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.
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Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.
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