Superconductors — Question 9

Section 26-8 · Physics pages 4308–4309

Question 9 A lossless conductor can still require substantial refrigeration power A cryogenic device has zero DC resistive loss but a constant heat leak ˙Qc = 12.0 W at Tc = 4.00 K. Ambient temperature is Th = 300 K. The refrigerator achieves 10.0 percent of the Carnot coefficient of performance, COPC = Tc/(Th −Tc). cold device ambient heat rejection W Schematic; labels, not drawing lengths, specify the data. (a) Find minimum reversible input power and actual refrigerator input power. (b) Compare actual input with a room-temperature alternative dissipating 6.00 kW, and find the largest heat leak that still uses less power than that alternative.
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Question 9 - Solution A lossless conductor can still require substantial refrigeration power cold device ambient heat rejection W (a) Method and calculation COPC = 4/296 = 1/74, Wmin = ˙Qc/COPC = 888 W. COP = .10/74, Wactual = 8880 W. (b) Interpretation and further calculation Wactual/6000 = 1.48. ˙Qc < 6000(.10/74) = 8.10811 W for a net power saving. Physical check Zero electrical resistance does not imply zero system power; maintaining the cold environment has an energy cost.

Question 9

A lossless conductor can still require substantial refrigeration power

A cryogenic device has zero DC resistive loss but a constant heat leak Q̇c=12.0\dot Q_c=12.0 W at Tc=4.00T_c=4.00 K. Ambient temperature is Th=300T_h=300 K. The refrigerator achieves 10.0 percent of the Carnot coefficient of performance, COPC=Tc/(Th−Tc)\mathrm{COP}_C=T_c/(T_h-T_c).

Diagram shown in the original SVG worksheet.

  1. Find minimum reversible input power and actual refrigerator input power.

  2. Compare actual input with a room-temperature alternative dissipating 6.00 kW, and find the largest heat leak that still uses less power than that alternative.

Question 9 - Solution

A lossless conductor can still require substantial refrigeration power

Diagram shown in the original SVG worksheet.

(a) Method and calculation

COPC=4/296=1/74,Wmin=Q̇c/COPC=888W.\mathrm{COP}_C=4/296=1/74,\quad W_{\min}=\dot Q_c/\mathrm{COP}_C=888\,\mathrm W.COP=.10/74,Wactual=8880W.\mathrm{COP}=.10/74,\quad W_{\mathrm{actual}}=8880\,\mathrm W.

(b) Interpretation and further calculation

Wactual/6000=1.48.W_{\mathrm{actual}}/6000=1.48.Q̇c<6000(.10/74)=8.10811Wfor a net power saving.\dot Q_c<6000(.10/74)=8.10811\,\mathrm W\quad\text{for a net power saving}.

Physical check

Zero electrical resistance does not imply zero system power; maintaining the cold environment has an energy cost.