Chapter 10, Problem 34EP

To determine

(a)

The value of convergence $\text{'}\alpha \text{'}$, and alsoto verify that $\tan \alpha \cong \alpha$ for the case.

Answer to Problem 34EP

Convergence $\text{'}\alpha \text{'}$ is −0.0005 and also $\tan \alpha$ is approx. equal to $\text{'}\alpha \text{'}$.

Explanation of Solution

Given:

  $\begin{array}{l}{h}_0=sleeperpad=\frac{1}{1000}\\ h_1=finalheightofsleeper=\frac{1}{2000}\\ lenght=1.0\end{array}$

Concept Used:

Convergence of the gap expression which is non-dimensional expression.

Also $h_0=\frac{1}{1000}in{h}_1=\frac{1}{2000}inandl=1.0in$ (1 inch 2.54 cm)

Calculation:

As non-dimensional convergence of the gap $\text{'}\alpha \text{'}$ is given by

  $\alpha =\frac{h_1-h_0}{L}$

Where,

  $\begin{array}{l}{h}_0=sleeperpad\\ h_1=finalheightofsleeper\\ l=totallenghtoftheslipper\end{array}$

Putting

  $\begin{array}{l}\frac{1}{1000}in{h}_0,\frac{1}{2000} in{h}_{1}and1forL\\ \alpha =\frac{\frac{1}{2000}-\frac{1}{1000}}{1}\\ =-0.0005\end{array}$

Conversion for convert angle from radians to degree

  $\begin{array}{l}1rad=\frac{{360}^0}{2\pi }\\ -0.0005rad=-0.0005\times \frac{{360}^0}{2\pi }\\ =-{0.0286}^0\end{array}$

Comparing $\alpha$ we get,

  $\begin{array}{l}\tan \alpha =\tan (-{0.0286}^0)\\ =-0.0005\end{array}$

As $\alpha$ is very small, approx. equal to $\alpha$

  $\tan \alpha =\alpha$

Conclusion:

Hence, the convergence of the gap $\alpha$ is -0.0005 and also the $\tan \alpha$ is approximately equal to $\tan \alpha$.

To determine

(b)

The gauge pressure half way along with slipper-pad $(atc\sim 0.5in)$ and to comment on the magnitude of gauge pressure.

Answer to Problem 34EP

The gauge pressure halfway along the slipper pad is 229.7 atm which is more than 200 atm. large value and this large force is act on small slipper pad bearing.

Explanation of Solution

Given:

Dynamic viscosity of the engine oil from table

  $\begin{array}{l}A-7at{40}^{0}c\\ \mu =0.2177kg/m.s\end{array}$

  $\begin{array}{l}{h}_0=sleeperpad=2.54\times {10}^{-5}m\\ h_1=finalheightofsleeper=1.27\times {10}^{-5}m\\ convergenceofgap\alpha =-0.0005\\ x=0.0127m\end{array}$

Concept Used:

Expression of gauge pressure halfway along the slipper pad.

Calculation:

Now, the pressure function of distance 'x' the expression is given by

  $p=p_{atm}+6\mu V_x[\frac{h_0-h_1+a_x}{(h_0+h_1){(h_0-a_x)}^2}]$

Where,

  $\begin{array}{l}{p}_{atm}=atmosphericpressure\\ \mu =vis\cos ityofoil\\ V=speedofplate\\ h_0=initialheightofslipperpad\\ h_1=heightofslipperpad\\ \alpha =non-\dim ensionalconvergenceofgap\\ z=dis\tan ce\end{array}$

Now the gauge pressure halfway along slipper pad is given by,

  $p_{guage}=p-p_{atm}$

  $p_{guage}=p_{atm}+6\mu V_x[\frac{h_0-h_1+a_x}{(h_0+h_1){(h_0-a_x)}^2}]$

Now, substituting the value in the givenexpression, we get,

  $\begin{array}{l}{p}_{guage}=2.32\times {10}^{7}Pa\\ =229.7atm(1Pa=\frac{1}{1.01\times {10}^5}atm)\end{array}$

  $p_{guage}=229.7atm$ is greater than 200 atm which is very large value and a large force acts on the small slipper pad bearing.

Conclusion:

Hence, the gauge pressure halfway along the slipper pad is 229.7 atm.

As the pressure is more than 200 atm which is very large,large force acts on small slipper pad bearing.

To determine

(c)

The plotting of P* as a function of x*

Answer to Problem 34EP

The expression of 'P' is given by

  $P=6N{o}^{6}x[\frac{5\times {10}^{-4}\times (1-x)}{(1.5\times {10}^{-3}){({10}^{-3}-5\times {10}^{-4}x)}^2}]$

Tabulating value of x from 0 to 1,

We get 'P's value and it plotted on a graph.

Explanation of Solution

Given:

Dynamic viscosity of the engine oil from table

  $\begin{array}{l}A-7at{40}^{0}c\\ \mu =0.2177kg/m.s\end{array}$

  $\begin{array}{l}{h}_0=sleeperpad=2.54\times {10}^{-5}m\\ h_1=finalheightofsleeper=1.27\times {10}^{-5}m\\ convergenceofgap\alpha =-0.0005\\ x=0.0127m\end{array}$

Concept Used:

Expression of gauge pressure halfway along sleeper pad.

Calculation:

The non-dimensional equation for the pressure exerted by slipper pad and also distance equations is given by,

  $\begin{array}{l}P=(P-P_{atm})\frac{h_0{}^2}{\mu VL}\\ x=\frac{x}{L}\end{array}$

We know that,

  $p=(p_{atm}+6\mu V_x[\frac{h_0-h_1+a_x}{(h_0+h_1){(h_0-a_x)}^2}]-P_{atm})\frac{h_0{}^2}{\mu VL}$

Substituting p and x,L for x, and also $\frac{1}{1000}$ for $h_0$, $\frac{1}{2000}$ for $h_1$, 1 in L and -0.0005 for $\alpha$,

We get

  $P=6\times {10}^{-6}x[\frac{5\times {10}^{-4}\times (1-x)}{(1.5\times {10}^{-3}){({10}^{-3}-5\times {10}^{-4}x)}^2}]$

Putting the value of from 0 to 1 we get the p's various values which is further plotted on the graph.

Conclusion:

Hence, we find out the expression of p in terms of x so that by various values of x (0 to 1), we get graph of p v/sx.

To determine

(d)

The pounds (lbf) of weight (load) this slipper pad bearing can support, if it is b = 6.0 in deep.

Answer to Problem 34EP

The load carrying capacity of slipper pad is 14460.345/bf.

Explanation of Solution

Given Information:

  $f={\displaystyle {\int }_0^{H}Pguagebdx}$

  $\begin{array}{l}initialheightoftheslipperpad=2.54\times {10}^{-5}m=h_0\\ finalheightofslipper=1.27\times {10}^{-5}m=h_1\\ velocityofplate=3.048m/s=V\\ dynamicvis\cos ity=0.2177km/s=\mu \\ length=0.0254m=l\\ convergenceofgap=-0.0005=\alpha \\ bredth=0.1524m=b\end{array}$

Concept used:

Expression of load carrying capacity of slipper pad

The load carrying capacity of slipper pad by math software is

  $f={\displaystyle {\int }_0^{H}Pguagebdx}$ ------(a)

But p gauge

  $6\mu vx[\frac{h_0-h_1+a_x}{(h_0+h_1){(h_0-a_2)}^2}]$

Substituting all P gauge value in equation (a)

We get

  $F=64268.2N$

But,

  $\begin{array}{l}1N=0.225lbf\\ F=14460.345lbf\end{array}$

Conclusion:

From the expression $f={\displaystyle {\int }_0^{H}Pguagebdx}$

We get,

The load carrying capacity of the slipper pad is $F=14460.345lbf$.