Thermodynamics -- Internal Energy of Water

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Homework Help Overview

The problem involves calculating the change in internal energy (ΔU) and enthalpy (ΔH) of water as it is pumped from an initial state at 1 bar and 20°C to a final state at 10 bar and 25°C. The subject area is thermodynamics, focusing on properties of water under varying pressure and temperature conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the state of water, identifying it as a compressed liquid and referencing tables for thermodynamic properties. There are considerations of using specific equations for internal energy and enthalpy, including the Compressed Liquid Approximation. Questions arise about the constancy of specific heat (C) across different states of water and the implications for calculations involving volume and pressure changes.

Discussion Status

The discussion is active, with participants exploring various methods to approach the problem. Some have proposed using specific heat and volume calculations, while others are verifying assumptions about the properties of water. There is no explicit consensus on a single method, but multiple lines of reasoning are being examined.

Contextual Notes

Participants note constraints related to the provided tables, which start at higher pressures, and the need to clarify the physical state of water to apply the appropriate thermodynamic equations. There are also discussions about the temperature change and its impact on calculations.

dlacombe13
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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.
 
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OK. It's compressed liquid.
 
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) ≅ uf(T)
v(T,P) ≅ vf(T)
h(T,P) ≅ hf(T) + vf(T) * (P - Psat(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?
 
What about using ##\Delta U=mC\Delta T## and ##\Delta H=\Delta U+\Delta (PV)=\Delta U+V\Delta P##?
 
Right, I was originally thinking that. So is C for water the same in all of its physical states?
 
dlacombe13 said:
Right, I was originally thinking that. So is C for water the same in all of its physical states?
No
 
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 ?
 
dlacombe13 said:
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.
 
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
 

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