transformed into usable power. The great benefit of using a dry cylinder sleeve is that by quickly installing new sleeves, the cylinder block can be quickly restored to its original specifications. Vacuum drying can be used to get rid of small amounts of water. A nozzle is used to spray the fuel into the vacuum chamber of engines. Air and unsolved free water are taken out of the oil. The fuel is evenly dispersed, which facilitates efficient drying.
Learn more about injectors here:
brainly.com/question/27969202
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Answer:
I can help but I need to know what it looking for
Answer:
B. Acid rain.
C. Photochemical smog.
Explanation:
Oxides of nitrogen contribute to the formation of photochemical smog and acid rain. Photochemical smog is a type of smog produced when ultraviolet light from the sun reacts with nitrogen oxides in the atmosphere while on the other hand, when nitrogen oxide react with the water vapor in the atmosphere forming nitric acid which falls on the earth surface with the help of precipitation.