Entropy in the Real World: Engines — Question 9

Section 20-4 · Physics pages 3308–3309

Question 9 Startup energy storage prevents treating a transient interval as a cycle During startup, an engine assembly receives 10000 J of heat, rejects 6000 J and delivers 3000 J of external work. The working fluid, metal and other internal stores are included in the system boundary. No matter crosses it. engine assembly Qh W Qc energy storage Schematic; labels, not drawing lengths, specify the data. (a) Find the assembly energy change and the measured work-to-heat-input ratio. (b) If a later complete cycle has the same input and rejected heats but zero storage change, find its work and efficiency. Explain why the startup ratio cannot be substituted for the steady cyclic efficiency.
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Question 9 - Solution Startup energy storage prevents treating a transient interval as a cycle engine assembly Qh W Qc energy storage (a) Method and calculation ∆U = Qh −Qc −W = 10000 −6000 −3000 = 1000 J. W/Qh = .300. (b) Interpretation and further calculation ∆Ucycle = 0 ⇒Wcycle = 10000 −6000 = 4000 J. ηcycle = .400. During startup, part of the net heat warms the assembly. Performance comparisons must specify an interval with a complete energy balance and distinguish storage from continuous output. Physical check The missing 1000 J is stored energy, not automatically an unmeasured heat loss.

Question 9

Startup energy storage prevents treating a transient interval as a cycle

During startup, an engine assembly receives 10000 J of heat, rejects 6000 J and delivers 3000 J of external work. The working fluid, metal and other internal stores are included in the system boundary. No matter crosses it.

Diagram shown in the original SVG worksheet.

  1. Find the assembly energy change and the measured work-to-heat-input ratio.

  2. If a later complete cycle has the same input and rejected heats but zero storage change, find its work and efficiency. Explain why the startup ratio cannot be substituted for the steady cyclic efficiency.

Question 9 - Solution

Startup energy storage prevents treating a transient interval as a cycle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔU=Qh−Qc−W=10000−6000−3000=1000J.\Delta U=Q_h-Q_c-W=10000-6000-3000=1000\,\mathrm J.W/Qh=.300.W/Q_h=.300.

(b) Interpretation and further calculation

ΔUcycle=0⇒Wcycle=10000−6000=4000J.\Delta U_{\mathrm{cycle}}=0\ \Rightarrow\ W_{\mathrm{cycle}}=10000-6000=4000\,\mathrm J.ηcycle=.400.\eta_{\mathrm{cycle}}=.400.During startup, part of the net heat warms the assembly. Performance comparisons must specify an interval with a complete energy balance and distinguish storage from continuous output.

Physical check

The missing 1000 J is stored energy, not automatically an unmeasured heat loss.