/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 The power loss, \(\mathscr{P},\)... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 48

The power loss, \(\mathscr{P},\) in a journal bearing depends on length, \(l,\) diameter, \(D,\) and clearance, \(c,\) of the bearing, in addition to its angular speed, \(\omega\). The lubricant viscosity and mean pressure are also important. Obtain the dimensionless parameters that characterize this problem. Determine the functional form of the dependence of \(\mathscr{P}\) on these parameters.

Short Answer

Expert verified
The dimensionless parameters are \(\frac{\mu}{P_m D}\), \(\frac{l}{D}\), \(\frac{c}{D}\), and \(\frac{\mathscr{P}}{P_m D^3 \omega}\). The generic functional form of power loss can be written as: \(\Pi_4 = f(\Pi_1, \Pi_2, \Pi_3)\), where f is a general function.

Step by step solution

01

Identifying the Physical Dimensions

First, the physical dimensions of each variable need to be identified. Here is the list of variables along with their respective physical dimensions.\n\n1. Length, l : [L]\n2. Diameter, D : [L]\n3. Clearance, c : [L]\n4. Angular Speed, \(\omega\) : [T^-1]\n5. Lubricant viscosity, \(\mu\) : [M T^-1] (mass per time per distance)\n6. Mean pressure, P_m : [M L^-1 T^-2] (force per area)\n7. Power Loss, P : [M L^2 T^-3] (energy per time)
02

Applying the Buckingham Pi theorem

The Buckingham Pi theorem helps in identifying dimensionless groups. The theorem states that if there are n variables in a problem and these n variables have m fundamental dimensions, then there can be (n - m) independent dimensionless groups.\n\nIn our case, n is 7 (since we have 7 variables) and m is 3 (since our variables have dimensions in Mass [M], Length [L], and Time [T]). Therefore, through the Buckingham Pi theorem, we can find (7 - 3 = 4) independent dimensionless groups.
03

Forming the Dimensionless Parameters

Let's construct the four dimensionless groups now.\n\nDimensionless Group 1 (let's call it \(\Pi_1\))\nWe can use the viscosity \(\mu\), pressure P_m, and diameter D to form a dimensionless group in the form of \(\Pi_1 = \frac{\mu^a P_m^b}{D^c}\). By balancing the dimensions on both sides, we get a = 1, b = -1, c = -1, hence \(\Pi_1 = \frac{\mu}{P_m D}\). \n\nThe other dimensionless groups can be formed similarly and they will look like below:\n\nDimensionless Group 2 (\(\Pi_2\)): \(\Pi_2 = \frac{l}{D}\) \n\nDimensionless Group 3 (\(\Pi_3\)): \(\Pi_3 = \frac{c}{D}\) \n\nDimensionless Group 4 (\(\Pi_4\)): \(\Pi_4 = \frac{\mathscr{P}}{P_m D^3 \omega}\)
04

Determining the functional form

Lastly, once the dimensionless groups are obtained, a generic functional form of power loss can be written as: \(\Pi_4 = f(\Pi_1, \Pi_2, \Pi_3)\) where f denotes a general function. If there is a specific dependence, then f could be replaced with the specific functional form.

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Most popular questions from this chapter

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