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Laminar flow between fixed parallel plates

Short answer

Velocity distribution u = -(1/(2μ))(dp/dx)(hy - y^2). Shear stress τ = -(dp/dx)(h/2 - y), linear. Maximum velocity u_max = -(h^2/(8μ))(dp/dx). Average velocity ū = (2/3)u_max = -(h^2/(12μ))(dp/dx). Discharge per unit width q = -(h^3/(12μ))(dp/dx). Pressure drop Δp = (12μūL)/h^2.

Derive the expressions for velocity distribution, shear stress distribution, maximum velocity, average velocity and discharge per unit width for steady laminar flow of a viscous incompressible fluid between two fixed parallel plates a distance hh apart. Also obtain the pressure drop in a length LL.

Assumptions

Steady, laminar, fully developed flow of a Newtonian incompressible fluid between two fixed wide plates, gap hh. Flow is in the xx-direction; yy is measured from the lower plate. There is no body force effect other than hydrostatic, and dp/dxdp/dx is constant.

   y ^ ---------------------------  upper plate (u = 0)
     |        ______
     |  h   /        \   u(y)
     |      \________/
     +---------------------------> x   lower plate (u = 0)

Force balance on an element

Consider a fluid element of length dxdx, thickness dydy and unit width. Net force in xx is zero:

p dy−(p+dpdxdx)dy−τ dx+(τ+dτdydy)dx=0p\,dy - \left(p + \frac{dp}{dx}dx\right)dy - \tau\,dx + \left(\tau + \frac{d\tau}{dy}dy\right)dx = 0 dτdy=dpdx\frac{d\tau}{dy} = \frac{dp}{dx}

With τ=μ du/dy\tau = \mu\,du/dy:

μd2udy2=dpdx\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}

Velocity distribution

Integrating twice, u=12μdpdxy2+C1y+C2u = \dfrac{1}{2\mu}\dfrac{dp}{dx}y^2 + C_1y + C_2. Boundary conditions u=0u = 0 at y=0y = 0 and at y=hy = h give C2=0C_2 = 0 and C1=−h2μdpdxC_1 = -\dfrac{h}{2\mu}\dfrac{dp}{dx}.

u=−12μdpdx(hy−y2)u = -\frac{1}{2\mu}\frac{dp}{dx}\left(hy - y^2\right)

The profile is parabolic. The pressure falls in the flow direction, so dp/dx<0dp/dx < 0 and u>0u > 0.

Shear stress distribution

τ=μdudy=−dpdx(h2−y)\tau = \mu\frac{du}{dy} = -\frac{dp}{dx}\left(\frac{h}{2} - y\right)

It is zero at the centre line and maximum at the walls: τw=h2∣dpdx∣\tau_w = \dfrac{h}{2}\left|\dfrac{dp}{dx}\right| (linear variation).

Maximum velocity

At y=h/2y = h/2:

umax=−h28μdpdxu_{max} = -\frac{h^2}{8\mu}\frac{dp}{dx}

Discharge and average velocity

q=∫0hu dy=−12μdpdx[hy22−y33]0h=−h312μdpdxq = \int_0^h u\,dy = -\frac{1}{2\mu}\frac{dp}{dx}\left[\frac{hy^2}{2} - \frac{y^3}{3}\right]_0^h = -\frac{h^3}{12\mu}\frac{dp}{dx} uˉ=qh=−h212μdpdx=23umax\bar u = \frac{q}{h} = -\frac{h^2}{12\mu}\frac{dp}{dx} = \frac23 u_{max}

Pressure drop

For a length LL, dp/dx=−Δp/Ldp/dx = -\Delta p/L:

Δp=12μuˉLh2=12μqLh3\Delta p = \frac{12\mu \bar u L}{h^2} = \frac{12\mu q L}{h^3}

The head loss is hf=Δp/ρg=12μuˉLρgh2h_f = \Delta p/\rho g = \dfrac{12\mu\bar u L}{\rho g h^2}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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