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Hi Folks,
I have a couple of questions regarding steam calcs.
1. How do you calculate steam quality? (Hg-Hf)/Hfg * 100?
2. When calculating power of a turbine by P=m(H1-H2)where m is mass flow rate of steam and H is enthalpy, which enthalpy do you use? Hf, Hg, Hfg?
3. How do you calculate specific volume of a steam that is less than 100% quality? For example 95% steam quality at 100.4F and Vf=.01613ft^3/lbm and Vfg=350.3ft^3/lbm.
Also, is there any good resources available of this?
Many Thanks,
Jim
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Steam Calculation Questions
Started by Guest_Jim_*, Apr 27 2003 10:25 PM
5 replies to this topic
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#2
Diederik Zwart
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Junior Member
Posted 28 April 2003 - 03:52 AM
On www.techwareeng.com you can download an evaluation version of a program called Winsteam. This can calculate all relevant steam properties.
#3
Guest_Jim_*
Guest_Jim_*
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- guestGuests
Posted 29 April 2003 - 09:34 PM
Thanks,
I guess that doesnt anwser my question, thanks for the info though. I need to know how to do this by hand. Anyone?
Many Thanks!
I guess that doesnt anwser my question, thanks for the info though. I need to know how to do this by hand. Anyone?
Many Thanks!
#4
Guest_Guest_Art Montemayor_*
Guest_Guest_Art Montemayor_*
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- guestGuests
Posted 30 April 2003 - 02:02 PM
Jim:
To answer your questions, we have to revisit basic thermodynamics:
Steam quality is defined as the weight % of a "wet" steam condition that is pure steam vapor (as opposed to the liquid condensate content).
Hg = Enthalpy of saturated steam vapor
Hf = Enthalpy of saturated condensate (liquid)
Hfg = Latent Heat (enthalpy difference) required to vaporize the saturated condensate into saturated steam
Now, look at what happens in a throttling calorimeter:
The expansion process of the wet steam is an irreversible, adiabatic steady flow in which no work is done. This, by thermodynamic principles, is an ISENTHALPIC process (enthalpy is constant through the expansion process).
If a sample of the high-pressure wet steam is throttled to a lower pressure, then, within limitation, it will become superheated. This can be seen in a Mollier diagram. The lower pressure, resultant steam will be at point 2 and you can locate this precisely because you can measure its pressure and temperature - and you know its enthalpy which can be obtained from the superheat steam tables. As a result,
h2 = (hg - y1hfg) , btu/lb
from which x1 = (1-y1) and this can be found by looking up hg and hfg corresponding to the known mainline steam pressure, p1.
You can visualize this system by looking at the Mollier (I like the T-S better) diagram. Drawing a horizontal line denotes the latent heat between the saturated vapor curve and the saturated liquid curve. Somewhere, on the right side of this horizontal line, is the point of the wet steam mixture being expanded. If this point is called the point 1, then by the "lever" rule (actually, a heat & material balance around the calorimeter) the quality of the steam at point 1 is:
x1 = (h1 - Hf) 100/Hfg
of course, h1 = h2
The process of a steam turbine is an ISENTROPIC one ( S1 = S2). The difference in enthalpy is that between the inlet and outlet steam conditions while the ENTROPY is constant. On a T-S diagram, this is the vertical distance that you drop in going from the inlet pressure to the outlet pressure (that's why I like the T-S diagram better). The inlet enthalpy is usually in the superheated region, outside the "dome"; you normally want to stay out of the dome, unless you have a condensing turbine. So, you see, Hfg has nothing to do with the expansion. The enthalpy that you have to use depends on the steam quality that you are feeding. And you never go down to Hf.
I don't understand your question. Steam quality is steam quality. If you want the specific volume of the steam, you must state that it is the STEAM (vapor) that you are trying to identify with the specific volume. There is no practical value (in my experience) to consider the specific volume of a 2-phase mixture. The steam tables only give you the specific volumes of either the saturated steam or the saturated liquid. However, if you wish, you can factor in the liquid quantity by using the quality value of a wet steam - what for, I can't guess.
I hope this explains the steam processes and properties you mentioned.
Art Montemayor
To answer your questions, we have to revisit basic thermodynamics:
Steam quality is defined as the weight % of a "wet" steam condition that is pure steam vapor (as opposed to the liquid condensate content).
Hg = Enthalpy of saturated steam vapor
Hf = Enthalpy of saturated condensate (liquid)
Hfg = Latent Heat (enthalpy difference) required to vaporize the saturated condensate into saturated steam
Now, look at what happens in a throttling calorimeter:
The expansion process of the wet steam is an irreversible, adiabatic steady flow in which no work is done. This, by thermodynamic principles, is an ISENTHALPIC process (enthalpy is constant through the expansion process).
If a sample of the high-pressure wet steam is throttled to a lower pressure, then, within limitation, it will become superheated. This can be seen in a Mollier diagram. The lower pressure, resultant steam will be at point 2 and you can locate this precisely because you can measure its pressure and temperature - and you know its enthalpy which can be obtained from the superheat steam tables. As a result,
h2 = (hg - y1hfg) , btu/lb
from which x1 = (1-y1) and this can be found by looking up hg and hfg corresponding to the known mainline steam pressure, p1.
You can visualize this system by looking at the Mollier (I like the T-S better) diagram. Drawing a horizontal line denotes the latent heat between the saturated vapor curve and the saturated liquid curve. Somewhere, on the right side of this horizontal line, is the point of the wet steam mixture being expanded. If this point is called the point 1, then by the "lever" rule (actually, a heat & material balance around the calorimeter) the quality of the steam at point 1 is:
x1 = (h1 - Hf) 100/Hfg
of course, h1 = h2
The process of a steam turbine is an ISENTROPIC one ( S1 = S2). The difference in enthalpy is that between the inlet and outlet steam conditions while the ENTROPY is constant. On a T-S diagram, this is the vertical distance that you drop in going from the inlet pressure to the outlet pressure (that's why I like the T-S diagram better). The inlet enthalpy is usually in the superheated region, outside the "dome"; you normally want to stay out of the dome, unless you have a condensing turbine. So, you see, Hfg has nothing to do with the expansion. The enthalpy that you have to use depends on the steam quality that you are feeding. And you never go down to Hf.
I don't understand your question. Steam quality is steam quality. If you want the specific volume of the steam, you must state that it is the STEAM (vapor) that you are trying to identify with the specific volume. There is no practical value (in my experience) to consider the specific volume of a 2-phase mixture. The steam tables only give you the specific volumes of either the saturated steam or the saturated liquid. However, if you wish, you can factor in the liquid quantity by using the quality value of a wet steam - what for, I can't guess.
I hope this explains the steam processes and properties you mentioned.
Art Montemayor
#5
Guest_Jim_*
Guest_Jim_*
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- guestGuests
Posted 30 April 2003 - 05:49 PM
Hi Art,
Thanks very much, this info was very helpful! I'm studying for the FE exam (1st part of PE). In any case, here's how I arrived at my questions, it was from 2 questions that I had trouble with in the text:
1. Steam (1000Kg/s) at 400C anf 400kPa enters a turbine and leaves at 30C.
a. What is the quality of the esxiting steam?
b. Assuming turbine efficiency of 85%, what power can be obtained?
Given: Steam at 400C and 400kPa
h=3273.4 kJ/kg
s=7.8992 kJ/kg K
at 30C:
hf=125.8kJ/kg
hfg=2430.4 kj/kg
hg=2556.2 kJ/kg
sf=0.4367kJ/kg K
sfg=8.0174 kJ/kg K
sg=8.4541 kJ/kg K
Question 2: What is the specific volume (m^3/kg) of a 37.8C steam mixture with a quality of 95%? Given
vf=0.01613 ft^3/lbm
vfg=350.3 ft^3/lbm
Im not certain if I did this right, but took a weithed average as .05*vf+.95*vfg
I guess theyre still a little confusing to me seeing that the entropy dropped a little in the first and, with the second, im not certain.
Many Thanks Again!
Thanks very much, this info was very helpful! I'm studying for the FE exam (1st part of PE). In any case, here's how I arrived at my questions, it was from 2 questions that I had trouble with in the text:
1. Steam (1000Kg/s) at 400C anf 400kPa enters a turbine and leaves at 30C.
a. What is the quality of the esxiting steam?
b. Assuming turbine efficiency of 85%, what power can be obtained?
Given: Steam at 400C and 400kPa
h=3273.4 kJ/kg
s=7.8992 kJ/kg K
at 30C:
hf=125.8kJ/kg
hfg=2430.4 kj/kg
hg=2556.2 kJ/kg
sf=0.4367kJ/kg K
sfg=8.0174 kJ/kg K
sg=8.4541 kJ/kg K
Question 2: What is the specific volume (m^3/kg) of a 37.8C steam mixture with a quality of 95%? Given
vf=0.01613 ft^3/lbm
vfg=350.3 ft^3/lbm
Im not certain if I did this right, but took a weithed average as .05*vf+.95*vfg
I guess theyre still a little confusing to me seeing that the entropy dropped a little in the first and, with the second, im not certain.
Many Thanks Again!
#6
mslei
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- Members
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- 1 posts
Brand New Member
Posted 07 May 2003 - 04:53 AM
Hi, my friends/classamates and I are currently working on designing and fabricating a miniature/portable steam generator that can be used in the laboratory. We're looking for websites where we can get good and useful information on steam, its properties, uses and related matters. We've already consulted and read a lot of books but we find the informations we gathered not enough for our review of literature and documentation! I hope you can help us with this! Thanks and God Speed!
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