Thermodynamics Enthelpy of water

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Discussion Overview

The discussion revolves around the thermodynamic properties of water, specifically focusing on the calculation of changes in internal energy and enthalpy during the transformation of liquid water at 212ºF and 1 atm to steam at 400ºF and 100 psi. Participants explore the relevant equations and unit conversions necessary for these calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the specific internal energy of liquid water and attempts to calculate the specific internal energy of steam using the equation h=u+Pv, arriving at a value of 1194.80 Btu/lbm.
  • Another participant suggests that the units for the pv term need to be expressed in BTU/lb and points out the need to use the pressure and specific volume of liquid water at 212ºF for calculations.
  • There is a discussion about converting specific volume from ft³/lb to BTU, with one participant claiming to have made corrections to the units but receiving feedback that the conversion was incorrect and irrelevant.
  • Participants discuss the calculation of initial specific enthalpy and emphasize the importance of ensuring that units are consistent, specifically in BTU/lbm.
  • One participant expresses confusion about how to achieve the correct units in BTU/lbm and seeks clarification on the conversion factors between different units of energy.
  • Another participant provides a conversion factor between ft-lbf and BTU, indicating a collaborative effort to resolve unit conversion issues.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to unit conversions and calculations for the enthalpy and internal energy changes. Multiple viewpoints and methods are presented, indicating ongoing uncertainty and debate.

Contextual Notes

There are limitations regarding the assumptions made about unit conversions and the dependence on specific definitions of pressure and volume. The discussion reflects unresolved mathematical steps and the need for clarity in unit handling.

Who May Find This Useful

This discussion may be useful for students and professionals interested in thermodynamics, particularly those dealing with phase changes and energy calculations in fluid systems.

ruiwp13
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Homework Statement


Liquid water at 212ºF and 1atm has a specific internal energy of 180.02Btu/lbm. The water is transformed into steam at 400ºF and the pressure of 100psi where it's specific volume is 4.937ft^3/lb and it's specific enthalpy is 1228.4Btu/lb. Determine the variation of internal energy and the variation of enthalpy of the process.


Homework Equations


ΔH=ΔU+(ΔPV)
h=u+Pv


The Attempt at a Solution



So I know the specific internal energy in the liquid state (180.02Btu/lbm), and I can calculate the specific internal energy in steam (h=u+Pv ⇔ 1228.4=u+6.804*4.937) which gives me 1194.80. But now I'm stuck... I have the variation of the specific volume (vsteam-vliquid) , the variation of pressure (6.804atm-1atm) and the variation of specific internal energy. So, I can get the variation of the specific enthalpy(h). But how do I get the mass to get the variation of enthalpy(ΔH)? (if what I said is correct of course).

Thank you for your time and patience
 
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You had the right idea in the first part, but you need to express the pv part in BTU/lb. Right now, the units are handled incorrectly.

As far as the second part is concerned, you need to make use of the pressure and specific volume of liquid water at 212 F. The density is about 62.4 lbm/ft3 and the pressure is 14.7 psi.
 
Chestermiller said:
You had the right idea in the first part, but you need to express the pv part in BTU/lb. Right now, the units are handled incorrectly.

As far as the second part is concerned, you need to make use of the pressure and specific volume of liquid water at 212 F. The density is about 62.4 lbm/ft3 and the pressure is 14.7 psi.

I've corrected the units, I think. Converted 4.937ft^3/lb to Btu. 1ft^3=1.027Btu. It gave me 5.070299.
So, instead of using the 1atm I'll use the 14.7 psi and the specific volume I calculate it from the density you gave me to use in the (ΔPV) part? But I still don't get it how will I found the mass to convert it to ΔH and ΔU
 
Last edited:
ruiwp13 said:
I've corrected the units, I think. Converted 4.937ft^3/lb to Btu. 1ft^3=1.027Btu. It gave me 5.070299.

This conversion is incorrect and also irrelevent.

So, instead of using the 1atm I'll use the 14.7 psi and the specific volume I calculate it from the density you gave me to use in the (ΔPV) part? But I still don't get it how will I found the mass to convert it to ΔH and ΔU

The initial specific enthalpy is 180 + pv, where the pv is expressed in units of BTU/lbm. You know the final enthalpy already, so you can get the change in enthalpy.

The final internal energy is calculated the way you did it, but make sure that the units of the pv are, again, BTU/lbm.

chet
 
Chestermiller said:
This conversion is incorrect and also irrelevent.



The initial specific enthalpy is 180 + pv, where the pv is expressed in units of BTU/lbm. You know the final enthalpy already, so you can get the change in enthalpy.

The final internal energy is calculated the way you did it, but make sure that the units of the pv are, again, BTU/lbm.

chet

I'm not seeing how will I get the units in BTU/lbm.
 
ruiwp13 said:
I'm not seeing how will I get the units in BTU/lbm.
p=14.7 psi = 14.7x144 lbf/ft2
v=1/62.4 ft3/lbm

pv = units of ft-lbf/lbm

What is the conversion factor between ft-lbf and BTU?

Chet
 
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Chestermiller said:
p=14.7 psi = 14.7x144 lbf/ft2
v=1/62.4 ft3/lbm

pv = units of ft-lbf/lbm

What is the conversion factor between ft-lbf and BTU?

Chet

1.285x10-3 :p thank you for your patience and time.

Best Regards
 

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