Chapter 1, Problem 1.39P

To determine

(a)

Estimate air viscosity at 500°C by the power-law.

Answer to Problem 1.39P

Air viscosity at 500°C by the power-law is $\approx 3.55E-5\frac{\text{kg}}{\text{m}\text{.s}}$

Explanation of Solution

Given:

$\mu \approx 1.80E-5Pa·s\text{ for air at }20°\text{C}$

Concept Used:

Formula used

$\mu ={\mu }_o{\left(T/T_o\right)}^n$

Calculation:

First change T from ${500}^o\text{C}$ to $\text{773}\text{K}$

for the power- law for air, $\text{n=0}\text{.7}$ and from equation.

$\mu ={\mu }_o{\left(T/T_o\right)}^n\approx \left(1.80E-5\right){\left(\frac{773}{293}\right)}^{0.7}\approx 3.55E-5\frac{\text{kg}}{\text{m}\text{.s}}$

This is less than $1\%$ low.

Conclusion:

Thus, air viscosity at 500°C by the power-law is $\approx 3.55E-5\frac{\text{kg}}{\text{m}\text{.s}}$.

To determine

(b)

The air viscosity at 500°C by the Sutherland law.

Answer to Problem 1.39P

Air viscosity at 500°C by the Sutherland law $=3.52E-5\frac{\text{kg}}{\text{m}\text{.s}}$

Explanation of Solution

Given:

$\mu \approx 1.80E-5Pa·s\text{ for air at }20°\text{C}$

Concept Used:

Formula used

$\mu ={\mu }_o\left[\frac{{\left(T/T_o\right)}^{1.5}\left(T_o+S\right)}{\left(T+S\right)}\right]$

Calculation:

For the Sutherland law, for air, $S\approx 11\text{0}\text{K}$ and from eq(1.30b)

$\begin{array}{c}\mu ={\mu }_o\left[\frac{{\left(T/T_o\right)}^{1.5}\left(T_o+S\right)}{\left(T+S\right)}\right]\\ =\left(1.80E-5\right)\left[\frac{{\left(773/293\right)}^{1.5}\left(293+110\right)}{\left(773+110\right)}\right]\\ =3.52E-5\frac{\text{kg}}{\text{m}\text{.s}}\end{array}$

This is only $1.7\%$ low

Conclusion:

Thus, air viscosity at 500°C by the Sutherland law $=3.52E-5\frac{\text{kg}}{\text{m}\text{.s}}$.

To determine

(c)

The air viscosity at 500°C by the Law of Corresponding States,.

Answer to Problem 1.39P

Air viscosity at 500°C by the Law of Corresponding States $=3.5E-5\frac{\text{kg}}{\text{m}\text{.s}}$

Explanation of Solution

Given:

$\mu \approx 1.80E-5Pa·s\text{ for air at }20°\text{C}$

Concept Used:

From fig 1.5

Calculation:

Finally use fig 1.6 critical values for air from Ref. 3 are:

Air: ${\mu }_c=1.93E-5\text{Pa}\text{.s}$

$T_c=13\text{2}\text{K}$ (‘mixture’ estimates)

At $\text{773}\text{K}$, the temperature ratio is $T/T_c=773/132=5.9$ From fig 1.5read $\mu /{\mu }_c=1.8$.

Then our critical-point-correlation estimate of air viscosity is only $3\%$ low:

$\mu =1.8{\mu }_c=\left(1.8\right)\left(1.93E-5\right)=3.5E-5\frac{\text{kg}}{\text{m}\text{.s}}$

Conclusion:

Thus, air viscosity at 500°C by the Law of Corresponding States $=3.5E-5\frac{\text{kg}}{\text{m}\text{.s}}$.