Liquid water at 200 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 200 kPa and 150°C. Liquid water enters the mixing chamber at a rate of 2.5 kg/s, and the chamber is estimated to lose heat to the surrounding air at 25°C at a rate of 1200 kJ/min. If the mixture leaves the mixing chamber at 200 kPa and 60°C, determine... (a) and (b) .
a.) The mass flow rate of the superheated steam
b.) The rate of entropy generation during this mixing process
c.) The lost work in kW
12 comments:
I found the first part ok, then for the second part I'm using sum(mdot*deltaS hat) = Qact/Tsurr + Sgen. I then use each flow rate individually and its delta S hat, and the fact that Qact=-20 and T surround is given. Unfortunately, I'm getting an answer of 0.31 not 0.33. Is this approach totally off?
Washburn 5:24 PM
Other students have demonstrated to me that the final answer in this problem is VERY sensitive to round-off errors and small variations in the property values from different tables. I think you did the problem correctly and got the right answer given the values you got from the tables. I am not sure, but I think you got it. Youmight try going back and carrying many significant figures through your calculations to see if that makes a difference. I doubt it, to be honest. I think the difference is caused by slightly different values in the steam tables.
Can someone explain how to get the first part to this problem? It doesn't seem as though we have enough information
When calculating part a, how do we calculate the kinetic energy for the 1st law. thanks
Anon 9:22PM
Start by looking up H and S for all 3 streams. NIST will save you some time here.
Apply a mass balance and the 1st Law. This gives you 2 equations in 2 unknowns: m2 and m3, where m3 is the outlet stream and m2 is the 150 degC stream.
In part (b), consider an extended system and apply the MIMO S-balance equation. Solve for the only unknown: Sgen.
Your answer may not be particularly close to my answer because of variations in tables and round-off error.
The version of the 1st law I am using has kinetic energy in it, but I don't understand how to find kinetic energy from our givens. Or am I uing the wrong equation?
Ok, a different system for the same process has kinetic energy as zero. They are both labeled as the same process, so I am confused as to how this conclusion is reached.
for part (b), I am still very shaky on the theory. I have the general equation but it seems like there is more than 1 unknown. is d/dt Ssys 0? Do I use the temperature of the surroundings as Tsys?
From Lan
To Carina:
There is no kenetic enery term in this question. There is no velocity provided.
Yes, Tsurr is the Tsystem as you imagine it's a whole system
Carina 2:14 PM
Assume changes in kinetic and potential energies are negligible.
Carine 2:35 PM
Assume that the process is operating at steady-state. Steady-state means that nothing changes with respect to time, so all time derivatives, including dS/dt, are zero.
Use an extended system so that the T at the boundary of your system is the same as Tsurr. This will let you calculate the total Sgen.
iv almost got this one figured out... except How did washburn get deltaS^ for all those mdots? i know i can get S1^,s2^,s3^ from table, but what do i substract from them to make them deltaS^?
Also Qact is confusing to me so far. where does it come from?
thanks~
Adamant 10:24 PM
DeltaS doesn't make much sense when we have 2 inlet streams and one outlet stream. So, don't try to follow Washburn. Just apply the MIMO form of the 2nd Law, the 1st Eqn on pg 196 of the TCD Notebook. Solve for Sgen.
Qact is tjust the rate of heat transfer into the system in the actual process. So, in this case, Qact = -1200 kJ/min.
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