Difficulty: Easy
Correct Answer: eta_th = 1 - 1 / r^((gamma - 1)/gamma)
Explanation:
Introduction / Context:
The Brayton (Joule) cycle models the ideal gas turbine. Its thermal efficiency relates closely to the compressor pressure ratio r and the specific heat ratio gamma for an ideal gas with isentropic compression/expansion and constant-pressure heat addition and rejection. Understanding this dependence helps with preliminary cycle sizing and estimating benefits of increasing pressure ratio.
Given Data / Assumptions:
Concept / Approach:
For an ideal Brayton cycle, temperature ratios across isentropic devices depend only on pressure ratio and gamma. The compressor exit to inlet temperature ratio is T2/T1 = r^((gamma - 1)/gamma) and the turbine exit to inlet temperature ratio is its reciprocal for the same r. The thermal efficiency simplifies to eta_th = 1 - (T4 - T1)/(T3 - T2). With the isentropic relations and equal pressure ratios, this reduces to a function of r and gamma alone.
Step-by-Step Solution:
Use isentropic relation: T2/T1 = r^((gamma - 1)/gamma).For the matched turbine: T3/T4 = r^((gamma - 1)/gamma) → T4/T3 = 1 / r^((gamma - 1)/gamma).Ideal Brayton efficiency simplifies to eta_th = 1 - (T1/T2) = 1 - 1 / r^((gamma - 1)/gamma).Therefore the correct expression is eta_th = 1 - 1 / r^((gamma - 1)/gamma).
Verification / Alternative check:
As r increases, r^((gamma - 1)/gamma) increases, making 1 / r^((gamma - 1)/gamma) smaller; hence efficiency rises, which matches known Brayton behavior (up to metal temperature limits).
Why Other Options Are Wrong:
(b) and (d) do not include gamma and are dimensionally and physically inconsistent. (c) would predict negative efficiency for r > 1. (e) represents the loss term alone, not 1 minus that term.
Common Pitfalls:
Confusing r^((gamma - 1)/gamma) with r^(gamma - 1), or forgetting the reciprocal in the final expression.
Final Answer:
eta_th = 1 - 1 / r^((gamma - 1)/gamma)
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