Monday, November 03, 2008

TE 303 - HW #6, P8 - Ammonia Carnot Vapor Refrigeration Cycle - 15 pts

Three kg of ammonia executes a Carnot vapor refrigeration cycle. During the isothermal compression (cooling) step, the ammonia begins as a saturated mixture at 10 bar with a quality of 97% and it is cooled until it is a saturated liquid. The adiabatic compression step requires 128 kJ/kg of work to increase the pressure from 1.1 bar to 10 bar.

a.) Sketch the cycle on a PV diagram.
b.) Evaluate the heat and work for each process in kJ.
c.) Evaluate the COP for this cycle.

7 Old Comments

4 comments:

Anonymous said...

For this problem I have solved everything for the adiabatic compression(3-2) and isothermal compression (2-1) but am having issues finding how to fill in the rest of the chart. To find another value to look up numbers, all the equations that come to mind you need either a Temperature or a Volume.

Can you give me any hints?

Anonymous said...

(Thanks to Ryland for pointing out to me that my hints/answers somehow didn't show up for this problem. Here is (hopefully) what I had tried to post previously.)

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HINTS:
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* Be sure to check out the comments from Dr. B's course!

* Please refer to TCD-6E1 for an excellent example.

* The trick to solve this problem is to balance what properties you know at each state with what you know for each process, so I recommend that you set up dual tables for each. (The Q and W table vs. each process is essential.)

* You may need to use the first law energy balance to find the quantities that you do not know. So, for instance, for state 3, all you know is the pressure, but you also know W32 and Q32. So using the first law, you can find U3 and thus use the steam tables to find T3 and other properties that you may need. [Please refer to my notation for the states below, as you may have defined them differently.]

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ANSWERS:
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(Note that I defined my Carnot cycle where state one is saturated liquid, and state 2 is on the same isotherm.)

T4 = 241.46 K

W21 = -123.1 kJ/kg

Q43 = 915.8 kJ/kg

COP = 4.3

Anonymous said...

Dear Dr. P's biggest fan,

Please let me know if the hints that I just posted didn't help.

Anonymous said...

Another HINT:

Once you determine Q21, you can determine Q43 from either the Kelvin principle OR the Coefficient of Performance.

From Q43, you can then determine another property for state 4 (such as enthalpy) so now you have two, independent, intensive properties...