Thermodynamics Fun: Is Net Work = Change in Q for Brayton Cycle?

[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?

 

[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

 

[so-crates]

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

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).