# Fluid Mechanics - Water, Ridiculous Answer

eurekameh
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.

eurekameh
Anyone?

TimSmith
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.

Mentor
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.