Very good question. Let's think for a minute how changing sign could alter the answer. You are right, Qh is leaving and so it would be negative. Ql is entering the system and should be positive. The work of the compressor is input work and so is also positive. Let's now write the first law for the system: Ql - Qh + W = 0 moving things around you get: Ql - Qh = - W now if we get rid of the negative sign: Qh - Ql = W as you can see, we got the same result. As long as you are confident about what is entering and what is leaving your system, you can set up the first law either way. Hope this helps.
keep posting content man
Thanks for your support.
For question 9, why cant we use m(h2 - h1) to find Qh for the water? Why do we use the mc∆T? Moreover, is the c, Cp or Cv?
Hi there. Great work man!
Thanks man :)
For problem 5 shouldn’t Qh be negative because it’s leaving the system?
Very good question. Let's think for a minute how changing sign could alter the answer. You are right, Qh is leaving and so it would be negative. Ql is entering the system and should be positive. The work of the compressor is input work and so is also positive. Let's now write the first law for the system:
Ql - Qh + W = 0
moving things around you get:
Ql - Qh = - W
now if we get rid of the negative sign:
Qh - Ql = W
as you can see, we got the same result. As long as you are confident about what is entering and what is leaving your system, you can set up the first law either way.
Hope this helps.
@@WannabeProfessor86 yes it does thank you!