**1 Answer**

written 5.4 years ago by |

**Using Buckingham's $\pi$ obtain expression $\Delta P$**.

**Solution:**

$\Delta P$ is a function of $D,l,V,$ $\mu$,$\rho ,K$,

$\Delta P$=$f(D,l,V,\mu,\rho,k)$.....(1)

Total variables n=7

writing dimensions of each variable,

Dimension of $\Delta P$=Dimension of pressure

=$ML^{-1}T^{-2}$

D=L, l=L, $VLT^{-1}$

$\mu=ML^{-1}T^{-1}$

$\rho=ML^{-3}$

K=L

Number of fundamental dimensions M=3

No of $\pi$ terms =n-m

=7-3=4

Now eqn (1) can be grouped in 4-terms as

f$_{1}(\pi_{1},\pi_{2},\pi_{3},\pi_{4})$=0...(2)

Each $\pi$ term contains m+1 or 3+1=4 variables out of four variables three are repeating variables choosing D,V,P as the repeating variables, We have four $\pi$ terms as

$\pi_{1}=D^{a_{1}}.v^{b_{1}}.\rho^{c_{1}}.\Delta P$

$\pi_{2}=D^{a_{2}}.v^{b_{2}}.\rho^{c_{2}} .l$.

$\pi_{3}=D^{a}_{3}.v^{b_{3}}.\rho^{c_{3}}.\mu$

$\pi_{4}=D^{a}_{4}.v^{b_{4}}.\rho^{c_{4}} .K$

**First $\pi$ term:-**

$\pi_{1}=D^{a_{1}}.V^{b_{1}}.\rho^{c_{1}} \Delta P$

Substituting dimensions on both sides

$M^{0}L^{0}T^{0}=L^{a_{1}}.(LT^{-1})^{b_{1}}.(ML^{-3})^{l_{1}}.ML^{-1}T^{-2}$

Equating the power of M,L,T on both sides

Power of M, $0=C_{1}$+1 $C_{1}$=-1

Power of L, $0=a_{1}+b_{1}-3c_{1}-1$

a$_{1}=-b_{1}+3c_{1}+1$

=2-3+1=0

Power of T, $0=-b_{1}-2 b_{1}$=-2

Substituting the values of a$_{1} b_{1}$ and C$_{1}$ in $\pi_{1}$

$\pi_{1}=D^{0}.v^{-1}.\rho^{-1} \Delta P$

= $\frac{\Delta P}{\rho v^{2}}$

**Second $\pi$ term:-**

$\pi_{2}=D^{a_{2}}.v^{b_{2}}.\rho^{c_{2}}.l$

Substituting dimensions on both sides

$M^{0}L^{0} T^{0}=L^{a_{2}}.(LT^{-1})^{b_{2}}.(ML^{-3})^{c_{2}}$

equating powers of M,L,T on both sides

Power of M, $0=c_{2} C_{2}$=0

Power of L, $0=a_{2}-b_{2}-3c_{2}+1 a_{2}=b_{2}+3c_{2}-1$ = -1

Power of T, $a=-b_{2}$ $ \ \ \ $ $b_{2}$=0

Susbtituting the values of $a_{2},b_{2},c_{2}$ in $\pi_{2}$

$\pi_{2}=o^{-1}.V^{0}.\rho^{0}.l=\frac{l}{D}$

**Third $\pi$ term**

$\pi_{3}=D^{a_{3}}.v^{b_{3}}.\rho^{c_{3}} . \mu$

Substituting dimensions on both sides

$M^{0}L^{0}T^{0}=L^{a_{3}}.(LT^{-1})^{b_{3}}.(ML^{-3})^{c_{3}}ML^{-1}T^{-1}$

equating the power of M,L,T on both sides

Power of M, $0=c_{3}+1 c_{3}$= - 1

power of L, $0=a_{3}+b_{3}+C_{3}(3)-1$ $a_{3}=-b_{3}+3c_{3}+1$ =1-3+1 = -1

power of T, $0=-b_{3}-1$ $ \ \ \ $ $b_{3}$=-1

Substituting the values of $a_{3}b_{3} $ and $c_{3}$ in $\pi_{3}$

$\pi_{3}= D^{-1}.V^{-1}.\rho^{-1}.\mu$

=$\frac{\mu}{DV \rho}$

**Fourth $\pi$ term:-**

$\pi_{4}=D^{a4}.v^{b4}.\rho^{c4}.K$

$M^{0}L^{0}T^{0}=L^{a4}.(LT^{-1})^{b4}.(ml^{-3})^{c4}$.L

equating the power of M,L,T on both sides

Power of M, 0=c4 c4=0

power of L, 0= a4-b4-3c4+1 a4=b4+3c4-1 = -1

power of T, 0 = -b4 b4=0

Susbtituting the value of a4,b4,c4 in $\pi_{4}$

$\pi_{4}=D^{-1}.y^{0}.\rho^{0}$.K

=$\frac{k}{D}$

Substituting the values of $\pi_{1}, \pi_{2}, \pi_{3}$ and $\pi_{4}$ in eqn (2) we get

f$(\frac{\triangle P}{pv^{2}}, \frac{l}{D}. \frac{\mu}{DVP}, \frac{k}{D}$)=0

or

$\frac{\triangle P}{PV^{2}}=\phi[\frac{l}{D}, \frac{\mu}{Dvp} \frac{K}{D}]$

the expression for hf, from experiments it was observed that pressure difference $\Delta P$ is a linear function of $\frac{l}{D}$ and hence it is taken out

$\frac{\Delta P}{\rho V^{2}}=\frac{l}{D} \phi[\frac{\mu}{DV \rho},\frac{K}{D}]$

$\frac{\Delta P}{p\rho}=V^{2}.\frac{l}{D} \phi[\frac{\mu}{DV \rho}, \frac{K}{D}]$

Dividing by g to both sides we have

$\frac{\Delta P}{\rho g}=\frac{V^{2}.l}{g.D} \phi[\frac{\mu}{DV \rho},\frac{K}{D}]$

Now $\phi[\frac{\mu}{DV \rho} \frac{K}{D}]$ contains two terms

First one is $\frac{\mu}{DV |rho}$ which is $\frac{1}{Re}$ and second one is $\frac{K}{D}$ which is called roughness factor

$\frac{\Delta P}{\rho g}=\frac{4f}{2}. \frac{V^{2}l}{gD} \ Since \ ----{f=\phi(\frac{\mu}{DV \rho},\frac{K}{D})}$

multiplying or dividing by any constant does not change the character of $\pi$ terms

$\frac{\Delta P}{\rho g}=hf=\frac{4flV^{2}}{D \times 2g}$