Chapter 10, Problem 28P

To determine

Examine the variation in the air viscosity and plot it in a graph.

Explanation of Solution

Determine the following properties for the corresponding temperatures from Table A-22.

$T\left(\text{K}\right)$	Table A-9	Power law	Sutherland	PL-Error %	Suth Error %
373	$2.181\times {10}^{-5}$	$2.12848\times {10}^{-5}$	$2.17277\times {10}^{-5}$	2.41	0.38
393	$2.264\times {10}^{-5}$	$2.20382\times {10}^{-5}$	$2.25649\times {10}^{-5}$	2.66	0.33
413	$2.345\times {10}^{-5}$	$2.27789\times {10}^{-5}$	$2.33802\times {10}^{-5}$	2.86	0.30
433	$2.42\times {10}^{-5}$	$2.35078\times {10}^{-5}$	$2.41752\times {10}^{-5}$	2.86	0.10
453	$2.504\times {10}^{-5}$	$2.42255\times {10}^{-5}$	$2.4951\times {10}^{-5}$	3.25	0.36
473	$2.577\times {10}^{-5}$	$2.49326\times {10}^{-5}$	$2.57089\times {10}^{-5}$	3.25	0.24
523	$2.76\times {10}^{-5}$	$2.66583\times {10}^{-5}$	$2.75316\times {10}^{-5}$	3.41	0.25
573	$2.934\times {10}^{-5}$	$2.83297\times {10}^{-5}$	$2.92627\times {10}^{-5}$	3.44	0.26
623	$3.101\times {10}^{-5}$	$2.9953\times {10}^{-5}$	$3.09135\times {10}^{-5}$	3.30	0.31
673	$3.261\times {10}^{-5}$	$3.15332\times {10}^{-5}$	$3.24935\times {10}^{-5}$	3.30	0.36
723	$3.414\times {10}^{-5}$	$3.30748\times {10}^{-5}$	$3.40104\times {10}^{-5}$	3.12	0.38
773	$3.563\times {10}^{-5}$	$3.45811\times {10}^{-5}$	$3.54707\times {10}^{-5}$	2.94	0.45
873	$3.846\times {10}^{-5}$	$3.74996\times {10}^{-5}$	$3.82427\times {10}^{-5}$	2.50	0.56
973	$4.111\times {10}^{-5}$	$4.03082\times {10}^{-5}$	$4.08449\times {10}^{-5}$	1.95	0.64
1073	$4.362\times {10}^{-5}$	$4.30219\times {10}^{-5}$	$4.33038\times {10}^{-5}$	1.37	0.73
1173	$4.6\times {10}^{-5}$	$4.56524\times {10}^{-5}$	$4.56397\times {10}^{-5}$	0.76	0.78
1273	$4.826\times {10}^{-5}$	$4.82088\times {10}^{-5}$	$4.78688\times {10}^{-5}$	0.11	0.81

Plot the accuracy of both models as in Figure (1).