Does decreasing pressure of fluid in a fixed volume lower temperature?

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Sorry if the formulation is imprecise, I'm a biology major trying to study some physics. The question is:
569. When comparing two points of fluid flowing through the same horizontal pipe, if the fluid velocity is greater, then
A. the temperature is less
B. the temperature is greater
C. the temperature is unchanged
D. the pressure is greater
So far, my reasoning is:
From Bernoulli's equation, you have

(1/2)(\rho)v^2 + \rho*g*h + P = constant

where \rho = density, v = velocity, g = acceleration due to gravity, h = height and P = pressure.

If the first term increases, one or both of the other two terms must decrease in order to satisfy the relationship. How can this happen?

(1) \rho*g*h decreases. This means either \rho, g, or h decreases. Since the fluid is incompressible (assumption I'm making), \rho is constant and cannot decrease. g is a constant and cannot decrease. So it's possible that the height decreased. But that's not one of the answer choices.

(Note: If the fluid were not incompressible, a possible answer would be that \rho, the density, decreases. Since volume is constant, this would mean the mass of the fluid decreased. This is not an answer choice.)

(2) P decreases. Notice that the question says "same horizontal pipe", which means the volume doesn't change to compensate. So the pressure in a fixed volume is being decreased, which means it can be reasonably be assumed (too lazy to go through the math of the relationship between pressure/volume and temperature for fluids, I'm assuming it's similar to gases) that the temperature decreases. That's why I would say it's A.
I make an assumption in Step (2) that decreasing the pressure will decrease the temperature. But I'm not sure what the justification is. I only know of the ideal gas law, and I'm not sure if the behavior is the same for incompressible fluids. Maybe incompressibility means fluid is similar to ideal gas behavior, or maybe it holds in general and the incompressibility assumption can be removed? Or maybe it's dependent on the fluid?

Thanks for any clarification!
 

Answers and Replies

  • #2
boneh3ad
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The bottom line is it's a bad question. The answer depends on what the fluid is (or more mathematically, the equation of state), so it will be have differently I it is a gas or liquid. Further, it would depend on the flow properties of the fluid as well since gases behave differently at different pressures and velocities.

The ideal gas law is the equation of state for an ideal gas and it only really applies to compressible gas flows.
 
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My reasoning is completely different from yours.

The answer cannot be D because A, B, and C cover all possible situations regarding temperature, therefore 1 of them must be true.

We are told nothing about the nature of the fluid or the reason it is flowing at different rates through the pipe, Only that it flows at different rates. Given that the authors of the question felt this was the only information necessary I believe the answer they are after is B. Increased velocity of the fluid results in increased friction between the fluid and the pipe resulting in higher temperature.
 
  • #4
boneh3ad
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mrspeedybob,

That isn't necessarily the case even then, though. Fluids are not like solids. Viscous dissipation of kinetic energy into thermal energy has a very weak effect on the temperature of the fluid except at high speeds compared to the effects of conduction and convection from the wall. In other words, without any information on the heat transfer boundary condition at the wall, it is impossible to make the conclusion you made.
 

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