# Thermodynamics fun!

1. Apr 10, 2005

### sportsrules

For a brayton cycle on a pressure and temperature diagram, using helium gas, is it reasonable to have the net work for the cycle and the change in Q for the cycle to come out to be equal to each other?

2. Apr 10, 2005

### audi476

i thought the first law of thermodynamics said "change in work is equal to change in heat"

so yeah it is reasonable. and i just worked some brayton cycle problems for homework, and yeah that's what i had

3. Apr 11, 2005

### so-crates

First law of Thermodynamics

$$\Delta U = q + w$$

Or in differential form

$$dU = dq + dw$$

In an ISOLATED system, $$\Delta U = 0$$. A Brayton cycle, however, is not isolated. Actually its not even closed. So offhand I would not expect the net work to be the same as the net heat evolved. I would however, calculate the work done and heat evolved through all 4 steps of the cycle for an ideal gas. You have two isobars and two isoentropes (constant entropy).

Last edited: Apr 11, 2005