Variation of gas viscosity with temperature can be represented by Sutherland\'s

Question

Variation of gas viscosity with temperature can be represented by Sutherland's equation 0.55570 + SV3/2 0.555 T + S [2.11 where gas viscosity (in centipoise, cP) at temperature T gas viscosity (in centipoise, cP) at reference temperature T0 temperature in degrees Rankine (R) To = reference temperature in degrees Rankine (°R) S= Sutherland's constant (unitless) For air: 110=0.01 827 cp at 70 = 524.07 "R and S=120 If the temperature is measured to be 50 °C ± 0.1 °C, calculate the viscosity of air and its uncertainty using Sutherland's equation (equation [2.1]) and the formula for propagation of uncertainty: Y = f(X1, X2, , x.) Assume there is no uncertainty in 110, lo and S Useful information: Conversion of units of temperature: ["R] = ([°C] + 273.15) x 1.8

Explanation / Answer

Sutherland's equation of viscosity describes a relationship between the temperature T(absolute) and the dynamic viscosity of a fluid.

Now from the aforesaid equation,

= 0 (0.555 T0 + S/ 0.555 T + S)(T / T0)^ (3/2)

= 0.01827 {(0.555 X 524.07 + 120) / (0.555 X 581.67 + 120)} (581.67/524.07)^ (3/2)

= 0.01970 cP

Therefore, = viscosity of air = 0.01970 cP

To determine uncertainty of propagation of air :

Since is function of T only. We use the formila for a function of single variable.

= (T)

del = | (d / d T) del T |

=| [ { - 0 (0.555 T0 + S) } / (0.555 T + S )^ 2 ] X (1/ T0) ^ (3/2) X (3/2) (T) ^ (1/2) del T |

Since del T = 0.1 degree C = 491.85 degree R

Therefore,

del = | - 1.139 |

= 1.139

Propagation uncertainty of air is 1.139 .

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