Question

If the viscosity of water at 68 °F is 0.01008 poise, compute its absolute viscosity (m) in pound-seconds per square
foot. If the specific gravity at 68 °F is 0.998, compute its kinematic viscosity (v) in square feet per second​

Answer 1

Answer:

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Answer 2

Answer:

The absolute viscosity of the water is equal to 1500 pdl s/ft².

The kinematic viscosity of water is equal to 1.086×10⁻⁵ ft/s.

Explanation:

Given, the viscosity of water = 0.01008 poise = 1.008 centipoise

We know that 1 centipoise = 1488.16 plb.s.ft⁻²

Absolute viscosity $= 1.008\times 1488.16\;plb\;s \; ft^{-2}$

                              $=1500\;pdl\; .s\;. ft^{-2}$

Let's convert poise (Pa) into Poise second (Pa s);-

Viscosity $= 0.01008\times 0.1 = 0.001008 Pa.s$

Mass density = Density of water × Specific density

Mass density $=1000Kgm^{-3}\times 0.998 = 998Kg/m^3$

Kinematic viscosity = Absolute viscosity/Mass density

Kinematic viscosity of water $=\frac{0.001008 Pa\;s}{998Kgm^{-3}}$

We know that 1 Pa = 1 kgm⁻¹s⁻²

Kinematic viscosity $= 1.01\times 10^6m^2s^{-1} = 1.086\times 10^{-5}ft^2/s$    [ ∵ 1m = 3.28ft]

Therefore, the kinematic viscosity of water is equal to 1.086×10⁻⁵ ft/s.

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