tutorial3.rst

Example: Journal Bearing

If one were interested in studying the frictional coefficient of the bearing, then we must consider the variables/parameters that may influence it.

• bearing length,
• bearing diameter,
• bearing load,
  • The load on the bearing is represented in terms of the average bearing pressure where is the actual load of bearing.
• rotating speed,
  • Assume that the resulting rotating speed of the bearing is the constant average speed N.
• viscosity of lubricating oil,
  • This is the viscosity at equillibrium temperature — the bearing rotating at an average of produces heat which is conducts and convects.
• clearance between bearing and journal,
• bearing moment,
  • Load applied to the shaft passing through the bearing results in bearing moment.

We can therefore start our study with the assumption that the frictional coefficient of the bearing is a function of the above seven variables resulting in the function value

But, , and have the same dimensions. Then, and are dimensionless. Therefore, if we temporarily disregard the variables and , then we can reduce seven variables to five.

Hence, the tentative function for the derivation is such that its value is given by

Since,

quantity symbol	quantity name	unit (say, SI)	dimensions
	pressure
	moment
	diameter
	viscosity
	speed

the dimensional system for the problem is MLT-system.

We can now proceed with the steps (four) for deriving the dimensionless products.

1. Generate Dimensional Formula for All the Terms

The derivation of the dimensionless products will be based on the reduced where the parent function depends on the independent five variables. The generation of the dimensional matrix is follows some preceding setup steps.

1.1. Setup for Generation

1.1.1. Definitions setup

Since our problem uses MLT dimensional system

(def varpars [{:symbol "x", :quantity "mass"}
              {:symbol "y", :quantity "length"}
              {:symbol "t", :quantity "time"}])

1.1.2. Expressions and equation

We express the variables for the unknown function as

(def manifold_eqn [{:name "term-P", :eqn {:term1 "x^(1)*y^(-1)*t^(-2)"}}
                   {:name "term-M", :eqn {:term1 "x^(1)*y^(2)*t^(-2)"}}
                   {:name "term-D", :eqn {:term1 "y^(1)"}}
                   {:name "term-mu", :eqn {:term1 "x^(1)*y^(-1)*t^(-1)"}}
                   {:name "term-N", :eqn {:term1 "t^(-1)"}}])

1.2. Getting the Dimensional Formula

The dimensional formula all the terms are

=> (pprint (formula-eqn-side-manifold varpars manifold_eqn))
  [{:quantity "term-P", :dimension "[M^(1)*T^(-2)*L^(-1)]"}
   {:quantity "term-M", :dimension "[M^(1)*T^(-2)*L^(2)]"}
   {:quantity "term-D", :dimension "[L^(1)]"}
   {:quantity "term-mu", :dimension "[M^(1)*T^(-1)*L^(-1)]"}
   {:quantity "term-N", :dimension "[T^(-1)]"}]

1.3 Standardize All the Generated Dimensional Formula

We add the above dimensional formulae into the standard_formula

=> (update-sformula (formula-eqn-side-manifold varpars manifold_eqn))
  [{:quantity "volume", :dimension "[L^(3)]"}
   {:quantity "frequency", :dimension "[T^(-1)]"}
   {:quantity "velocity", :dimension "[L^(1)*T^(-1)]"}
   {:quantity "acceleration", :dimension "[L^(1)*T^(-2)]"}
   {:quantity "force", :dimension "[M^(1)*L^(1)*T^(-2)]"}
   ...
   {:quantity "term-N", :dimension "[T^(-1)]"}
   {:quantity "term-mu", :dimension "[M^(1)*T^(-1)*L^(-1)]"}
   {:quantity "term-D", :dimension "[L^(1)]"}
   {:quantity "term-M", :dimension "[M^(1)*T^(-2)*L^(2)]"}
   {:quantity "term-P", :dimension "[M^(1)*T^(-2)*L^(-1)]"}]

1.4. Definitions setup for dimensional matrix

(def varpars2 [{:symbol "P", :quantity "term-P"}
               {:symbol "M", :quantity "term-M"}
               {:symbol "D", :quantity "term-D"}
               {:symbol "mu", :quantity "term-mu"}
               {:symbol "N", :quantity "term-N"}])

2. Generate Dimensional Matrix

=> (view-matrix (generate-dimmat varpars2))
  [-1N 2N 1N -1N 0]
  [-2N -2N 0 -1N -1N]
  [1N 1N 0 1N 0]
  Size -> 3 x 5

3. Get the Homogeneous equation of the Dimensional Matrix

3.1. Get the augmented matrix of the dimensional matrix

=> (view-matrix (get-augmented-matrix (generate-dimmat varpars2)))
  [1N -1N 0 1N -2N]
  [0 -1N -1N 2N 2N]
  [0 1N 0 -1N -1N]
  Size -> 3 x 5

3.2. Solve the augmented matrix

=> (view-matrix (solve (get-augmented-matrix (generate-dimmat varpars2))))
  [1N 0N 0N 0N -3N]
  [0 1N 0N -1N -1N]
  [0 0N 1N -1N -1N]
  Size -> 3 x 5

3.3. Get the solution matrix

=> (view-matrix (get-solution-matrix (solve (get-augmented-matrix (generate-dimmat varpars2)))))
  [1 0 0N -1N -1N]
  [0 1 -3N -1N -1N]
  Size -> 2 x 5

This is a 2 × 5 matrix. Therefore, two dimensionless products will be derived.

We can put all these individual steps involving matrix into one coding step such that it returns the solution matrix.

=> (def solution_matrix (get-solution-matrix
                            (solve
                                (get-augmented-matrix
                                    (generate-dimmat varpars2)))))
=> (view-matrix solution_matrix)
  [1 0 0N -1N -1N]
  [0 1 -3N -1N -1N]
  Size -> 2 x 5

4. Get Dimensionless Products

=> (def all_dimless (get-dimensionless-products solution_matrix varpars2))

=> (pprint all_dimless)
  [{:symbol "pi0", :expression "P^(1)*mu^(-1)*N^(-1)"}
   {:symbol "pi1", :expression "M^(1)*D^(-3)*mu^(-1)*N^(-1)"}]

=> (get-pi-expression all_dimless "pi0")
  "P^(1)*mu^(-1)*N^(-1)"

Note that these two dimensionless products are derived from the tentative function where we temporarily disregarded and .

But, and are dimensionless. Therefore, the number of products in the complete set of dimensionless products is four. Hence, the frictional coefficient has the value