Engineering Example 4

6 Engineering Example 4

6.1 Crank used to drive a piston

Introduction

A crank is used to drive a piston as in Figure 20.

Figure 20 :

{ Crank used to drive a piston}

Problem

The angular velocity of the crankshaft is the rate of change of the angle $θ, ω = d θ ∕ d t .$ The piston moves horizontally with velocity $v_{p}$ and acceleration $a_{p}$ ; $r$ is the length of the crank and $l$ is the length of the connecting rod. The crankpin performs circular motion with a velocity of $v_{c}$ and centripetal acceleration of $ω^{2} r$ . The acceleration $a_{p}$ of the piston varies with $θ$ and is related by

$a_{p} = ω^{2} r (cos θ + \frac{r cos 2 θ}{l})$

Find the maximum and minimum values of the acceleration $a_{p}$ when $r = 150$ mm and $l = 375$ mm.

Mathematical statement of the problem

We need to find the stationary values of $a_{p} = ω^{2} r (cos θ + \frac{r cos 2 θ}{l})$ when $r = 150$ mm and $l = 375$ mm. We do this by solving $\frac{d a_{p}}{d θ} = 0$ and then analysing the stationary points to decide whether they are a maximum, minimum or point of inflexion.

Mathematical analysis .

$a_{p} = ω^{2} r (cos θ + \frac{r cos 2 θ}{l})$ so $\frac{d a_{p}}{d θ} = ω^{2} r (- sin θ - \frac{2 r sin 2 θ}{l})$ .

To find the maximum and minimum acceleration we need to solve

$\frac{d a_{p}}{d θ} = 0 \Leftrightarrow ω^{2} r (- sin θ - \frac{2 r sin 2 θ}{l}) = 0.$

$sin θ + \frac{2 r}{l} sin 2 θ = 0 \Leftrightarrow sin θ + \frac{4 r}{l} sin θ cos θ = 0$

$\Leftrightarrow sin θ (1 + \frac{4 r}{l} cos θ) = 0$

$\Leftrightarrow sin θ = 0 or cos θ = - \frac{l}{4 r}$ and as $r = 150$ mm and $l = 375$ mm

$\Leftrightarrow sin θ = 0$ or $cos θ = - \frac{5}{8}$

CASE 1: $sin θ = 0$

If $sin θ = 0$ then $θ = 0$ or $θ = π$ . If $θ = 0$ then $cos θ = cos 2 θ = 1$

so $a_{p} = ω^{2} r (cos θ + \frac{r cos 2 θ}{l}) = ω^{2} r (1 + \frac{r}{l}) = ω^{2} r (1 + \frac{2}{5}) = \frac{7}{5} ω^{2} r$

If $θ = π$ then $cos θ = - 1, cos 2 θ = 1$ so

$a_{p} = ω^{2} r (cos θ + \frac{r cos 2 θ}{l}) = ω^{2} r (- 1 + \frac{r}{l}) = ω^{2} r (- 1 + \frac{2}{5}) = - \frac{3}{5} ω^{2} r$

In order to classify the stationary points, we differentiate $\frac{d a_{p}}{d θ}$ with respect to $θ$ to find the second derivative:

$\frac{d^{2} a_{p}}{d θ^{2}} = ω^{2} r (- cos θ - \frac{4 r cos 2 θ}{l}) = - ω^{2} r (cos θ + \frac{4 r cos 2 θ}{l})$

At $θ = 0$ we get $\frac{d^{2} a_{p}}{d θ^{2}} = - ω^{2} r (1 + \frac{4 r}{l})$ which is negative.

So $θ = 0$ gives a maximum value and $a_{p} = \frac{7}{5} ω^{2} r$ is the value at the maximum.

At $θ = π$ we get $\frac{d^{2} a_{p}}{d θ^{2}} = - ω^{2} r (- 1 + \frac{4}{l}) = - ω^{2} r (\frac{3}{5})$ which is negative.

So $θ = π$ gives a maximum value and $a_{p} = - \frac{3}{5} ω^{2} r$

CASE 2: $cos θ = - \frac{5}{8}$

If $cos θ = - \frac{5}{8}$ then $cos 2 θ = 2 {cos}^{2} θ - 1 = 2 {(\frac{5}{8})}^{2} - 1$ so $cos 2 θ = - \frac{7}{32}$ .

$a_{p} = ω^{2} r (cos θ + \frac{r cos 2 θ}{l}) = ω^{2} r (- \frac{5}{8} + - \frac{7}{32} \times \frac{2}{5}) = \frac{57}{80} ω^{2} r .$

At $cos θ = - \frac{5}{8}$ we get $\frac{d^{2} a_{p}}{d θ^{2}} = ω^{2} r (- cos θ - \frac{4 r cos 2 θ}{l}) = ω^{2} r (\frac{5}{8} + \frac{4 r}{l} \frac{7}{32})$ which is positive.

So $cos θ = - \frac{5}{8}$ gives a minimum value and $a_{p} = - \frac{57}{80} ω^{2} r$

Thus the values of $a_{p}$ at the stationary points are:-

$\frac{7}{5} ω^{2} r$ (maximum), $- \frac{3}{5} ω^{2} r$ (maximum) and $- \frac{57}{80} ω^{2} r$ (minimum).

So the overall maximum value is $1.4 ω^{2} r = 0.21 ω^{2}$ and the minimum value is

$- 0.7125 ω^{2} r = - 0.106875 ω^{2}$ where we have substituted $r = 150$ mm ( $=$ 0.15 m) and $l = 375$ mm ( $= 0.375$ m).

Interpretation

The maximum acceleration occurs when $θ = 0$ and $a_{p} = 0.21 ω^{2}$ .

The minimum acceleration occurs when $cos θ = - \frac{5}{8}$ and $a_{p} = - 0.106875 ω^{2}$ .

Exercises

Locate the stationary points of the following functions and distinguish among them as maxima, minima and points of inflection.
1. $f (x) = x - ln | x |$ . [Remember $\frac{d}{d x} (ln | x |) = \frac{1}{x}$ ]
2. $f (x) = x^{3}$
3. $f (x) = \frac{(x - 1)}{(x + 1) (x - 2)} - 1 < x < 2$
A perturbation in the temperature of a stream leaving a chemical reactor follows a decaying sinusoidal variation, according to
$T (t) = 5 exp (- a t) sin (ω t)$

where $a$ and $ω$ are positive constants.
1. Sketch the variation of temperature with time.
2. By examining the behaviour of $\frac{d T}{d t}$ , show that the maximum temperatures occur at times of $({tan}^{- 1} (\frac{ω}{a}) + 2 π n) ∕ ω$ .

1. $\frac{d f}{d x} = 1 - \frac{1}{x} = 0$ when $x = 1$
  $\frac{d^{2} f}{d x^{2}} = \frac{1}{x^{2}} \frac{d^{2} f}{d x^{2}} |_{x = 1} = 1 > 0$
  
  $∴ x = 1, y = 1$ locates a local minimum.
2. $\frac{d f}{d x} = 3 x^{2} = 0$ when $x = 0$ $\frac{d^{2} f}{d x^{2}} = 6 x = 0$ when $x = 0$
  However, $\frac{d f}{d x} > 0$ on either side of $x = 0$ so $(0, 0)$ is a point of inflection.
3. $\frac{d f}{d x} = \frac{(x + 1) (x - 2) - (x - 1) (2 x - 1)}{(x + 1) (x - 2)}$
  This is zero when $(x + 1) (x - 2) - (x - 1) (2 x - 1) = 0$ i.e. $x^{2} - 2 x + 3 = 0$
  
  However, this equation has no real roots (since $b^{2} < 4 a c$ ) and so $f (x)$ has no stationary points. The graph of this function confirms this:
  
  Nevertheless $f (x)$ does have a point of inflection at $x = 1, y = 0$ as the graph shows, although at that point $\frac{d y}{d x} \neq 0.$
2. $\frac{d T}{d t} = 0$ implies $tan ω t = \frac{ω}{a}$ , so $tan ω t > 0$ and
  $ω t = {tan}^{- 1} (\frac{ω}{a}) + k π$ , $k$ integer
  
  Examination of $\frac{d^{2} T}{d t^{2}}$ reveals that only even values of $k$ give $\frac{d^{2} T}{d t^{2}} < 0$ for a maximum so setting $k = 2 n$ gives the required answer.

6 Engineering Example 4

6.1 Crank used to drive a piston

Exercises

Answer