Answer:
![\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp](https://tex.z-dn.net/?f=%5Cmu%20_j%3D%5Cdfrac%7B1%7D%7BC_p%7D%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p-v%5Cright%5Ddp)
Explanation:
Joule -Thompson effect
Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.
Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.
Now lets take Steady flow process
Let
Pressure and temperature at inlet and
Pressure and temperature at exit
We know that Joule -Thompson coefficient given as
Now from T-ds equation
dh=Tds=vdp
So
![Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp](https://tex.z-dn.net/?f=Tds%3DC_pdt-%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p%5Cright%5Ddp)
⇒![dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp](https://tex.z-dn.net/?f=dh%3DC_pdt-%5Cleft%20%5BT%5Cleft%28%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%5Cright%29_p-v%5Cright%5Ddp)
So Joule -Thompson coefficient
This is Joule -Thompson coefficient for all gas (real or ideal gas)
We know that for Ideal gas Pv=mRT
So by putting the values in
For ideal gas.
So what is the main question? I Need to know what the main question is to solve it.
Answer:
1.265 Pounds
Explanation:
Data provided:
Tire outside diameter = 49"
Rim diameter = 22"
Tire width = 19"
Now,
1" = 0.0254 m
thus,
Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m
Rim radius, r₂ = 22" / 2 = 11" = 0.2794 m
Tire width, d = 19" = 0.4826 m
Now,
Volume of the tire = π ( r₁² - r₂² ) × d
on substituting the values, we get
Volume of air in the tire = π ( 0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³
Also,
Density of air = 1.225 kg/m³
thus,
weight of the air in the tire = Density of air × Volume air in the tire
or
weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg
also,
1 kg = 2.204 pounds
Hence,
0.5742 kg = 0.5742 × 2.204 = 1.265 Pounds
To solve this problem, we must basically count the total energy lost converting all the values given in the international system.
The energy loss is given by both 124KJ and 124food heats.
Since the energy conversion we know that 1 food calories is equal to 4,184J. So:
Therefore the total energy lost will be
Therefore the total energy lost to the surroindings in form of heat is 124.518kJ
