Fluid Mechanics - Water, Ridiculous Answer

AI Thread Summary
Increasing the temperature of water in a sealed, rigid container by 50°C will result in an increase in pressure, contrary to the initial calculation suggesting a negative pressure change of -92.3 MPa. The coefficients of compressibility and thermal expansion were used to derive this pressure change, but the sign of the result was identified as incorrect. The discussion emphasizes that an increase in temperature leads to an increase in pressure, referencing practical examples like refrigerators and aerosol cans. A correction is suggested to properly set the change in volume equal to zero in the relevant equations. Understanding the relationship between temperature and pressure is crucial in fluid mechanics.
eurekameh
Messages
209
Reaction score
0
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.
 
Physics news on Phys.org
Anyone?
 
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.
 
Back
Top