- #1
Mary2100
- 12
- 0
Hi guys,
It is not some homework, so I think it is better to post here. I would like to know if my method is correct for the work of the walls x and y.
I suppose no mass, no friction and the gas keeps its temperature constant.
An external device modify the shape of a triangle but it keeps constant the area. The external device turns the walls x clockwise around the dot O and the wall y counterclockwise around the dot O. The wall x rotates at the same angle as the wall y. At the same time, the external device moves the wall z up. Inside the triangle there is a gas and the pressure of the gas is higher than outside. There are gaskets between x/z and y/z: the gas doesn't escape.
I noted:
At start, a=b=√2/2 m and c=1 m
Difference of pressure P=1 Pa
Area of the triangle = 1*1/2 = 0.5 m² = constant
Calculation for an angle of 1e-1 rad
Depth of the device = 1 mEnergy from x and y :
The area is constant so a*b=0.5, so a=0.5/b
Pythagoras's theorem: c²=a²+b²
b=c*cos(α):
c²=0.25/(c²*cos²(α))+c²*cos²(α)
<=> (c²-c²*cos²(α))*(c²*cos²(α))=0.25
<=> c⁴*(1-cos²(α))cos²(α)=0.25
<=> c=(0.25/((1-cos²(α))*cos²(α)))^0.25
There are two walls but I need to divide by two (the moment), so the work is:
## \int_{\frac{pi}{4}}^{\frac{pi}{4}+0.1} (\frac{0.25}{(cos^2(x)*(1-cos^2(x)))^{0.25}} dx ##
Is it ok for you ?
It is not some homework, so I think it is better to post here. I would like to know if my method is correct for the work of the walls x and y.
I suppose no mass, no friction and the gas keeps its temperature constant.
An external device modify the shape of a triangle but it keeps constant the area. The external device turns the walls x clockwise around the dot O and the wall y counterclockwise around the dot O. The wall x rotates at the same angle as the wall y. At the same time, the external device moves the wall z up. Inside the triangle there is a gas and the pressure of the gas is higher than outside. There are gaskets between x/z and y/z: the gas doesn't escape.
I noted:
At start, a=b=√2/2 m and c=1 m
Difference of pressure P=1 Pa
Area of the triangle = 1*1/2 = 0.5 m² = constant
Calculation for an angle of 1e-1 rad
Depth of the device = 1 mEnergy from x and y :
The area is constant so a*b=0.5, so a=0.5/b
Pythagoras's theorem: c²=a²+b²
b=c*cos(α):
c²=0.25/(c²*cos²(α))+c²*cos²(α)
<=> (c²-c²*cos²(α))*(c²*cos²(α))=0.25
<=> c⁴*(1-cos²(α))cos²(α)=0.25
<=> c=(0.25/((1-cos²(α))*cos²(α)))^0.25
There are two walls but I need to divide by two (the moment), so the work is:
## \int_{\frac{pi}{4}}^{\frac{pi}{4}+0.1} (\frac{0.25}{(cos^2(x)*(1-cos^2(x)))^{0.25}} dx ##
Is it ok for you ?