Tuesday, April 24, 2007

HW #6, P8 - Ammonia Carnot Vapor Refrigeration Cycle - 10 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 comments:

Anonymous said...

The answers I copied down last week don't match the ones online presently. The numbers I have been getting match the earlier answers, not the present ones. Why have the answers changed?

Dr. B said...

George:
The answers changed because the units changed. Before, I had posted Qhat and What in kJ/kg. Then, I noticed the problem actually asks for Q &W in kJ. So, I multiplied all the old answers by 3 kg NH3 and got the answers I currently have posted.

No worries !

Dr. B said...

TCD Notebook Ammonia Tables:

Today, it came to my attention that the U, H and S values in the 1st 6 rows of the saturation pressure table are WRONG. There is a reference state problem ! I will give you a replacement page to correct this error ASAP.

In the meantime (for HW #6) please use the NIST Webbook or the TFT Excel plug-in to get the properties of ammonia.

Anonymous said...

Hi Professor,
I wasn't sure how to sketch the PV diagram. I know the first point is somewhere inside the camel hump, and lies very close to the sat. vapor curve (x=97%). After that, I'm pretty much lost. Please help! Thank you!

Unknown said...

How do I get the properties for point 4? Specifically the volume at point 4 so I can find work from 4-3. It seems like the only data points I have for 4 so far are the pressure and temperature which are the same as point 3 but the volume is less than point 3. Without a VLE quality I don't know how to find the volume.

Dr. B said...

Anon:
Point 2 is at the higher P and does lie in the envelope near the sat vapor curve. Point 1 is sat liq at the same P as 2.

Points 3 &4 are on the lower P isobar and they both lie at undetermined qualities.

I wish you had taken some time during the week to ask this before 8:41 PM the night before the HW is due.

Dr. B said...

Cary:
The key is that this is a Carnot cycle: Q43 / -Q21 = T3 / T1. Solve this for the unknown Q43. Then, appply the 1st Law to step 4-3 to determine H4 from H3 and Q43. After that, you should be in good shape.