Thermodynamics Enthelpy of water

In summary, the water is transformed into steam at 400 degrees and the pressure of 100 psi where it's specific volume is 4.937 ft^3/lb and it's specific enthalpy is 1228.4 BTU/lb.
  • #1
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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  • #2
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.
 
  • #3
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:
  • #4
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
 
  • #5
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.
 
  • #6
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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  • #7
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
 

1. What is the definition of thermodynamics enthalpy of water?

Thermodynamics enthalpy of water refers to the total amount of heat energy present in a given amount of water at a specific temperature and pressure. It includes the internal energy of the water molecules as well as the energy associated with any changes in volume or pressure.

2. How is thermodynamics enthalpy of water calculated?

The thermodynamics enthalpy of water can be calculated using the equation H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume. This equation takes into account the energy associated with the water's temperature and any changes in volume or pressure.

3. What factors affect the thermodynamics enthalpy of water?

The thermodynamics enthalpy of water can be affected by changes in pressure, temperature, and the amount of water present. Additionally, the presence of other substances, such as solutes, can also impact the enthalpy of water.

4. How is the thermodynamics enthalpy of water used in practical applications?

The thermodynamics enthalpy of water is used in various industrial processes, such as power generation and chemical reactions. It is also used in thermodynamic calculations to determine the efficiency of energy conversion systems and to predict the behavior of water in different environments.

5. What is the significance of studying thermodynamics enthalpy of water?

Studying the thermodynamics enthalpy of water is important for understanding the behavior of water in various systems and environments. It also plays a crucial role in the development and optimization of energy conversion technologies, as well as in the study of chemical reactions and processes involving water.

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