Bulk Modulus Problem - Two solutions?

AI Thread Summary
When water freezes, it expands by approximately 9%, leading to a pressure increase in a confined space like an automobile engine. The bulk modulus of ice is given as 2x10⁹ N/m², and the correct calculation for pressure increase when considering the engine's constraints results in 165 MPa. The initial incorrect approach calculated the pressure based on an expansion from V to 1.09V, which misinterpreted the engine's role in preventing expansion. A balloon scenario was discussed, clarifying that the balloon's volume does not increase with the water's expansion, thus invalidating the first calculation method. Understanding the context of volume changes is crucial for accurate pressure calculations in these scenarios.
najd
Messages
9
Reaction score
0

Homework Statement


Question:
When water freezes it expands about 9%. What pressure increase would occur inside your automobile engine if the water froze. The bulk modulus of the ice is 2x10⁹N/m².


Homework Equations


B = - Δp/(ΔV/V)


The Attempt at a Solution


My 1st solution:
Volume was originally V. Then increased by 9%, so that Vf = 0.09V + V = 1.09V.
ΔV/V = 0.09.
Δp = 0.09*2x10⁹= 180MPa.

Answer should be 165MPa.

My 2nd solution:
I tried treating it in reverse.
Volume was originally 1.09V. Then decreased by 9% to V.
Surprisingly, ΔV/V = 0.09V/1.09V.
Δp = (0.09*2x10⁹)/1.09 = 165MPa.

Can someone please explain the error in my first attempt?
 
Physics news on Phys.org
Hi najd, welcome to PF!

The engine is not forcing the ice to expand from V to 1.09V. The engine is forcing the ice, which would otherwise expand on its own from V to 1.09V, to remain at V. This is equivalent to applying pressure to reduce the volume from 1.09V to V. Does this make sense?
 
Hey! Thanks!

So you're saying my second approach is correct because the pressure caused by the engine, which can only hold V of water, is preventing the ice from expanding to 1.09V, whereas if it the engine weren't there, the ice would expand normally.

Okay. I understand. Thank you!

Hmm, okay, another situation. Let's pretend that water of volume V is inside a balloon which has no effect except for occupying the water. As it freezes, it will expand by 9%, right? The balloon is NOT stopping it from expanding because it's merely occupying it. If I were to calculate the pressure done by the water on the balloon, would my first attempt at solving the problem be correct?
 
najd said:
Hmm, okay, another situation. Let's pretend that water of volume V is inside a balloon which has no effect except for occupying the water. As it freezes, it will expand by 9%, right? The balloon is NOT stopping it from expanding because it's merely occupying it. If I were to calculate the pressure done by the water on the balloon, would my first attempt at solving the problem be correct?

Not really, because the balloon's volume doesn't increase by 9%.
 
I think you misunderstood me because I forgot to mention that the balloon's initial volume is V as well. The water of volume V was filling it, entirely, so the balloon's volume would increase by 9%, too.
 
najd said:
I think you misunderstood me because I forgot to mention that the balloon's initial volume is V as well. The water of volume V was filling it, entirely, so the balloon's volume would increase by 9%, too.

Certainly, but you asked whether you could apply your first approach to the balloon. The volume of the balloon material (rubber, for example) doesn't increase by 9%. So you couldn't take the bulk modulus of the balloon material and argue that the pressure on the balloon material is \Delta P=-B(\Delta V/V). The pressure that a stretched balloon exerts on its contents is a different type of calculation.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top