Cold water (cp = 4.18 kJ/kg · °C) leading to a shower enters a well-insulated, thin-walled, double-pipe, counter-flow heat exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water (cp = 4.19 kJ/kg · °C) that enters at 100°C at a rate of 3 kg/s. Determine ...
a.) the rate of heat transfer
b.) the rate of entropy generation in the heat exchanger
c.) The lost work in kW, assuming the temperature of the surroundings is 20oC
22 comments:
I'm kind of lost here. Do the two water streams mix and then exit at a temperature of 45C? How do we apply the entropy balance law for an open system, and is Qexchanged just Cp for cold water(delta T of the cold water)?
Washburn 5:00 PM
No, the water streams do not mix. This heat exchanger consists of two concentric tubes...an inner tube and an outer tube with one fluid flowing through the annular space between the tubes. They exchange heat, but they do not mix.
You can apply the 1st law to either stream to determine Q exchanged. Keep in mind that heat LEAVES the hot stream and ENTERS the cold stream.
In order to determine Sgen, you will need to to apply the MIMO entropy balance equation, 1st eqn on page 196 of the TCD Notebook. This heat exchanger operates at steady-state, so the left side of this equation is zero. Use T2 from the 1st part of the problem and solve for Sgen !
ok im getting stuck, im working on part b and I started with the first equation on page 196. I know the left side is zero (SS), I know min from problem statement, I know mout = sum min. I know Q dot form part a. But how do I know entropy and how can I find tsys? I thought it could be delta S equals m cp ln (t2/t2 ) but this gives me the wrong answer when I plug in numbers. Is it safe to assume incompressable fluid?? confusion..
i do not under why the left side of the equation is zero if the heat exchanger operates at steady-state. can someone explain this?
I too am having difficulties with part b. I assumed it was a incompressable liquid because all quarter we have been assuming that water is incompressable. However, when you plug values into detaS= integral(Cp/T dT) you do not get the correct units. (Cp is in KJ/Kg*C not Kelvin) This leads me to believe that either I am missing something or I am going the wrong way. Could you give me some guidance.
I'm getting something like .025KW/K- is that acceptable Dr. B?
I'm a little confused on this problem- that was my post above. My answer to part b is .025 kW/K. I don't understand where the rate of 3kg/s comes into the problem. I just have Q/Tc+ -Q/Th + Sgen = 0. I found my rate of heat transfer by integrating Cp (for cold stream) from T1 to T2 which is delta H. Then I multiplied this number by .025 kg/s. Any help would be great!
Chris Lee 6:49 PM
See my previous comment at 6:49 PM.
The two streams do not mix, they just exchange heat.
Yes, use deltaS for EACH stream is Cp Ln[T2/T1] because we can assume the liquid water is incompressible.
I think you are messed up because you treated this as a mixer instead of a heat exchanger.
Anon 8:45 PM
At steady-state, nothing changes with respect to time. So, all derivatives with respect to time are zero. The left side of the eqn is just dS/dt and at steady-state, this is zero.
Noah 9:25 PM
A change in T of 1 degC is the same as a change of 1 degK. So it is equally trrue that the units of Cp are kJ/kg-K.
I think you are doing the problem correctly. What did you get for T2 in part (a) ? You must use T in Kelvin when you solve deltaS = Cp Ln[Tout/Tin].
Anon 11:02 PM
On this problem there is nothing to lookup, so the only reason your answer would be different from mine would be round-off error...or something more serious. I suggest you check your calculations carrying may digits. If you don't get closer to my answer (0.019 kW/K) then I think there is a more serious error in your solution.
Shooter 11:36 PM
In order to determine Sgen, you will need to to apply the MIMO entropy balance equation, 1st eqn on page 196 of the TCD Notebook. This equation includes mass flow rates. In general, if the problem isn't SISO, use the MIMO equations for both the 1st Law and the S-balance.
I think I should have given a diagram with this problem statement. Would that have helped ?
I seem to still be doing something wrong on part b. When I use the incompressable liquid equation to find delta S for both hot and cold. (I solved the Q from part A to get T2 for R-134a). Then I multiplied these delta S values by their respective mass flow rates. This however is still not correct. I think I should not have canceled Qsys/Tsys however I don't know how to find Tsys. Any help would be wonderful.
Meredith 12:57 PM
Sgen is the sum of the mdot Cp Ln[Tout/Tin] for the two fluids. I think that is what you did. So, I think there must be a problem in part (a).
Tin and Tout are different for the two fluids. What did you get for Twater,out in part (a) ? That is my only/best guess at what is going wrong for you. I am sorry I could not be more helpful.
I will check the blog again in about 2 hours.
I don't understand what Tsys is in the part of the equation that is Sum(Qdot/Tsys),
From Lan:
Sgen should be the sum of all the generated S for each part, which is cold water, hot water, and maybe Ssurr.
In part b,the Tsurr is unknown, so you have to neglect the term Ssurr, whereas part c you have the Tsurr so you can include that term too.
Graham 3:05 PM
This is the 1st Eqn from page 196 solved for Sgen. There are 2 streams entering the HEX and 2 streams leaving the HEX. Mdot * deltaS = Mdot*Sout - Mdot*Sin. We can neglect the sum(Qdot/T) because Q = 0. No heat leaves the HEX. All the heat leaving the hot stream goes into the cold stream.
Greenpepper7
Tsys in the entropy balance equation is always the temperature within the system at which the heat transfer occurs. As long as this T is constant, we can pull the 1/Tsys out of the integral and integrate dQ to get Q. The result is Q/Tsys.
In this problem, you want to use the entire heat exchanger as your system. Then, if we assume all of the heat that leaves the hot stream goes into the cold stream, then Q crossing the boundary of the system (the entire heat excahnger) is ZERO.
I do not understand the first in part a) they want the rate of heat transfer
I found it id equal to 691.35 KW
Hi To0ot,
You must apply the 1st Law for open systems to determine the heat transfer rate in part (a). Assume that all the heat that leaves the how water stream enters the cold water stream. Assuming the system operates at steady state and taking the cold water as our system or control volume, the 1st Law for open systems reduces to Qdot = Mdot * deltaHhat, where the "dot" implies a rate and the "hat" means per unit mass (specific enthalpy). If we further assume that Cp is constant, deltaHhat = Cp * deltaTwater. We know everything we need to know to evaluate deltaHhat and then Q water. This is the answer to part (a). I got about 31 kW.
Best of luck !
Can u solve the problems with all the STEPs and the solution
To0ot:
I will not post the full solution here. I will be happy to answer questions and help you work through the problem.
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