Work, Energy, and Emf — Question 10

Section 27-2 · Physics pages 4350–4351

Question 10 A thermal charge pump has an emf ceiling imposed by its heat engine A heat engine operating between 600 K and 300 K drives a lossless electrical generator. It absorbs heat at 120 W and drives current 2.00 A. Ignore all dissipation except the thermodynamic efficiency limit. low V high V positive charge non-electrostatic pump heat engine drives generator Schematic; labels, not drawing lengths, specify the data. (a) Find the largest possible generator emf. (b) If the measured emf is 24.0 V, find engine efficiency and heat-rejection rate, and compare with the reversible limit.
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Question 10 - Solution A thermal charge pump has an emf ceiling imposed by its heat engine low V high V positive charge non-electrostatic pump heat engine drives generator (a) Method and calculation ηC = 1 −Tc/Th = 1 −300/600 = .500. EI ≤ηC ˙Qh ⇒E ≤(.500)(120)/2.00 = 30.0 V. (b) Interpretation and further calculation Pe = EI = 48.0 W, η = 48.0/120 = .400 < .500. ˙Qc = ˙Qh −Pe = 120 −48 = 72.0 W. Physical check Even a lossless generator cannot overcome the efficiency limit of the engine supplying its mechanical work.

Question 10

A thermal charge pump has an emf ceiling imposed by its heat engine

A heat engine operating between 600 K and 300 K drives a lossless electrical generator. It absorbs heat at 120 W and drives current 2.00 A. Ignore all dissipation except the thermodynamic efficiency limit.

Diagram shown in the original SVG worksheet.

  1. Find the largest possible generator emf.

  2. If the measured emf is 24.0 V, find engine efficiency and heat-rejection rate, and compare with the reversible limit.

Question 10 - Solution

A thermal charge pump has an emf ceiling imposed by its heat engine

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ηC=1−Tc/Th=1−300/600=.500.\eta_C=1-T_c/T_h=1-300/600=.500.ℰI≤ηCQ̇h⇒ℰ≤(.500)(120)/2.00=30.0V.\mathcal E I\le\eta_C\dot Q_h\Rightarrow\mathcal E\le(.500)(120)/2.00=30.0\,\mathrm V.

(b) Interpretation and further calculation

Pe=ℰI=48.0W,η=48.0/120=.400<.500.P_e=\mathcal EI=48.0\,\mathrm W,\quad\eta=48.0/120=.400<.500.Q̇c=Q̇h−Pe=120−48=72.0W.\dot Q_c=\dot Q_h-P_e=120-48=72.0\,\mathrm W.

Physical check

Even a lossless generator cannot overcome the efficiency limit of the engine supplying its mechanical work.