Wednesday, March 29, 2006

HW #2 - P#1 - Steam Table Fundamentals

Please post any questions about this problem as comments on this posting.

34 comments:

Anonymous said...

Hi Dr. B.

I'm having issues with figuring out which phase its in. Obviously, we need to compare it to the saturation at a given T or P, but what does it tell you when the values for say, U or H are above or below? And do we compare it to the sat liquid or sat vapor?

Also, how can you calculate Q? I know Q= mass vap/ mass total, but doesn't that require you to know the system volume or # of moles?

I'm probably missing something simple here on both questions... Thanks.

Dr. B said...

Good question !
The key here is that the specific volume of the saturated vapor is greater than the specific volume of the saturated liquid. So if the overall or average specific volume of the system lies BETWEEN the specific volume of the saturated liquid and the saturated vapor, then a two-phase mixture exists in the system and you must calculate the quality. Quality is not defined for subcooled liquids or superheated vapors.

The same logic applies for specific internal energy, specific enthalpy and specific entropy.

Dr. B said...

On 1 a at 30C, 200Kpa is more preasure than the saturation preasure I think that means in is a compressed liquid. But I can not find a table for 200Kpa or .200MPa where am I going wrong

On b 130C abd 270.3Kpa it is at saturation but no additional information is given on the quality how do I get the V, U, and Q.

Dr. B said...

Part (a)
This problem starts out with a tricky one !

You are correct that because P > Psat@T the water is a subcooled or compressed liquid.

The problem is that there is not a subcooled liquid table for P = 200 kPa. The lowest P for subcooled liquid water in the tables is 5 MPa...and 30 degC does NOT appear in this table !

This means you need to interpolate once on the 5 MPa subcooled liquid table to determine U^ and V^ at 5 MPa and 30 degC and then interpolate AGAIN between the saturated liquid at 30 degC and Psat to get the properties at 30 degC and 200 kPa.

Dr. B said...

Part (b)
You hit the nail on the head here. You cannot determine U^ and V^. When 2 phases are present you need to know the value of some other intensive variable before you can use the tables to determine all the remaining properties. The presence of the extra phase messes things up. This is a lead-in to the Phase Rule, which we will learn about next week.

Dr. B said...

graham 5:30 PM:
You have almost got it. The last step is to interpolate for V^ & U^ between the values at Psat(~4.3 kPa) 5 MPa to get the values at 200 kPa. For this interpolation all values of V^ and U^ are at 30 degC.

Anonymous said...

I must be missing something as well for b. I Interpolated between 20 and 40 degrees to get the values of V and U at 30 degrees and at 5MPa. I can't seem to get the next interpolation. I know that I can use these values and some other value I know to interpolate to get .2MPa. But where do I get those other values? If U and V are the same in both cases I don't get a change since Y2-Y1 will be zero. I am very confused...thanks

Dr. B said...

Katie 7:33 PM:
I think you are talking about part (a).
Part (b) is a sat'd vapor.

You get the other values of V^ and U^ from the saturated steam tables at 30 degC. The P associated with these values is 4.247 kPa. So, 200 kPa falls between 4.247 kPa and 5000 kPa (5 MPa). Now, you can do the 2nd interpolation to get the answer.

Anonymous said...

Part D) Is it ok to assume that the temperature is the temperature at 300kPa as given in the tables? This would make it so that we'd have a saturated mix and could find the quality and such using the V^ given. Otherwise I have no idea how to solve the problem as it seems like T could be sort of arbitrary.

Anonymous said...

Whats the definition of quality and how do I calculate it?

Dr. B said...

jo @ 2:42 PM
Quality is the fraction of the mass in the gas phase.
You can calculate it directly from the mass of liquid and mass of vapor. But you don't know those in this problem.

So, instead use
x = (Vsatmix-Vsatliq) / (Vsatvap-Vsatliq)

V is specific volume here. This eqn works with any specific quantity: V^, U^, H^ or S^

Dr. B said...

Anon @ 11:52 AM
T = 300 kPa ???
I don't think that is a good assumption. The units do not work.

I think you will need to interpolate on the 500 kPa superheated steam table to answer this question.

tritron said...

i have a very basic questino to ask.why does the specific volume of saturated vapor decreases and a that of saturated liquid increases with temperature?what is the physical reasoning behind it?

Anonymous said...

it is very simple. as temperature of a saturated liquid increases two things happen simultaneously, one more and more vapour generates which decreases the V^ of vapour in equilibrium with it, second temperature of the liquid increases which decreases the density of the saturated liquid thus increasing the V^. i think now u must have got it.

rodz said...

water at 2.5MPa and 200C is heated at constant temperature up to a quality of 80%. Find(a) the quantity of heat receive by the water, (b) the change in internal energy.


I dont know what to do in here,. how can i find the first s and u in this problem,..

tnxs,..

Dr. B said...

Hi Rodz,
Since Tsat at 2.5MPa is 224 degC, the initial state is a subcooled liquid (also known as a compressed liquid). You could use the NIST website to determine s and u. If you want to use printed tables, then you will most likely need to interpolate between the saturated liquid state at 200 degC and Psat=1.554 MPa and whaterver P values for which you have a subcooled liquid table (probably 5 MPa). The key is that you interpolate on P with T always at 200 degC. Best of luck.

Steam Tables said...

The comment posting is very informative. It would be great if you provide more details about steam tables. Thanks a lot....

Dr. B said...

Hello Steam Tables,
You can find a lot more information about how to use the stem tables on my website, LearnThermo.com. Lesson 2C and its example problems might be of particular interest to you:
http://www.learnthermo.com/T1-tutorial/ch02/lesson-C/pg01.php

Best of luck!

Sam Neupane said...

my steam table don't have a value greater than 110 bar but i need the value of 220 bar. What do i do? my teacher said me to only use the table..Help

Dr. B said...

Hi Sam,
You are in a tough spot. There are advanced methods to predict properties of fluids at very high pressures, but I doubt that is what your instructor has in mind.
I suspect you made an error somewhere and the pressure is not really 220 bar. That is a very high pressure. Perhaps there is a problem with the units. I would double and triple check.

Dr. B said...

Hi Sam,
If you have to solve this problem with P=220 bar, I suggest you use the NIST Webbook at:
http://webbook.nist.gov/chemistry/fluid/
Best of luck,
Dr. B

Cast Iron Table Base said...

Hello,

A steam table comprising an open-top tank, said tank having a double bottom comprising an upper base and a lower base defining a water tank integral with steam table beneath said higher base, water tank having a water inlet and a water outlet, and said water tank for eliminating air bubbles in the water tank which enter the water tank when water is fed into the water tank through a said water inlet. Thanks.

Dr. B said...

Hello Cast Iron,
I am not sure what you want to ask. Maybe you can clarify the question.

Anonymous said...

Hello Dr. B
"water at 2.5MPa and 200C is heated at constant temperature up to a quality of 80%. Find(a) the quantity of heat receive by the water, (b) the change in internal energy."

I have gained the proper values of s and u.. but when computing for the quantity of heat i get the wrong answer. q=t(delta s) q=(200+273)(5.61202-2.3294).

In my textbook.. the answer is 2025.7

Dr. B said...

Assuming the process is internally revesible, I obtain results similar to yours (also assuming your units are kJ/kg). What units apply to the answer given in your textbook? It would be very helpful if you supplied the complete problem statement.

Unknown said...

Water @2.5Mpa and 200°C is heated at constant temperature up to a quality of 80%. Find (a)the quantity of heat received by the water, (b)the change in internal energy, (c)the work of a nonflow process. So far, i managed to solve for the u and s, but i got a different result from what is the correct answer, which is 2025.kj/kg. My solution for Q is t*∆s.

Dr. B said...

Assuming the process is internally reversible, Q=T ΔS. See my previous comment.

Unknown said...

Have you find the value of enthalpy?

Unknown said...

How do you find the change in enthalpy?

Dr. B said...

I can help if you will post the problem statement here.
Cheers,
Dr. B

Unknown said...

A piston-cylinder containing steam at 700 kPa and 250oC undergoes a constant pressure process until the quality is 70%. Determine per kilogram (a) the work done, (b) the heat transferred, (c) the change of internal energy, and (d) the change of enthalpy.

Dr. B said...

Hi Unknown,
Do you want me to do your homework for you? I don't think so. But you can ask me a question. Better yet, visit LearnThermo.com!
Best regards,
Dr. B

Unknown said...

Helloo anonymous what's the name of your text book?

Dr. B said...

The book is available online on my website at www.LearnThermo.com