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where h is the actual clearance and hd is the design clearance However, the value for the thrust force and the resistance ratio to be applied in the preceding equation remains as the values for the design condition Previously, h = hd The comparison between the values is made in Table 610 From Table 610, it is proven that the maximum stiffness is at the design condition, where the design clearance and the actual clearance are equal It can also be seen that the total pumping power increases with the increase in the clearance or the film thickness This can also be observed for the Table 610 Comparison of the design parameters due to different clearance values

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In view of the limited ways that tnesses can be assigned, only a limited number of selection models are possible. Table 20.4 lists all possible selection models if we assume that tnesses are constants and the highest tness is one. (You might now go through the list of models and determine the equilibrium conditions for each.) Note that two possible tness distributions are missing.There is no model in which tnesses are 1 s, 1, and 1 for the A1A1, A1A2, and A2 A2 genotypes, respectively (remembering that p f[A1] and q f[A2]).That model is for selection against the A1A1 homozygote. Some re ection should show that this is the same model as model 1 of table 20.4, except that the A1 allele is acting like a recessive allele. In other words, natural selection acts against A1A1 homozygotes, but not against the A1A2 and A2 A2 genotypes. Thus, the model reduces to model 1 if we treat A1 as the recessive allele and A2 as the

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Figure 1623 depicts the con guration of a magnetic structure with two air gaps Determine the equivalent circuit of the structure

Known Quantities: Structure geometry Find: Equivalent circuit diagram Assumptions: All magnetic ux is linked by the coil; the ux is con ned to the magnetic core; the ux density is uniform The reluctance of the magnetic structure is negligible Analysis:

q(1 (1

sq ) sq2)

Calculation of reluctance Knowing the magnetic path length and cross sectional area we can calculate the equivalent reluctance of the two gaps: Rgap 1 = Rgap 1 = lgap 1 lgap 1 = gap 1 Agap 1 0 Agap 1 lgap 2 lgap 2 = gap 2 Agap 2 0 Agap 2

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h = 0001 in (Maximum load) Inflow Resistance, Ri Outflow Resistance, Ro The Resistance Ratio, z Pocket Pressure, PR Virtual Bearing Area, Av Hydrostatic Thrust, T Bearing Stiffness, dT/dh Volumetric flow, Q Total pumping power, Pp Friction force, FL Friction Power, Pf Temperature Rise, T 553 lbf s in 5 4420 lbf s in 5 0125 200 lbf in 2 535 in2 10700 lbf 357 106 lbf in 1 0045 in3 s 1 00154 hp 159 lbf 012 hp 117 deg F h = hd = 0002 in (Design condition) 553 lbf s in 5 553 lbf s in 5 1000 112 lbf in 2 535 in2 6020 lbf 451 106 lbf in 1 0203 in3 s 1 007 hp 795 lbf 06 14 deg F h = 0004 in (Minimum load) 553 lbf s in 5 69 lbf s in 5 8000 25 lbf in 2 535 in2 1340 lbf 892 105 lbf in 1 0362 in3 s 1 012 hp 397 lbf 03 hp 5 deg F

Calculation of magnetic ux and ux density Note that the ux must now divide between the two legs, and that a different air-gap ux will exist in each leg Thus: 1 = 2 = N i 0 Agap 1 Ni = Rgap 1 lgap 1 N i 0 Agap 2 Ni = Rgap 2 lgap 2

Equation 20.20 is the general form of equation 20.18 for any value of s. The change in allelic frequency due to mutation can be found by using equation 20.4: q p q

f Ni + _

and the total ux generated by the coil is = 1 + 2 The equivalent circuit is shown in the bottom half of Figure 1622

where and are the rate of forward and back mutation, respectively. When equilibrium exists, the change

1 2

volumetric flow This is understandable since an increase in clearance will increase the flow of the fluid into the gap

The McGraw Hill Companies, 2001

Determine the inductance and the magnetic stored energy for the structure of Fig 1617(a) The structure is identical to that of Example 162 except for the air gap Assume that the ux density in the air gap varies sinusoidally as B(t) = B0 sin( t) Determine the induced voltage across the coil, e

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