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.

Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter28: Special Refrigeration Applications
Section: Chapter Questions
Problem 4RQ
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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
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
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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