Solutions to Chapter8

Chapter 8

CALCULUS OF VARIATION

Problems 8.1 to 8.5 | Problems 8.6 to 8.8

8-6(9). For steady flow of an incompressible fluid in a square duct, the equation of continuity and motion simplify to a partial differential equation which requires an elaborate analytical solution involving the sum of an infinite series. An approximate solution can be obtained using the calculus of variation which gives a simple equation that predicts the volumetric flow rate within 1% of the exact solution. The equation that describes the flow at a point z along the axis of the duct is:

where v is the axial velocity; µ is the viscosity of the fluid; dP/dz is the pressure gradient, a constant; and a is the length of one-half of the side of the duct. The boundary conditions are that there is no slip at the wall, i.e. v = 0 at x = a for 0 < y < a and at y = a for 0 < x < a.

a. Show that the integral to be minimized corresponding to the differential equation is:

b. A simple approximation to v is given by the following equation which satisfies the boundary conditions.

v = A(a² - x²)(a² - y²)

Using this equation perform the integration of the equation in part a to obtain the following result.

I[v] = (64/45)A²a⁸ - (8/9µ)(dP/dz)Aa⁶

c. Find the value of A that minimizes I.

d. If the mass flow rate, w, through the duct is given by the following equation:

Show that the following result is obtained:

w = 0.556 r(dP/dz) a⁴/µ

The analytical solution has the same form, but the coefficient is 0.560. Thus, the approximate solution is within 1% of the exact solution.

Solution:

a. For

The extension of Euler equation for this case of two independent variables was given by equation (8-55).

This equation can be written as:

The integrand for F in the above equation is:

F = (¶v/¶x)² + (¶v/¶y)² - (2/µ)(dP/dz) v

The partial derivatives appearing in the Euler equation are:

¶F/¶v_x = 2(¶v/¶x) , ¶F/¶v_y = 2(¶v/¶y)

¶F/¶v = - (2/µ)(dP/dz)

Substituting the above expressions into the Euler equation gives:

This is the differential equation required.

v = A(a² - x²)(a² - y²)

v_x = A(-2x)(a² - y²)

v_x² = 4A²x²(a⁴ - 2a²y² + y⁴)

v_y = A(-2y)(a² - x²)

v_y² = 4A²y²(a⁴ - 2a²x² + x⁴)

Performing the integration in steps as follows, gives:

= 4A²[8a⁸/45]

Similarly:

= - (2/µ)(dP/dz) A(4a⁶/9) = - (8a⁶A/9µ)(dP/dz)

I[v] = (64/45) A²a⁸ - (8a⁶A/9µ)(dP/dz)

A = (5/16)(dP/dz)(1/µa²)

d. To compute the mass flow rate, the following integration is required:

w = 4 A 4a⁶/9

Using the value of A determined in part c gives:

w = 4r(4/9)a⁶ [5/16 (dP/dz)(1/µ a²)] =(5/9)(ra⁴/µ)(dP/dz)

w = 0.556 (r a⁴/µ)(dP/dz)

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8-7. In a production schedule problem, the production rate is to be changed from 100 units per unit time to 300 units per unit time in ten time units, i.e. p(0) = 100 and p(10) = 300. The costs as a function of time are associated with changes in machines, personnel and raw materials. For this simple problem this cost is given as:

C(t) = 2(p')² + 4tp'

where p' = dp/dt. Determine the production rate as a function of time that maximizes the cost over the time period.

Solution:

The Euler equation for this problem is:

4p" + 4 = 0

p" = -1

Integrating this simple second-order ordinary differential equation gives:

p' = -t + C₁

p = -t²/2 + C₁t + C₂

Solving for C₁ and C₂ with the boundary conditions give:

100 = 0 + C₁(0) + C₂ and C₂ = 100

300 = -100/2 + 10C₁ + C₂ and C₁ = 25

C₁ = 25

Substituting, the particular solution is:

p = -t²/2 + 25t + 100

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8-8(1). Determine the deflection in an uniformly-loaded, cantilever beam, y(x), where y is the deflection as a function of distance down the beam from the wall (x = 0)to the end of the beam (x = L). The total potential energy of the system to be minimized is given by:

where E is the bending rigidity and q is the load. The boundary conditions at the wall end are y(0) = y'(0) = 0 and at the unsupported end of y"(L) = y'''(L) = 0.

Solution:

The Euler-Poisson equation for m = 2 is: