Problem 1 A block of metal, maintained at 850 K, is mounted on a pedestal above
ID: 2250112 • Letter: P
Question
Problem 1 A block of metal, maintained at 850 K, is mounted on a pedestal above a platform at 350 K. The experiment being conducted requires that there be absolutely no heat transfer from the metal block through the pedestal. To achieve this, the pedestal was made into a thermocouple and the appropriate current was driven through it. The thermocouple has a heat conductance of 1 W/Kand a resistance of 1 milliohm. Its figure of merit is 0.001K-1. All these data are average values over the temperature range of interest. a. What current(s) must be driven through the thermocouple? b. What is the voltage across its terminals? c. What is the electric power required? Problem 2 We want to pump 100 W of heat from 210 K to 300 K. Two stages of thermocouples must be used. The second stage pumps 100 W from 210 K to TH2. The rst stage pumps the necessary energy from Ta = TH2 to 300 K. All thermocouples are made of the same materials whose combined characteristics are: -0.001 VK-1 The geometry of each thermocouple is optimized. The current through the thermoelec- tric pair is adjusted for maximum cooling. The current through the rst stage is not neces- sarily the same as that through the second. Within each stage, all elements are connected electrically in series. The total electric power input to the system depends critically on the choice of TH2 a. What value of TH2 minimizes the electric power consumption, PE? b. What is the value of PE? c. What is the voltage applied to each stage?Explanation / Answer
Problem 1.
a.Current driven through the thermocouple:-
Given the temperature of metal block T1=850k
Temperature of the pedestal=350k
In case of thermocouple,the voltage is called as Seebeck voltage and it is proportional or equal to the temperature difference between two surfaces. It is given in the order of millivolts.
Therefore Voltage=Temperature difference between the two surfaces
Voltage =850-350=500mV
Current=Voltage/resistance
=500/1
=500A
b.Voltage across the terminals:-
The voltage across the terminals is equal to the difference in temperature between the surfaces.
Voltage=850-350=500mV
c.Electrical power required=V*I
=500*10^-3*500
=250W
Problem 2:-
a.Value of TH2:-
The value of TH2 in order to minimize PE,power consumption is,
TH2=a//R *300
=0.0005/0.001*300
=150k
b.Value of PE=P*t
Here P=100W and t is considered to be 1.
PE=100*1=100J
c.Voltage applied in each stage=Difference between the temperature levels.
Voltage=300-210=90mV
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