Derive on the basis of dimensional analysis suitable parameters to present the thrust developed by propeller.Assume that the thrust P depends upon the angular velocity speed of advance V,Diameter D dynamic viscosity $\mu$ mass density P, elasticity of fluid medium which can be denoted by the speed of sound in medium c given that the thrust p developed by the propeller is the function of angular velocity w, speed of advance v, diameter D, dynamic viscosity $\mu$. mass density p and the velocity of sound c therefore,

p-f(w,v,d,$\mu$,p,c)

or f1=(p,w,v,D,$\mu$,p,c)=0

Dimensions of variables involved are

Number of fundamental dimensions m=3

Number of dimensionless x-terms n-m=7-3=6

Hence Equation(i) can be written as:

f1$(\pi_{1},\pi_{2},\pi_{3},\pi_{4})$=0

Each $\pi$ term contains (m+1)=3+!=4 variables out of these three are repeating, selecting these repeating variables as D,v and p then we get the $\pi$-term as follows:

$\pi_{1}=D^{a_{1}}.v^{b_{1}}.p^{c_{1}}.p$

$\pi_{2}=D^{a_{2}}.v^{b_{2}}.p^{c_{2}}.w$

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

$\pi_{4}=D^{a_{4}}.v^{b_{4}}.p^{c_{4}}.c$

For $\pi_{1}$term:

$\pi_{1}=D^{a_{1}}.v^{b_{1}}.p^{c_{1}}.p$

on writing the dimensions on both sides we get

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

on equating the powers of M.L.T on both sides we get

power of M:0=$C_{1}+1$; c1=-1

power of L:0=$0_{1}+b_{1}-3c_{1}+1; \ \ \ \ \ a=a_{1}+b_{1}-3(-1)+1$

$a_{1}+b_{1}$=-4

power of T;0=-$b_{1}-2; \ \ \ \ \ b_{1}=-2$

from equation (ii); $a_{1}=-2$

$\pi_{1}=0^{-2}.v_{1}^{2}.p=\frac{p}{pv^{2}D^{2}}$;

for $\pi_{2}$ term

$\pi_{2}=D^{o}_{2}.v^{b_{2}}.p^{c_{2}}.w$

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

power of M;o=$c_{2}$

power of L:o =$a_{2}+b_{2}-3_{c_{2}}; \ a_{2}+b_{2}=0 \ \ \ (c_{2}=0)$

power of t:0=-$b_{2}-1; \ \ \ \ \ \ \ \ \ \ b_{2}=-1 \ and \ a_{2}=1$

for $\pi_{3}$ term

$\pi_{3}=D^{o_{3}}.v^{b_{3}}.p^{c_{3}}.\mu$

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

power of M:0=$C_{3}+1; \ \ \ \ \ \ \ \ C_{3}=-1$

power of L:0= $a_{3}+b_{3}-3c_{3}-1 \ \ \ \ \ \ \ a=a_{3}+b_{3}-3(-1)-1$

$a_{3}+b_{3}=-2$

power of T o=$-b_{3}-1; \ \ \ \ b_{3}=-1$

From equation (iv) $a_{3}=-2-b_{3}=-2-(-1)=-1$

$\pi_{3}=D^{-1}.v^{-1}.p^{-1}.\mu=\frac{\mu}{DVP}$

for $pi_{4}$ term

$\pi_{4}=D^{a_{4}}.v^{b_{4}}.p^{c_{4}}.c$

$M^{o}L^{o}T^{o}=L^{o_{4}}.(LT^{-1})^{b_{4}}.(ML^{-3})^{c_{4}}.LT^{-1}$

power of M;0=$C_{4}; \ \ \ i.e c_{4}=0$

power of L;0=$a_{4}+b_{4}-3_{a_{4}+1}=a_{4}+b_{4}-(3\times 0)+1$

$a_{4}+b_{4}$=-1

power of T 0=$-b_{4}$-1;

from equation (v);$a_{4}+b_{4}=-1 \ \ \ a_{4}=-1-b=-1-(-1)=0$

$\pi_{4}=D^{o}.v^{-1}.p^{0}.c=\frac{c}{v}$

substituting the values of $\pi_{1}.\pi_{2}.\pi_{3}$ and $\pi_{4}$ in equation (i) we get

f1$(\frac{p}{D^{2}V^{2}P},\frac{wD}{v},\frac{\mu}{Dvp},\frac{c}{v})$=0

or $\frac{p}{D^{2}v^{2}p}=f_{2}(\frac{wD}{v}\frac{\mu}{DVp},\frac{c}{v})$

p=$D^{2}.v^{2}.p.f_{2}(\frac{wD}{v}\frac{\mu}{Dvp},\frac{c}{v})$