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

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In a Brayton cycle using helium gas, the relationship between net work and change in heat (Q) is debated. While some argue that the first law of thermodynamics implies net work equals change in heat, others clarify that a Brayton cycle is not an isolated system, making this equality unlikely. The first law states that the change in internal energy (ΔU) equals the heat added (q) plus the work done (w), but this does not directly apply to non-isolated systems. Calculating work and heat through the cycle's four steps is essential for understanding the dynamics involved. Overall, the net work and change in heat are not expected to be equal in a Brayton cycle.
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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?
 
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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
 
audi476 said:
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).
 
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