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enter image description hereI'm a high school student, and I am interested in fluid viscosity. I was doing an experiment, and I got some data points. Following are some of them. Here,x1 is for temperature(in kelvin), and x2 is a viscosity(in pa/s).

Using these data, I was trying to figure out the specific equation representing all data points. Since I've seen many graphs of viscosity and temperature in curved form, I assumed that it would be a logarithmic graph. Here's my question.

  1. Do I need more data points to correctly "assume(or predict)" the shape of the graph?

  2. Is there any equation that can be used to express the relationship between viscosity and temperature?

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    $\begingroup$ Usually liquid viscosity dependence on temperature is modeled as $\mu = \alpha e^{-\beta T}$ $\endgroup$
    – Agnius Vasiliauskas
    Commented Jul 1, 2022 at 14:46
  • $\begingroup$ Your data are all centered around 300 K. Note that it is inadvisable to extrapolate very far outside the range of your data, as such predictions become more and more uncertain the farther away you are from the extremes of your data points. $\endgroup$
    – David White
    Commented Jul 1, 2022 at 16:01
  • $\begingroup$ I note that there is something wrong with your first 3 data points. If possible, you may want to verify your equipment and method, and take those 3 measurements again. $\endgroup$
    – David White
    Commented Jul 1, 2022 at 19:30

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Here is how kinematic viscosity interpolates with temperature.

For engine oils two reference values are needed, at 40°C = 313.15°K and at 100° C = 373.15°K. Then the viscosity for any temperature between or outside this range is evaluated with

$$ \mu(T) = \exp \left( b\, T^m \right) $$

where $T$ is the temperature in Kelvin

The coefficients $b$, $m$ of interpolation are evaluated from $T_1$, $T_2$ are the reference temperatures in Kelvin, and $\mu_1$, $\mu_2$ the reference viscosities in the following manner

$$ \begin{aligned} m & = \frac{ \ln \left( \frac{\ln \left( \mu_1 + 0.6 \right)}{\ln \left( \mu_2 + 0.6 \right)} \right) }{ \ln \left( \frac{T_1}{T_2} \right) } \\ b & = T_1^{-m} \ln \left( \mu_1 + 0.6 \right) \\ \end{aligned} $$

I am not sure what the magic number 0.6 is, but it acts like a minimum viscosity. The units for viscosity is the same units used in $\mu_1$ and $\mu_2$, typically in centi-stokes ($\rm cSt$).

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