Thermodynamics -- Internal Energy of Water

[dlacombe13]

Homework Statement

Water is initially at P = 1 bar and T = 20°C. 100kg of water is pumped to a higher pressure at which P = 10 bar and T = 25°C. Find ΔU and ΔH

Homework Equations

H = m*h
du = c*dT
dh = c*dT + v*dP

The Attempt at a Solution

So far I have looked in my table and found that at P =1 bar, the boiling point is T = 99.63°C. Since my temperature is much lower than that, it can't be super heated steam. Is it compressed liquid? I think once I find this out, I can maybe get the specific volume use it to solve for dh. I'm a little lost on this one.

 

[Chestermiller]

OK. It's compressed liquid.

 

[dlacombe13]

OK thanks, that's what I thought. So my tables that I am provided start at P = 25 bar. I have been doing a lot of searching on the internet, since some of my other problems involve compressed liquid. I found something called Compressed Liquid Approximation. It states that:
u(T,P) ≅ u_f(T)
v(T,P) ≅ v_f(T)
h(T,P) ≅ h_f(T) + v_f(T) * (P - P_sat(T))

So I think my approach should be to look up u and h on the table for each of the states. Then:
100kg*(u₂ - u₁) = ΔU
100kg*(h₂ - h₁) = ΔH
How does this look?

 

[Chestermiller]

What about using $\Delta U=mC\Delta T$ and $\Delta H=\Delta U+\Delta (PV)=\Delta U+V\Delta P$?

 

[dlacombe13]

Right, I was originally thinking that. So is C for water the same in all of its physical states?

 

[Chestermiller]

Right, I was originally thinking that. So is C for water the same in all of its physical states?

No

 

[dlacombe13]

Oh okay I see now, so C = 4.187 kJ/kgK. So I found that ΔU = (100kg) * (4/187 kg/kgK) * (278.15K) = 116,461.41 kJ.
Now I am ready to use that in the equation for ΔH = ΔU + VΔP.
So in this case, volume does not change? And if so, would I have to find volume by using V = m/ρ ?

 

[Chestermiller]

Oh okay I see now, so C = 4.187 kJ/kgK. So I found that ΔU = (100kg) * (4/187 kg/kgK) * (278.15K) = 116,461.41 kJ.
Now I am ready to use that in the equation for ΔH = ΔU + VΔP.
So in this case, volume does not change? And if so, would I have to find volume by using V = m/ρ ?

The temperature change is only 5 C, not 278 C. Your equation for the volume is correct.

 

[dlacombe13]

Oh yes, oops. So then ΔU = 2093.5 kJ
And then:

V = m/ρ = 100kg / 1000kg/m³ ==> V = 0.1m³

ΔH = ΔU + VΔP = 2093.5 kN*m + (0.1m³)*(900 kN/m²) =2183.5 kN*m ==> ΔH = 2183.5 kJ