Modern Engineering Thermodynamics

120 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms
EXAMPLE 4.9 (Continued )
From Table 4.3, we find that, for water, χ e = 77.5. Then,
ð 1
W 2 Þ electric
= − ð8:85419 × 10 − 12 N/V 2 Þð77:5Þð1:00 × 10 − 4 m 3 Þ × ½ð1:20 × 10 4 Þ 2 − 0 2 V 2 /m 2 Š/2
polarization
= − 4:94 × 10 − 6 N.m ¼ − 4:94 × 10 − 6 J
The work is negative since it went into the capacitor (the system).
Exercises
19. How much voltage would be required to store 1.00 MJ of electrical polarization work in the capacitor of Example 4.9?
Answer: V = 3.82 × 10 7 V.
20. Determine the electrical polarization work in Example 4.9 when the gap between the capacitor plates is filled with air at
20.0°C. Answer: ( 1 W 2 ) polarization = −3.42 × 10 −11 J.
21. A capacitor is made from two concentric cylinders 0.100 m long. The diameter of the outer cylinder is 0.0200 m and the
diameter of the inner cylinder is 0.0100 m. The gap between the cylinders is filled with glycerine at 25.0°C. Determine the
electrical polarization work required to charge the capacitor when 120. V is applied. Answer: ( 1 W 2 ) polarization = −1.04 × 10 −10 J.
The polarization work is a small fraction of the total energy required to charge an entire capacitor. The total
work required to charge a capacitor is divided into two parts. The largest fraction goes into increasing the electric
field strength E ! itself, and the remaining goes into the polarization of the material exposed to the electric field.
Consequently, if the thermodynamic system you are analyzing is just the material between the plates of a capacitor,
then the only polarization work is done on the material and Eq. (4.54) gives the correct electrical work
mode value. On the other hand, if you are analyzing the entire capacitor (plates and dielectric), then Eq. (4.47)
must be used to determine the correct electrical work mode value.
4.7.3 Magnetic Work
Materials are classified as either diamagnetic, paramagnetic, or ferromagnetic. Diamagnetic materials have no permanently
established molecular magnetic dipoles. However, when they are placed in a magnetic field, their molecules
develop magnetic dipoles whose magnetic field opposes the applied field (the Greek prefix dia means “to
oppose”). Paramagnetic materials have naturally occurring molecular magnetic dipoles. When placed in a magnetic
field, these dipoles tend to align themselves parallel to the field (the Greek prefix para means “beside”). Ferromagnetic
materials retain some magnetism after the removal of a magnetic field. The thermodynamic state of
these materials depends not only on the present values of their thermomagnetic properties, but also on their magnetic
history. In this sense, ferromagnetic materials have a “memory” of their previous magnetic exposure.
As in the case of an electric field, the work associated with the initiation or destruction of a magnetic field consists
of two parts. The first part is the work required to change the magnetic field itself (as though it existed
within a vacuum), and the second part is the work required to change the magnetization of the material present
inside the magnetic field.
For calculating the total work of magnetization, the generalized force is the intensive property ! H (in A/m 2 ), the magnetic
field strength, and the generalized displacement is the extensive property V ! B , the product of the system volume V
(in m 3 ) and the magnetic induction ! B (in tesla or V·s/m 2 ). Thus, assuming the magnetic field is applied to the system,
ðdWÞ magnetic = − H ! .dðV B ! Þ (4.56)
and since H ! and B ! are always parallel and point in the same direction in magnetic materials, this reduces to
ðdWÞ magnetic = − H.dðVBÞ (4.57)
where H is the magnitude of ! H and B is the magnitude of ! B . The magnetic induction can be decomposed into two
vectors as
!
B
! = μ0 H + μ ! 0 M (4.58)
where M ! is the magnetization vector per unit volume of material exposed to the magnetic field (in a vacuum, M ! is
equal to the null vector 0 ! ), and μ 0 = 4π × 10 −7 V · s/(A·m) is a universal constant called the magnetic permeability.
Inserting this information into Eq. (4.57) gives
ðdWÞ magnetic
ðtotalÞ
= − μ 0 HdðVHÞ − μ 0 HdðVMÞ (4.59)