Assume that thrust 'P' depends upon angular velocity 'w' speed of advance 'V' diameter 'D' dynamic viscosity '$\mu$ mass density $\rho$ elasticity of fluid medium which can be denoted by the speed of sound in the medium 'c'.

Solution:

Thrust 'P' is a function of $w,V,D,\mu , \rho, c$

$P=f(w,V,D,\mu , \rho, c)$

or

$f_{1}=(P,w,V,D,\mu,\rho,c)=0$.....(1)

Total variables=7

writing dimensions of each variable,we have

$P=MLT^{-2}, w=T^{-1}, V=LT^{-1}, D=L$

$\mu=ML^{-1}T^{-1}, \rho=ML^{-3}, c=LT^{-1}$

No. of fundamentals dimensions m=3

No of $\pi$ terms = n-m

=7-3 = 4

Hence equation (1) can be written as

$f_1(\pi_{1},\pi_{2},\pi_{3}, \pi_{4})$=0....(2)

each $\pi$ term contains m+1 i.e 3+1 =4 variables out of four variables three are repeating

$\therefore$ Choosing $D,V,\rho$ as repeating variables

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

$\pi_{2}=D^{a_{2}}.V^{b_{2}}.\rho^{c_{2}}$.w

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

$\pi_{4}=D^{a}_{4}.V^{b_{4}}.\rho^{c_{4}}$. c

First $\pi$ term:-

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

writing dimensions on both sides

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

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

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

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

$a_{1}=-b_{1}+3c_{1}-1$

=2-3-1=0

$a_{1}=-2$

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

substituting the values of $a_{1} b_{1}$ and $c_{1}$ in $\pi_{1}$

$\pi_{1}=D^{-2}.V^{-2}.\rho^{-1} P$ $\frac{P}{D^{2}v^{2}\rho}$

Second $\pi$ term:-

$\pi_{2}=D^{a_{2}}.V^{b_{2}}.\rho^{c_{2}}.w$

writing dimensions on both sides

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

equating powers of M,L,T on both sides

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

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

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

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

$\pi_{2}=D^{-1}.V^{-1}.\rho^{0}. w$

=$\frac{Dw}{V}$

Third $\pi$ term

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

writing dimensions on both sides

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

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

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

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

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:-

$\pi_{4}=D^{a4}.V^{b4}.rho^{c4}.c$

writing dimensions of both sides

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

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

Power of $M, 0=c4 , \ c4=0$

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

power of T, $0 = -a4-1, \ b4=-1$

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

$\pi_{4}=D^{0}.V^{-1}.\rho^{0}.c$

=$\frac{c}{V}$

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

$f_1(\frac{P}{D^{2}v^{2}\rho}, \frac{Dw}{V}. \frac{\mu}{DV\rho}, \frac{c}{V}$)=0

$\frac{P}{D^{2}V^{2}\rho}=\Phi[\frac{Dw}{V}, \frac{\mu}{Dv \rho} \frac{c}{V}]$

$\therefore P=D^{2}v^{2}\rho \phi (\frac{Dw}{V} \frac{\mu}{DV \rho}. \frac{c}{V})$