$Nu= C.Re^m.Pr^n$

Step 1: Determination of dimension of all variables

Step 2: Number of variables (m) = 7

Step 3: Number of basic dimension (n) = 4

Step 4:Non-dimensional number that can be obtained=m-n=7-4=3

Step 5:

Selecting the reference variable as below

(1) l → geometrical property

(2) μ →fluid property

(3) v →flow property

(4) K →thermal property

By Buckingham π theorem

L0M0T0θ0=la μbνcKdρ

L0M0T0θ0 = La(ML-1T-1)b (LT-1)c (MLT-3θ-1)d(ML-3)

L0M0T0θ0=L(a-b+c+d-3) M(b+d+1) T(-b-c-3d) θ-d

∴ -d=0 →d=0

b+d+1=0 →b=-1

-b-c-3d=0 →c=1

a-b+c+d-3=0 →a=1

∴ First non-dimensional number

π1= la μbνcKdρ= l1 μ-1 ν1 K0 ρ

∴π1= ρvlμ→Re (Reynold's number)

Calculation for π2

L0M0T0θ0=la μbνcKd h

L0M0T0θ0=La(ML-1T-1)b (LT-1)c (MLT-3θ-1)d(MT-3θ-1)

L0M0T0θ0=L(a-b+c+d) M(b+d+1) T(-b-c-3d-3) θ-d-1

∴ -d-1=0 →d=-1

b+d+1=0 →b=0

-b-c-3d-3=0 →c=0

a-b+c+d=0 →a=1

∴ Second non-dimensional number

π2= la μbνcKd h= l1 μ0 ν0 K-1 h

∴π2= hlK→Nu (Nusselt's number)

Calculation for π3

L0M0T0θ0=la μbνcKdCp

L0M0T0θ0=La(ML-1T-1)b (LT-1)c (MLT-3θ-1)d(L2T-2θ-1)

L0M0T0θ0=L(a-b+c+d+2) M(b+d) T(-b-c-3d-2) θ-d-1

∴ -d-1=0 →d=-1

b+d=0 →b=1

-b-c-3d-2=0 →c=0

a-b+c+d+2=0 →a=0

∴ Third non-dimensional number

π3= la μbνcKdCp= l0 μ1 ν0 K-1 Cp

∴π2= μCpK→Pr (Prandtl's number)

By Buckingham’s π theorem

f (π1, π2, π3)=0

∴f (Re, Nu, Pr)=0

∴Nu=ϕ (Re.Pr)

OR

$Nu=C.Re^m.Pr^n$

Where,

m, n, c are the constants and can be determined by experimental method.