What is the Rise in Temperature of Molten Metal During Cylindrical Die Forging?

[RajarshiB91]

Homework Statement

[/B]
A molten metal is forced through a cylindrical die at a
pressure of 168000 Kg per square meter. Given that the density of
the molten metal is 2000 Kg per cubic meter and the specific heat of the
metal is 0.03, find the rise in temperature during this process. Assume
that the mechanical equivalent of heat is 420 Kg-meter per Kcal.

Homework Equations

H=MCΔT and V/T =constant (probably)[/B]

The Attempt at a Solution

I'm not able to understand which equations to use to solve the problem. The density is given but volume is missing, so I don't have the mass. Also, how do I find the total heat added to the molten metal in the process? Do we need more data to solve this or am I misunderstanding the question?
This is not homework but a question from a competitive exam so I'm not sure what theory to be used to solve this.

 

[Chestermiller]

Are you familiar with the open system version of the first law of thermodynamics from your thermo course? If you do, then you will know immediately what the change in enthalpy per unit mass is for the molten metal being pushed through the die.

Chet

 

[RajarshiB91]

Hi Chet, thanks for your reply. Yeah, I had a brief introduction to it a long time back. I read it again and I think I have solved the problem. Basically, all the terms cancel out and we are left with the following equation: Q=m(h₂-h₁) where enthalpy per unit mass h=u+Pv (v=specific volume;inverse of density)
Now, I think it is easily solvable with the given data.

 

[Chestermiller]

Hi Chet, thanks for your reply. Yeah, I had a brief introduction to it a long time back. I read it again and I think I have solved the problem. Basically, all the terms cancel out and we are left with the following equation: Q=m(h₂-h₁) where enthalpy per unit mass h=u+Pv (v=specific volume;inverse of density)
Now, I think it is easily solvable with the given data.

Actually, you can also expected to (reasonably) assume that Q = 0, since the die would be insulated. So you are left with Δh = 0. This will give you the answer you need.

Chet

 

[RajarshiB91]

Actually, you can also expected to (reasonably) assume that Q = 0, since the die would be insulated. So you are left with Δh = 0. This will give you the answer you need.

Chet

But if you put only Δh=0, so the equation is then Δu+PΔν=0. But Δv=0 assuming it is incompressible. So, we get only Δu=0. Then how do I solve the problem?

 

[Chestermiller]

But if you put only Δh=0, so the equation is then Δu+PΔν=0. But Δv=0 assuming it is incompressible. So, we get only Δu=0. Then how do I solve the problem?

h = u + Pv, so for an incompressible fluid, Δh = Δu + vΔP.

Chet

 

[RajarshiB91]

h = u + Pv, so for an incompressible fluid, Δh = Δu + vΔP.

Chet

So, Δu=C_vdT and ΔP=RΔT/v. The ΔT cancels out again if I put Δh=0. I still don't see how I can get ΔT from just Δh=0.

 

[Chestermiller]

So, Δu=C_vdT and ΔP=RΔT/v. The ΔT cancels out again if I put Δh=0. I still don't see how I can get ΔT from just Δh=0.

The equation in bold is incorrect. You are dealing with a liquid, not an ideal gas (even for an ideal gas, the equation is incorrect). You are given the value of ΔP in the problem statement.

Chet