Fluid Mechanics - Water, Ridiculous Answer

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Discussion Overview

The discussion centers on the relationship between temperature and pressure in a sealed, rigid container filled with water, specifically examining the effects of a 50°C temperature increase on pressure. The conversation involves theoretical considerations and mathematical reasoning related to fluid mechanics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation involving the coefficients of compressibility and thermal expansion to determine the pressure change resulting from a temperature increase.
  • Another participant asserts that the initial calculations are incorrect, suggesting that an increase in temperature should lead to an increase in pressure, referencing common examples such as refrigerators and deodorant cans.
  • A later reply acknowledges the magnitude of the calculated pressure change but claims that the sign is incorrect, suggesting a need to reformulate the equations used.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are competing views regarding the correctness of the initial calculations and the expected relationship between temperature and pressure.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations, particularly concerning the coefficients used and the implications of the sign in the pressure change equation.

eurekameh
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Water at standard atmospheric pressure and temperature fills a sealed, rigid container. If the temperature of the water is increased by 50°C, what is the pressure?

V = volume, P = pressure, T = temperature

Coefficient of compressibility, kappa = (-1/V)(dV/dP)
Coefficient of thermal expansion, beta = (1/V)(dV/dT)

I solved for dV/V from the coefficient of thermal expansion equation:
dV/V = beta*dT.

Plugging into coefficient of compressibility equation:
kappa = -(beta*dT)/dP.

Solved for dP = -(beta*dT)/kappa, changed the d's into deltas.

Plugging kappa = 5.18 * 10^-10 and beta = 1.48 * 10^-4 into delta P equation gave me delta P = -92.3 MPa. Is this reasonable at all? I mean, the pressure difference seems a bit large.
 
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Hi eurekameh,

Now without even looking through your equations I can tell you that they are wrong, or you have made a mistake somewhere.

An increase in temperature will cause an increase in pressure, vice versa... Look up how refrigerators and heat pumps work, also think of your deodorant can - it loses pressure and gets colder.

I'm afraid I can't help you much more as my brain is currently fried from work.
 
eurekameh said:
Water at standard atmospheric pressure and temperature fills a sealed, rigid container. If the temperature of the water is increased by 50°C, what is the pressure?

V = volume, P = pressure, T = temperature

Coefficient of compressibility, kappa = (-1/V)(dV/dP)
Coefficient of thermal expansion, beta = (1/V)(dV/dT)

I solved for dV/V from the coefficient of thermal expansion equation:
dV/V = beta*dT.

Plugging into coefficient of compressibility equation:
kappa = -(beta*dT)/dP.

Solved for dP = -(beta*dT)/kappa, changed the d's into deltas.

Plugging kappa = 5.18 * 10^-10 and beta = 1.48 * 10^-4 into delta P equation gave me delta P = -92.3 MPa. Is this reasonable at all? I mean, the pressure difference seems a bit large.

The magnitude of your answer is correct, but the sign is wrong. Write the equation of the change in V in terms of the changes in P and T. Then set the change in V equal to zero in the equation.
 

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