One type of capillary-tube viscometer is shown in Video V1.5 and in Fig. P1.57. For this device the liquid to be tested is drawn into the tube to a level above the top etched line. The time is then obtained for the liquid to drain to the bottom etched line.The kinematic viscosity, \(\nu,\) in \(\mathrm{m}^{2} / \mathrm{s}\) is then obtained from the equation \(\nu=K R^{4} t\) where \(K\) is a constant, \(R\) is the radius cf the capillary tube in \(\mathrm{mm}\), and \(t\) is the drain time in seconds. When glycerin at \(20^{\circ} \mathrm{C}\) is used as a calibration fluid in a particular viscometer, the drain time is 1430 s. When a liquid having a density of \(970 \mathrm{kg} / \mathrm{m}^{3}\) is tested in the same viscometer the drain time is 900 s. What is the dynamic viscosity of this liquid?

Step by step solution

01

- Determine the constant K

Put the known parameters of glycerin into the equation \(ν=K R^{4} t\) to solve for \(K\). The kinematic viscosity \(ν\) of glycerin at \(20^{\circ}C\) is approximately \(1.5 * 10^{-6} m^{2}/s\), and the drain time \(t\) is 1430 seconds. Thus, the equation can be rearranged to \(K = ν/ (R^{4} t)\). Assuming a constant radius since the same viscometer is used for both fluids, \(K\) is defined and can be used later in the analysis of the unknown liquid.

02

- Compute the kinematic viscosity for the unknown liquid

Once \(K\) is known, put it into the same equation, but now with the parameters of the unknown liquid to find its kinematic viscosity \(ν\). The drain time \(t\) for the unknown liquid is 900 seconds. So, putting these into the equation, we find \(ν = K R^{4}t\). The radius stays the same because the same viscometer is used.

03

- Find the dynamic viscosity

Using the known density of the unknown liquid, \(ρ = 970 kg/m^{3}\), determine the dynamic viscosity, \(μ\), using the equation \(μ = ρν\). Plug the previously computed \(ν\) and given \(ρ\) into the equation to solve for \(μ\).

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