In this problem, you will consider an "unusual" solid-state refrigeration device that achieves cooling by varying the magnetization of the working mate- rial (gadolinium sulphate octahydrate) at constant- pressure conditions. By contrast, most refrigeration cycles achieve cooling based on manipulating pres- sures of the working fluid. The device functions roughly as shown at the right wherein a paramag- netic wheel is rotated through different regions: i) a low-temperature region (12) ii) a low-magnetic field region (23) iii) a high-temperature region (34) iv) a high-magnetic field region (4 → 1) In the figure, the thermodynamic state at the point of entry/exit from each region is shown.
In this problem, you will consider an "unusual" solid-state refrigeration device that achieves cooling by varying the magnetization of the working mate- rial (gadolinium sulphate octahydrate) at constant- pressure conditions. By contrast, most refrigeration cycles achieve cooling based on manipulating pres- sures of the working fluid. The device functions roughly as shown at the right wherein a paramag- netic wheel is rotated through different regions: i) a low-temperature region (12) ii) a low-magnetic field region (23) iii) a high-temperature region (34) iv) a high-magnetic field region (4 → 1) In the figure, the thermodynamic state at the point of entry/exit from each region is shown.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Assuming that the refrigeration device can operate reversibly, using the data below, estimate the
amount of heat absorbed from the cold-temperature region qC and the heat rejected to the high-temperature region qH .
Hint: it may be helpful to note that the temperature changes for certain processes are small(e.g., 1 → 2 and
3 → 4). In such situations, one might presume qrev ≈ ∫sfs0 Tds where T is the mean temperature of the process.
![In this problem, you will consider an "unusual"
solid-state refrigeration device that achieves cooling
by varying the magnetization of the working mate-
rial (gadolinium sulphate octahydrate) at constant-
pressure conditions. By contrast, most refrigeration
cycles achieve cooling based on manipulating pres-
sures of the working fluid. The device functions
roughly as shown at the right wherein a paramag-
netic wheel is rotated through different regions:
• i) a low-temperature region (1 → 2)
ii) a low-magnetic field region (2 → 3)
iii) a high-temperature region (3 → 4)
iv) a high-magnetic field region (4 → 1)
In the figure, the thermodynamic state at the point of
entry/exit from each region is shown.
T₁ = 1.1 K
B₁ = 0.9 Tesla
Low temperature "reservoir"
Qc
T₂ = 0.9 K
B₂ = 0.0 Tesla
high
magnetic field
rotating
magnetic
wheel
low
magnetic field
T₁ = 9.5 K
B₁ = 6.4 Tesla
QH
high temperature "reservoir"
T3 = 8.0 K
B3 = 1.6 Tesla](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0ff916c-a9ca-4791-8d5c-b57cb6fa675f%2F9d3b559a-c6b0-41ee-a605-588eb6eb348a%2Fktrvom3_processed.png&w=3840&q=75)
Transcribed Image Text:In this problem, you will consider an "unusual"
solid-state refrigeration device that achieves cooling
by varying the magnetization of the working mate-
rial (gadolinium sulphate octahydrate) at constant-
pressure conditions. By contrast, most refrigeration
cycles achieve cooling based on manipulating pres-
sures of the working fluid. The device functions
roughly as shown at the right wherein a paramag-
netic wheel is rotated through different regions:
• i) a low-temperature region (1 → 2)
ii) a low-magnetic field region (2 → 3)
iii) a high-temperature region (3 → 4)
iv) a high-magnetic field region (4 → 1)
In the figure, the thermodynamic state at the point of
entry/exit from each region is shown.
T₁ = 1.1 K
B₁ = 0.9 Tesla
Low temperature "reservoir"
Qc
T₂ = 0.9 K
B₂ = 0.0 Tesla
high
magnetic field
rotating
magnetic
wheel
low
magnetic field
T₁ = 9.5 K
B₁ = 6.4 Tesla
QH
high temperature "reservoir"
T3 = 8.0 K
B3 = 1.6 Tesla
![T(K)
Temperature,
10
B= 10 Tesla
B8.Nesla
0.0R
R = 8.314 J/(mol K)
B
0.5R
iso-magnetic field curves
6.4 Tesla
B = 4,9 Tesla
B 3.6 Tesla
B 2.5 Tesla
B=1.6 Tesla
B=0.9 Tesla
1.0R
Entropy, S
1.5R
B=0.4 Tesla
2.0R
B = 0.0 Tesla](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0ff916c-a9ca-4791-8d5c-b57cb6fa675f%2F9d3b559a-c6b0-41ee-a605-588eb6eb348a%2Ffvvcbzh_processed.png&w=3840&q=75)
Transcribed Image Text:T(K)
Temperature,
10
B= 10 Tesla
B8.Nesla
0.0R
R = 8.314 J/(mol K)
B
0.5R
iso-magnetic field curves
6.4 Tesla
B = 4,9 Tesla
B 3.6 Tesla
B 2.5 Tesla
B=1.6 Tesla
B=0.9 Tesla
1.0R
Entropy, S
1.5R
B=0.4 Tesla
2.0R
B = 0.0 Tesla
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