I have a system with a fairly viscous fluid, 70,000 cP. The flow is induced by applying pressure to a plunger via compressed air. The error in viscosity for this material is +- 15% which can be significant in trying to control the flow of the material.

Given the measured viscosity of the "batch" of material, how could I adjust the pressure on the plunger accordingly to get the flow out of an exit hole with a known area?

The boundaries nor area change for the fluid path and I want a constant velocity:

$$F=A*\mu{\frac{u}{y}}$$ where $\mu$ is the viscosity, $A$ is the cross sectional area, $u$ is the velocity and $y$ is the boundary separation.

I'm thinking I need:


Which implies at first glance I can simply adjust the pressure by a factor of the ratio of optimum and actual viscosity values.

Let $p_0$ be optimum pressure, $p_a$ be actual pressure and what I'm trying to find, $v_0$ be the optimum viscosity and $v_a$ be the actual viscosity.

$$p_a = p_0*\frac{v_a}{v_0}$$

It seems too simplistic and I feel like I'm missing something.

  • $\begingroup$ Felix, what is the shape of the channel in which the fluid flows? Is it cylindrical? Can you also mention its dimensions? $\endgroup$
    – Amey Joshi
    Commented Dec 11, 2014 at 15:41
  • $\begingroup$ @AmeyJoshi The dispenser is a needle, so a cylinder at the end. The rest of the system is hosing with 3/8" tubing and a 2.5" reservoir where the pressure is applied. I'm assuming that without hardware changes this is all constant. $\endgroup$ Commented Dec 11, 2014 at 15:45

1 Answer 1


You are not missing anything.

$P = \frac{F}{A}$

so then your second equation becomes:

$P = \mu \frac{u}{y}$

For the actual pressure:

$P_a = \mu_a \frac{u_a}{y_a}$

Since $u_a = u_0$ because you require it to and $y_a = y_0$ because the boundary separation doesn't change, when you divide $P_a$ by $P_0$:

$\frac{P_a}{P_0} = \frac{\mu_a}{\mu_0}$

and as you stated

$P_a = P_0\frac{\mu_a}{\mu_0}$


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