Chapter 11, Problem 11.6CP

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To determine

(a)

Non-dimensionalize the given function into two pi where one is directly proportional to T.

Answer to Problem 11.6CP

The function that is dimensionless is $\frac{T}{\rho D^5{\omega }_P^2}=fcn(s)$

Explanation of Solution

Given Information:

$$

Diameter=1ft

Coupling rate=2500r/min

Density=56lbm/ft3

Concept Used:

${\pi }_1={\rho }^a{D}^b{\omega }_p^{c}T$

$\begin{array}{l}where,\\ \rho =density\\ D=diameter\\ {\omega }_{{}_p}=speed\\ T=torque\end{array}$

Calculation:

The number of pi in the groups are calculated.

$N=k-r$

Where,

k=total number of variables

r=total number if references

Calculating we have,

$N=k-r$

$=5-3\Rightarrow 2$

Dimensional analysis of the first group pi,

${\pi }_1={\rho }^a{D}^b{\omega }_p^{c}T$

$\to (1)$

Equating M coefficients

$a+1=0\Rightarrow a=-1$

Equating T coefficients

$-c-2=0\Rightarrow c=-2$

Equating T coefficients

$-3a+b+2\Rightarrow -3(-1)+b+2$

$b=-5$

From (1)

${\pi }_1={\rho }^a{D}^b{\omega }_p^{c}T$

${\pi }_1={\rho }^{-1}{D}^{-5}{\omega }_p^{-2}T\Rightarrow \frac{T}{\rho D^5{\omega }_p^2}$

$\to (2)$

For second pi group

${\pi }_1={\rho }^a{D}^b{\omega }_p^{c}s$

$\to (3)$

$M^0{L}^0{T}^0={\left(M{L}^{-3}\right)}^a{L}^b{\left(T^{-1}\right)}^c\left(M^0{L}^0{T}^0\right)$

Substituting we have,

$M^0{L}^0{T}^0={\left(M{L}^{-3}\right)}^a{L}^b{\left(T^{-1}\right)}^c\left(M{L}^2{T}^{-2}\right)$

$=\left(M^{a+1}{L}^{-3}{}^{a+b+2}{T}^{-c-2}\right)$

$=\left(M^a{L}^{-3}{}^{a+b}{T}^{-c}\right)$

By equating M and T coefficients we get

$a=0,c=0$

And so b=0

From (3) equation

${\pi }_2={\rho }^0{D}^0{\omega }_p^{0}s\Rightarrow s$

$\to (4)$

For the given function:

${\pi }_1=fcn({\pi }_2)$

From equation (3), (4):

$\frac{T}{\rho D^5{\omega }_P^2}=fcn(s)$.

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To determine

(b)

The slip from the table.

Answer to Problem 11.6CP

Slip = $15\%$

Explanation of Solution

Given Information:

$$

Coupling runs at 3600r/min and torque is given as 900ft.lbf

Concept Used:

$\begin{array}{l}\frac{T}{\rho D^5{\omega }_P^2}\\ where,\\ \rho =density\\ {\omega }_P^{}=angularvelocity\\ s=slip\end{array}$

Calculation:

Angular velocity unit conversion:

${\omega }_p=2500\times \frac{1}{60}\Rightarrow 41.7r/s$

Density unit conversion:

$\rho =56lbm/f{t}^3\Rightarrow 1.74slugs/f{t}^3$

With the equation:

$\frac{T}{\rho D^5{\omega }_P^2}=\frac{90}{1.74\times 1^5\times {41.7}^2}\Rightarrow \frac{90}{3025.67}$

$=0.0298$

From the table the slip is given as 5%:

s%	0	5	10	15	20	25
$\frac{T}{\rho D^5{\omega }_P^2}$	0	0.0298	0.0911	0.146	0.192	0.225

Calculating the given values:

$\frac{T}{\rho D^5{\omega }_P^2}=\frac{900}{1.74\times {58.3}^2\times 1^5}\Rightarrow \frac{900}{5914.069}$

=0.152

From the table we have, 15%.

Conclusion:

The slip using the table is calculated.

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To determine

(c)

The diameter for a same coupling that runs at 3000r/min.

Answer to Problem 11.6CP

Diameter = 1.36ft.

Explanation of Solution

Given Information:

$$

Power=3000r/min

s=5%

Torque=600ft-lbf

Concept Used:

$\begin{array}{l}\frac{T}{\rho D^5{\omega }_P^2}\\ where,\\ \rho =density\\ {\omega }_P^{}=angularvelocity\\ s=slip\end{array}$

Calculation:

With the equation:

$\begin{array}{l}\frac{T}{\rho D^5{\omega }_P^2}=\frac{600}{1.74*D^5*{50}^2}\\ 0.0298=\frac{600}{4350*D^5}\\ D=1.36ft\end{array}$

Conclusion:

Thus, the diameter is calculated.