Consider the Laminar boundary layer velocity distribution over a flat plate,
uU=2(Yδ)−2(Yδ)2+(yδ)3
For the Boundary layer thickness, we know the Von Karman’s Momentum Integral equation,
τ0ρU2=∂θ∂x
Where,
τ0→Shear stress in the fluid
ρ→Density of the fluid
U→Free stream velocity of the fluid
θ→Momentum thickness
We know that the Momentum thickness is given by the equation,

θ=∫δ0[2(Yδ)−2(Yδ)2+(yδ)3]1−[2(Yδ)−2(Yδ)2+(yδ)3]dY
θ=∫δ0[2Yδ−4Y2δ2+4Y3δ3−2Y4δ4−2Y2δ2+4Y3δ3−4Y4δ4+2Y5δ5+Y3δ3−2Y4δ4+2Y5δ5−Y6δ6]dY
θ=[2Y22δ−4Y33δ2+4Y44δ3−2Y55δ4−2Y33δ2+4Y44δ3−4Y55δ4+2Y66δ5+Y44δ3−2Y55δ4+2Y66δ5−Y77δ6]δ0

θ=[73δ420]
∴\frac{∂θ}{∂x}=\frac{73}{420} \frac{∂δ}{∂x}
Hence,
\frac{\tau_0}{ρU^2}=\frac{73}{420} \frac{∂δ}{∂x}
∴\tau_0=\frac{73ρU^2}{420}\frac{∂δ}{∂x}
Now, By Newton’s Law of Viscosity for Newtonian fluids,
∴\tau_0=μ\frac{du}{dY}
where, \frac{du}{dY}=\frac{2}{δ}-\frac{4Y}{δ^2} +\frac{3Y^2}{δ^3}
Hence, equating it to the above equation,
\frac{73ρU^2}{420} \frac{∂δ}{∂x}=μ \frac{du}{dY}
\frac{73ρU^2}{420}\frac{∂δ}{∂x}=μU[\frac{2}{δ}-\frac{4y}{δ^2} +\frac{3y^2}{δ^3}]_(Y=0)
\frac{73ρU^2}{420} \frac{ ∂δ}{∂x}=μU(\frac{2}{δ})
\frac{73ρU}{840} \frac{∂δ}{∂x}=(\frac{μ}{δ})
\frac{73}{840} \frac{∂δ}{∂x}=(\frac{μ}{ρU})
Integrating the above differential equation by variable separable method,

\frac{73δ^2}{1680}=(\frac{μ}{ρU})x
\frac{δ^2}{x^2} =\frac{1680}{73} (\frac{μ}{ρUx})
We know that, Local Reynolds number is given by the equation,
Re_x=\frac{ρUx}{μ}
∴\frac{δ^2}{x^2} =\frac{1680}{73} (\frac{1}{Re_x})
∴\frac{δ}{x}=\sqrt{\frac{1680}{73Re_x}}
∴\frac{δ}{x}=\frac{4.797}{\sqrt{Re_x}}
… This is the boundary layer thickness equation.
Shear Stress on one side of the plate is given by,
\tau_0=μ \frac{du}{dY}
∴τ_0=μU[\frac{2}{δ}-\frac{4y}{δ^2} +\frac{3y^2}{δ^3}]_(Y=0)
∴\tau_0=μU[\frac{2}{δ}-\frac{4}{δ}+\frac{3}{δ}]
∴\tau_0=\frac{μU}{δ}
Substituting the value of δ
∴\tau_0=\frac{μU}{x\sqrt{\frac{1680}{73Re_x}}}
∴\tau_0=\frac{μU}{x} \sqrt{\frac{73Re_x}{1680}}
∴\tau_0=\frac{μU}{x} \sqrt{\frac{73ρUx}{1680μ}}
∴\tau_0=ρU^2 \sqrt{\frac{73μ}{1680ρUx}}
∴\tau_0=ρU^2 \sqrt{\frac{73}{1680Re_x}}
∴\tau_0=\frac{0.2084ρU^2}{\sqrt{Re_x }}
… This is the Shear Stress equation.
Force per unit width on one side of the plate due to drag from leading to the trailing edge (length=l) is given by,
F_D=∫_0^l \tau_0 dx
F_D=∫_0^l ρU^2 \sqrt{\frac{73\mu}{1680Re_x}}dx
F_D=∫_0^l ρU^2 \sqrt{\frac{73μ}{1680ρUx}}dx
F_D=ρU^2 \sqrt{\frac{73μ}{1680ρU}} ∫_0^l\frac{1}{\sqrt{x}}dx
F_D=ρU^2 \sqrt{\frac{73μ}{1680ρU}} [2\sqrt{x}]_0^l
F_D=ρU^2 \sqrt{\frac{73μ}{1680ρU}} 2\sqrt{l}
F_D=ρU^2 \sqrt{\frac{292μl}{1680ρU}}
F_D=ρU^2 l\sqrt{\frac{292μ}{1680ρUl}}
We know that the Reynolds number for the flat plate is given by,
Re=\frac{ρUl}{μ}
∴F_D=ρU^2 l\sqrt{\frac{292}{1680Re}}
∴F_D=\frac{0.4169ρU^2 l}{\sqrt{Re}}
… This is the Drag force per unit width on one side of the plate