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Old Feb 12th 2017, 11:37 PM   #1
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Law of conservation of thrust

Is there some law of conservation of thrust. See my question is simple. Thrust= pressure x area. Now we all know that pressure increases as area decreases. Suppose pressure is 8 and area is 2 then thrust is 16. Now when pressure is 2 and area is 8 then also thrust is 16. Is there some kind of conservation law?
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Old Feb 13th 2017, 04:43 AM   #2
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we all know that pressure increases as area decreases
No, I don't agree.
Pressure and Area are pretty much independent.

Pressure and Volume in a closed vessel are related,
Pressure is Inversely Proportional to Volume (at constant temperature).
i.e. if you reduce the size of the vessel, thereby squeezing the gas, the pressure will go up.

If you define the Area as the Width times the Length of the vessel,
and the third dimension of the Volume as the Height,
then if you hold the Height constant the Pressure and Area are related.

However, there are a number of additional stipulations in the above statement
that are not apparent in your original post.
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Last edited by Woody; Feb 13th 2017 at 04:47 AM.
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Old Feb 13th 2017, 06:27 AM   #3
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I think what you're looking for, in an intuitive way, is what Bernoulli figured out back in 1738, now known as Bernoulli's Principle. It states that the energy content of a a flow with no external sources of energy is constant:

$\displaystyle \frac {v^2} 2 + gz + \frac P {\rho}$ = constant

Here the term 'z' is the change in vertical position, so for a horizontal pipe z = 0. Consider fluid flowing down a pipe. If the the cross-sectional area of the pipe increases then the velocity decreases (to maintain constant mas flow, i.e. $\displaystyle \rho Av = \dot m$ = Constant), and hence the factor $\displaystyle P/\rho$ must increase, which means some combination of P increasing and density decreasing must occur. For an incompressible fluid like water, where rho is constant, and combining with the constant mass flow equation, this simplifies to:

$\displaystyle \frac {v^2} 2 + \frac P {\rho} = \frac {\dot m ^2} {2 \rho^2 A^2} + \frac P {\rho}= $ Constant.

Rearranging, and shifting all the constants to the right hand side:

$\displaystyle \rho P A^2 = Constant $

(Note - this is not the same constant as above.) Hence for an incompressible flow, increasing the area reduces the pressure, but not linearly - note the exponent. It gets complicated for compressible fluids, such as air - the effect of density decreasing with decreasing pressure means it's difficult to come up with an exact correlation between P and A without understanding the thermodynamic effects in play - as pressure decreases temperature decreases, which has the effect of increasing density, though not in a linear fashion. Bottom line is: yes, there is a correlation between pressure and area, but no, it's not linear.
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Last edited by ChipB; Feb 13th 2017 at 06:29 AM.
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