Material Energy Balance / Thermodynamics Question

Click For Summary
SUMMARY

The discussion focuses on a thermodynamics problem involving a 10.0 m3 tank containing steam at 275°C and 15.0 bar, which is cooled to 1.2 bar, resulting in condensation. The key equations used include the energy balance equation for a closed, transient system: mi(hi+v2/2+gzi) - mo(ho+v2/2+gzo) + Q + W = ∆U. The user successfully determined the initial internal energy (U1) of superheated steam as 2739.25 kJ/kg but faced challenges in calculating the final internal energy (U2) due to the presence of both steam and liquid at the final state. The discussion emphasizes the need to apply steam tables effectively to find U2 and calculate the heat transferred (Q).

PREREQUISITES
  • Understanding of thermodynamic principles, specifically energy balance in closed systems.
  • Familiarity with steam tables for superheated and saturated steam properties.
  • Knowledge of latent heat of vaporization and its application in phase change calculations.
  • Basic algebraic skills for solving equations involving multiple variables.
NEXT STEPS
  • Study the use of steam tables for calculating properties of steam and water at various pressures and temperatures.
  • Learn about the latent heat of vaporization and its role in phase change thermodynamics.
  • Explore the concept of internal energy changes in thermodynamic processes.
  • Practice solving energy balance problems in closed systems with phase changes.
USEFUL FOR

Students in thermodynamics courses, engineers working with steam systems, and anyone involved in energy balance calculations in closed systems.

twiztidmxcn
Messages
43
Reaction score
0
I have this problem as a homework question for my MEB/Thermodynamics class and have been having some trouble with it.

A 10.0 m^3 tank contains steam at 275°C and 15.0 bar. The tank and its contents are cooled until the pressure drops to 1.2 bar. Some of the steam condenses in the process.

a) How much heat was transferred from the tank?
b) What is the final temperature of the tank contents?
c) How much steam condensed (kg)?


I started by doing the energy balance on the closed, transient system:

mi(hi+v2/2+gzi) - mo(ho+v2/2+gzo) + Q + W = ∆U

where v2 = v^2, i is subscript in, o is subscript out

Since the system is closed, mi = mo = 0, Ep = 0 (no height change), Ek = 0 (no movement), W = 0 (no work)

So I'm left with Q = ∆U = U2 - U1

I assume for part A that I am looking for Q, the heat transferred.

I started out knowing that at 275 C and 15 bar, we have a superheated steam. So using steam tables, I interpolated a value for U1 being 2739.25 kJ/kg.

My problem is with finding U2. Using the pressure of 1.2 bars, I used steam tables to find a value for U2, but since the tank's final contents are both steam and liquid, I am not quite sure which value to use.

In any case, U2 for liquid water is 439.2 kJ/kg and U2 for steam is 2512.1 kJ/kg.

Pretty much, I need to find out how to use this information to get a reasonable value and find the heat transferred.

As for B and C...any pointers in the right direction would be much appreciated.

Any help at all is appreciated (as usual)

Much thanks
-twiztidmxcn
 
Physics news on Phys.org
Ok. For the first part, since you have the pressure, volume and temperature, you can find the moles of steam using pv=nrt. Next, you know final pressure, final volume, but not the final moles or temperature.

Lets assume n1 moles were condensed during the process. If the latent heat of vapourisation is x per mol, then the total heat taken away from the system is n1*x=dQ. Since dQ=dU, then U_i=nfRt_i/2, U_f=n'fRt_f/2, dU=\frac{fR}{2}(nt_i-n't_f and n=n1+n'.
Where f is the degree of freedom of steam.

Two variables and two equations, solve for them and you have final temperature and number of moles condensed. From there the final part is easy.
 
Woah...I'll take a look at it and get back if I need any more help
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
23
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K