cnblue asks:

Temperature & the viscosity of a liquid

I am doing an experiment on this and I am not sure how I could heat up and measure the temperature of the liquid (oil).
Any suggestions?
In Topics: Science fair
> 60 days ago



Jun 10, 2011
Subscribe to Expert

What the Expert Says:

Heating oil, which is flammable, could be dangerous and should be supervised by an adult, especially if done over a gas flame. Temperature is measured with a thermometer but you will need one that is designed for high temperature readings. These might be available from a high school chemistry lab. Since microwaves work by agitating water molecules, you won't get far trying to heat oil in a microwave.

Did you find this answer useful?

Additional Answers (1)

seetasravya writes:
The temperature dependence of liquid viscosity is the phenomenon by which liquid viscosity tends to decrease (or, alternatively, its fluidity tends to increase) as its temperature increases. This can be observed, for example, by watching how cooking oil appears to move more fluidly upon a frying pan after being heated by a stove. It is usually expressed by one of the following models:Exponential model

    \mu(T)\,=\,\mu_0 \exp(-bT)

Engineering Colleges in Kolkata  

where T is temperature and μ0 and b are coefficients. See first-order fluid and second-order fluid. This is an empirical model that usually works for a limited range of temperatures.
[edit] Arrhenius model

The model is based on the assumption that the fluid flow obeys the Arrhenius equation for molecular kinetics:

    \mu(T)\,=\,\mu_0 \exp\left( \frac {E}{RT} \right)

where T is temperature, μ0 is a coefficient, E is the activation energy and R is the universal gas constant. A first-order fluid is another name for a power-law fluid with exponential dependence of viscosity on temperature.
[edit] Williams-Landel-Ferry model

The Williams-Landel-Ferry model, or WLF for short, is usually used for polymer melts or other fluids that have a glass transition temperature.

The model is:

    \mu(T)\,=\,\mu_0 \exp \left( \frac {-C_1 (T-T_r)} {C_2+ T -T_r} \right)

where T-temperature, C1, C2, Tr and μ0 are empiric parameters (only three of them are independent from each other).

If one selects the parameter Tr based on the glass transition temperature, then the parameters C1, C2 become very similar for the wide class of polymers. Typically, if Tr is set to match the glass transition temperature Tg, we get

    C_1 \approx17.44


    C_2 \approx 51.6 K.

Van Krevelen recommends to choose

    T_r\,=\,T_g+43 K, then

    C_1 \approx 8.86


    C_2 \approx101.6 K.

Using such universal parameters allows one to guess the temperature dependence of a polymer by knowing the viscosity at a single temperature.

In reality the universal parameters are not that universal, and it is much better to fit the WLF parameters from the experimental data.
[edit] Seeton Fit

The Seeton Fit[1] is based on curve fitting the viscosity dependence of many liquids (refrigerants, hydrocarbons and lubricants) versus temperature and applies over a large temperature and viscosity range:

    \ln \left( {\ln \left( {\nu + 0.7 + e^{ - \nu } K_0 \left( {\nu + 1.244067} \right)} \right)} \right) = A - B*\ln \left( T \right)

where T is absolute temperature in kelvins, ν is the kinematic viscosity in centistokes, K0 is the zero order modified Bessel function of the second kind, and A and B are liquid specific values. This form should not be applied to ammonia or water viscosity over a large temperature range.

For liquid metal viscosity as a function of temperature, Seeton proposed:

    \ln \left( {\ln \left( {\nu + 0.7 + e^{ - \nu } K_0 \left( {\nu + 1.244067} \right)} \right)} \right) = A - {B \over T}

Viscosity of water equation accurate to within 2.5% from 0 °C to 370 °C:

μ (Temp)= 2.414*10^-5 (N·s/m²) * 10^(247.8 K/(Temp - 140 K))

    N - newton
    s - second
    m - meter
    K - kelvin

> 60 days ago

Did you find this answer useful?
Answer this question


You are about to choose ${username}'s answer as the best answer.

Cancel | Continue

*You can change the best answer in the future if you think that you received a better answer

How likely are you to recommend to your friends and colleagues?

Not at all likely
Extremely likely