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Hi guys this question is not from me but if i can get the
answer it will most definitely be helpful for me
and the friend that requires the answer , or hopefully
reference material would be great.
Here it is.
I am trying to get hold of some formulas for enthalpy and entropy. I already have some but I don't know what the letters mean. ie Q=? h1=?
I'm also trying to find a legend for reading electrical and mechanical schematics for refrigeration.
If anyone can help please let me know.
Thanks.
Q = m.c.dt
or
Q = m.c.dh
h = enthalpy
h1 is a point on a p/t chart as is h2, h3, h4
Sorry Marc - you are correct. :)
Long day - been out for a 10 mile walk and brain not functioning at present
Q= H2 - H1
Frank I dont know where you got "c" from as Mark points out its all relative ie E=M x c x c a very famous quote as you may recallQuote:
Originally Posted by frank
regards
eve walt
In this equation the "c" stands for the Specific Heat Capacity of the fluidQuote:
Originally Posted by eve walt
Hi guys.
fridg has posted my thread from another site.
What I'm looking for are templates or legends if u like which are industry standard. eg. >< = expansion device etc etc. I've looked and asked just about everywhere for something I can print out which has the symbol and its meaning to use as a quik refrence when study'n.
If Q=m.dt what is Q if Q=m.c.dt?:eek:
ok. If Q=m.c.dt does this mean the same as q=m.c.dt
and is Q or q allways kJ:D
dt. now how do i find dt. dt from.......??
I know c and m and where to find that but I can't remember where I get dt.
So if Q=m.LH does this mean Q = m.q41 0r q23whichever, If h1-h4=q41 and h3-h2=q23 ????
Am I on the right track or............?:confused:
Now by this Wc = h2 - h1 yes?
sooooooo.. COPref = q41/Wc ???:confused:
Ok it 1.30am,
Q=m.c.dt
but Q can't = m.c.LH
so does LH = m/Q
whats LH?
is LH merely q41?
Can you guys see my delema?
crap, now it 2am
Ok now once I get COP can I convert this to kW/ tons/ btu/h?......mmmmmm maybe not.
So how do I go about finding what my kW cap will be?
I don't have the study reference to work all this out and my workplace can't provide the answers, so I'm in a bit of a pickle.
TAFE starts in 6weeks:eek:
Some of the notations such as Q, q, m, h, etc. can be confusing when you are just starting. ;)
Whenever these are used in a text it is good form to supply how these are being used.
In some cases you will see h meaning enthalpy. In other cases it can mean film coefficient for heat transfer.
When you are calculating a heat load for a fluid where only the fluid temperature can change this is a sensible heat transfer process. This is found by:
Q=m*cp*dT
Q = amount of heat transferred from one temperature to the other
m= kilograms per hour
cp = specific heat of the fluid (air. water, etc)
dT = temperature difference of the fluid (higher temp. minus lower temp.)
If you are calculating a heat load for a fluid where the fluid changes state (liquid to vapor or vapor to liquid) this is a latent heat transfer process. This is found by:
Q=m*dh
Q = amount of heat transferred during phase change
m= kilograms per hour
dh = enthalpy difference of the fluid when it changes from one state to the other
LH is probably the latent heat if I had to guess.
Almost any load calculation will be based on one of the two above equations. This is what Frank said earlier.
Hope this helps.
Ok thanks guys. I know whats going on now.
Just to clarify US iceman, yes LH is latent heat. At tafe we were shown Quality=mass x latent heat. Now I'm assuming Q=m*dh is the same thing no?
So my latent heat value will be h1-h4 or h2-h3 for a simple cycle.
This is my dilema. I don't know what text I should be refering to for my studies. I need some info to explain all this a bit more. Your damn straight it confusing. Especially because where I work, we don't do these calculations. We are reliant on the software.
This reminds me of a conversation some time ago with a fellow collegue about how the kids of today only use calculators in school and haven't a clue how to do mental arithmatic - if the battries go dead then they are lost. :)Quote:
we don't do these calculations. We are reliant on the software.
Dogma
Would it help if you posted one of your questions and then the members could explain the answer for you, i.e. how to apply the formulas?
Ok frank here we go.:)
R22, 254kPa, 1700kPa gauge
I have 100l of pure H2O at 55c and wish to cool it to 5c
Now Qh=2935kJ/kg
q41= approx 144kJ/kg
COPref= approx 3.6
(all these values are very rough);)
What I want to know is where to find the info and equations to work out flow rate required and pull down time. How many other variables need to be considered?:eek:
I want to work out kW capasity also.
240V/50HZ:o
NO... Q=m*dh is a measure of the heat transfer when the fluid is changing state. Q does not mean quality.Quote:
we were shown Quality=mass x latent heat. Now I'm assuming Q=m*dh is the same thing no?
You can find the quality of the refrigerant at one point by using the difference in enthalpies for three points, but this is for one kilogram of mass only. This is another way of finding the flash gas for the cycle.
Quality of the refrigerant is the percentage of the refrigerant mixture as a two-phase condition. A two-phase condition is one where liquid and vapor co-exist at the same time.
Again, to add to the use of notation, "x" is normally used to signify the quality of the refrigerant.
If x=1, then the refrigerant is all vapor. If x=0, then the refrigerant is all liquid. If x is equal to a decimal such as 0.2, 0.5, etc., this represents the portion of the refrigerant that exists as vapor.
Therefore, if x=0.5, 50% of the refrigerant is all vapor, and 50% of the refrigerant is liquid. Using another example of x=0.2, 20% of the refrigerant is vapor and 80% of the refrigerant is liquid.
Without meaning offense, this is exactly the reason why students should not be shown software as part of the training. After you can do this by calculation and understand what the calculations mean, then software is OK.
Without knowing which points you label as h1, h2, etc., I cannot comment. Although, you can assume the latent heat of the cycle is normally referred to as the net refrigeration effect, or NRE. This is used to describe the useful cooling of the cycle and is associated with the evaporator only.Quote:
So my latent heat value will be h1-h4 or h2-h3 for a simple cycle.
The more you do this, the better you will understand the process. It is a little confusing at first. Don't quit on us now.:)
Dogma,
Try this link. The article goes through the explanation of what I think you are trying to find. I have to warn you this article is in Inch-pound units, but if you can work through this, it should be very useful in explaining some of the details.
It even has some nice pictures to help.
http://www.sporlan.com/5-200.htm
There is a lot of very good information on this website you can review in your spare time.
Dogma
lets deal with this one step at a time so as not to get confused.
You have 100L of water to cool from 55C to 5C.
So lets convert this.
100L = 100kg (now we can input this into the formula) and the temperature difference is 55 - 5 = 50, so the dt = 50
Refering to the formula - Q = m.c.dt we now have two parts, the dt and the c (sensible heat capacity) = 4.19 for water.
With only two parts known in the equation we need one more before we can do a calculation.
The m refers to mass flow but we only have a mass so we need to decide how we are going to cool the fluid (water) is it going to sit in a tank? are you going to move it through a heat exchanger?
If you are going to move it through a heat exchanger then we now need to decide at what speed (kg/s). by introducing a time element we will end up with a Q(uantity) in j/s (watts) or kj/s (kw).
Lets assume that you decide to use a plate heat exchanger and you decide on a pump that will deliver 2kg/s through the exchanger.
Now we have 3 parts of the equation and we can find the 4th - the amount of refrigeration you will need to reduce 100kg of water by 50K.
So we have :
2kg/s
4.19kj/kg
50k
therefore Q = 2 x 4.19 x 50 =419kw.
This will complete your water cycle in 100/2 = 50 secs.
but - this is much too large a plant for such a small amount of water and would be mega expensive so lets look at why.
We have a large dt - abnormally high in the real world and a really short time period for the cycle (the system probably wouldn't even settle down in that time).
So to consider a smaller plant (Q) you have to balance this with a sacrifice on the other side of the equation - the m or/and the dt. The C is a constant and does not change.
With a smaller Q and the same mass flow rate the dt alters. Most chillers (either shell & tube or PHE operate with a dt of 6k so if we rework the formula allowing for a dt of 6 we get the following
Q = 2 x 4.19 x 6 =50.28kw.
Obviously the time factor to bring your 100kg down 50K is somewhat longer but you can't have it all ways.
50kw is still an expensive piece of plant to cool 100kg so next I would look at reducing the mass flow rate.
Is this a theoretical exercise or a project you are working on?
This was covered before.;)Quote:
Originally Posted by me
Ok thanks guys, you've all pointed me in the right direction cheers. I'll be back to this thread in the very near future to bombard your brains with more of my dilemas.
Cheers...
Q= H2 - H1
Q= Actual energy input (from compressor)
in kJ/kg yes? Now is there any real accurate way of measuring or calculating entropy?
This information has not been compiled by me. It is information I have located and is protected by copywrite. However the publisher has stated that the information can be used for educational purposes. So here it is.
Entropy?
How can this be measured?
It is often stressed that entropy be related to information. However, entropy is an objective quantity, only depending on the system and not on an observers information about it. How can this quantity be measured? A procedure is presented to get the entropy of, say, a glass of water. Like for potential energy, only entropy differences have physical meaning, and a reference state is needed to give absolute entropies. Based on this reference state, absolute entropies of a few substances are given, and a practical example shows how such values can be obtained.
--------------------------------------------------------------------------------
Measuring entropy
Entropy change within a system equals the heat added reversibly to it divided by the temperature of the system
dS = dQ(rev)/T
Like for potential energy, only changes can be determined, except if a reference state is defined. For potential energy, the reference state is arbitrarily fixed to zero Joules at infinite distance of the force center. The reference state of entropy is a perfect crystal at zero Kelvin. Entropy of this reference state is arbitrarily fixed to zero Joules per Kelvin.
With this definition, an absolute entropy can be attributed to every substance at specified conditions like temperature and pressure.
For instance, Moore (W.J. Moore, physical chemistry, Longman 1972) gives Third Law entropies for a bunch of substances like:
Table 1: Third Law Entropies
Substances in the standard state at 298.15 K
Substance Entropy Substance Entropy
H2 130.59 D2 144.77
HD 143.7 N2 191.5
O2 205.1 Cl2 223.0
HCl 186.6 CO 197.5
CO2 213.7 H2O (gas) 188.72
NH3 192.5 CH4 186.2
Water (liquid) 70.0 Methanol 127
Ethanol 161 Diamond 2.77
Graphite 5.694 Ag 42.72
Cu 33.3 Fe 27.2
NaCl 72.38 AgCl 96.23
Like potential energy changes, entropy changes can always be measured. How can you find out how much the entropy of that glass of water over there changes if it is heated by 13.4 K?
This is what you have to do, exactly following the above definition
set entropy equal to zero: S = 0
set temperature equal to actual temperature of the water: T = T(start)
set end temperature equal to temperature + 13.4 K: T(end) = T(start) + 13.4 K
add a tiny little bit of heat dQ, so tiny that temperature of the water does not rise appreciably
record the temperature T and divide the heat portion dQ by it
add the result to S
if temperature is larger than T(end) then end; S is the quantity you have been looking for
else proceed with the next tiny portion of heat as stated in the fourth step above
What you were actually doing here is taking the integral over dQ/T from T(start) to T(start) + 13.4 K. For this small temperature range, the heat capacity c of the water can be taken as constant, so that
dQ = c * dT
and
where the indentifier at the left hand side denotes DeltaS, a difference between two entropies. The integral can be evaluated to
DeltaS = c * ln (T(end) / T(start))
where
Delta
refers to a difference, S(end) - S(start) the entropy at temperature T(end) minus the entropy at temperature T(start) in this case
T
the respective temperatures, T(start) the lower one and T(end) = T(start) + 13.4 K
For larger temperature ranges, the heat capacity c cannot be taken as constant.
Since integrating is computing the area between lower and higher limits of temperature, the horizontal axis and the function line, the above procedure can be depicted as shown in the diagram. We are actually computing the shaded area:
If there are phase changes between T(start) and T(end), discontinuities in the curve are to be expected.
With increasing temperature, entropy increase becomes smaller and smaller, because the heat capacity does not increase as fast as temperature does. In fact, for an ideal gas, heat capacity is not dependend on temperature. In this latter case, c/T plotted against T is a hyperbola.
This fact is important since it enables heat engines. No car's engine, no airplane's jet propulsion, no nuclear power plant would work if entropy would not change like this with temperature.
Phase changes
To get the figures in the table 1 above, the integration was performed from T = 0 K to T = 298.15 K. In this temperature range, the heat capacity is not constant. The dependence of heat capacity on temperature has to be determined experimentally.
Liquids and gases undergo phase transition on cooling; what entropy change is to be expected on, say, melting?
While melting, if the melting system is in equilibrium, temperature stays constant at the melting point. The heat supplied, DeltaQ, is completely used up for melting, so nothing is left to rise temperature. From the definition of entropy, the heat supplied (reversibly) need only be divided by the (constant) temperature to get entropy change. Since both heat and temperature are positive, entropy change is posivite, too, thus entropy increases on melting. According to this, entropies of liquids are larger than those of solids, as a glance at table 1 above confirms.
An example
(Heat capacities)/T of argon (Ar) are given in the graph below (values are taken from Försterling and Kuhn, "Moleküle und Molekülanhäufungen", Springer, Berlin 1983)
As in the example graph above, entropy is the area confined by the curve, the horzontal axis and the perpendicular lines at zero and at 298.15 K. The graph was plotted only to 120 K. In the gas phase (above 87.29 K), argon behaves very close to ideal, and for ideal gases the heat capacity is independend of temperature. Note the jumps from solid to liquid and from liquid to gas. Phase changes are discontinuities, so physical properties are expected to change discontinually there.
With these data, and knowing that heat of melting of Ar is 1.18 kJ/mol and heat of vaporization 6.52 kJ/mol, the entropy change of argon from zero Kelvin to any temperature in this range can be calculated. Since at zero Kelvin, entropy (of a perfect Ar-crystal) is nil by definition, standard entropy of Ar at normal temperature and pressure is calculated to be approximately 154 J/K/mol
Dogma,
To find the entropy at a particular state point, the fastest and most accurate means is to use published data. All refrigerants have been thoroughly characterized, and their properties are widely available. To calculate (look up, really) the entropy of a particular state point, you either get a p-h diagram for the refrigerant (published by many sources including the refrigerant manufacturers), or for more accuracy use software that has the equations of state built in.
The Technical University of Denmark wrote a piece of software called Refrigeration Utilities that is very good, and free (you can google to find the website).
Hope this helps,
Erik
Hi guys, back again. Thank you to all whom contributed. My Tafe now starts in 3 weeks, and I am stressing out. Apparently there are 3 cap tube, Air balancing, cpr,epr, Prac, psychrometrics, p-h diagrams, fans, fault fnding etc.
At the same time I can't wait to to get there.
Wish me luck.
Ok guys I'm back with a new dilema.
I have 85kg of peaches at 18C and I wish to freeze them at -25C. I'm working on about 12hrs pull down time. My ambient is 36 and wish to condense at 45C.
Now as far as my calcs go I'm looking at 824Watts to cool in 12 hrs.
AM I RIGHT TO THIS POINT?
Now If I wish to size a system how do I go about it?
If I'm using R404a am I right by going straight to a capillary selection table and selecting 2m at 1.1mm dia?
Then I sized the compressor to 825watts, A Maneurope LTZ22JE.
Stop me guys if i'm off in the wrong direction.
How do I size my Evap and Condenser? And do I need to take into account the size of my cabinet as well.
Where do I find the info I need to size my coils? Oh once I size this I will then need to size my evap and condenser fans.
How much info do I need to be able to build a prototype?
:confused:
This is just a theoredical exercise.:)
Thanks for the insight guys.