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Problem No. 5: t=60 ℉ γ=0.0765lb/ft g=32.2 ft /s

This problem asks us to: 1) Use Grindley and Gibson's equation to calculate the viscosity of air in poises at 60°F. 2) Convert the viscosity to slug/ft-s and find the kinematic viscosity in ft2/s and cm2/s. 3) The viscosity is calculated to be 0.0857820841 slug/ft-s. The kinematic viscosities are 1161.705612 ft2/s in English units and 1079259.829 cm2/s in metric units.

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Joshua Ramirez
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165 views3 pages

Problem No. 5: t=60 ℉ γ=0.0765lb/ft g=32.2 ft /s

This problem asks us to: 1) Use Grindley and Gibson's equation to calculate the viscosity of air in poises at 60°F. 2) Convert the viscosity to slug/ft-s and find the kinematic viscosity in ft2/s and cm2/s. 3) The viscosity is calculated to be 0.0857820841 slug/ft-s. The kinematic viscosities are 1161.705612 ft2/s in English units and 1079259.829 cm2/s in metric units.

Uploaded by

Joshua Ramirez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Problem No.

Using Grindley and Gibson’s equation, determine the viscosity of air in poises for a

temperature of 60°F and convert this into English units. If the specific weight of air at this

temperature is 0.0765 lb/ft 3, determine the kinematic viscosity in both sets of units.

Given: t=60 ℉

γ =0.0765lb/ft 3

g=32.2 ft /s 2

Required: viscosity (μ)

kinematic viscosity (v)

Schematic Diagram:

t=60 ℉
Air

γ =0.0765lb/ft 3

Solution:

Using Grindley and Gibson’s equation, we have the equation,

μ=0.0001702 ( 1+0.00329 t+ 0.000007 t 2 )

Where μ is in poises and t is in degree celsius.

We need to convert the temperature into degree celsius


t=60 ℉

5
t=( 60℉ −32)×
9

t=15.5556 ℃

Substitute the temperature to the given Grindley and Gibson’s equation,

μ=0.0001702 ( 1+0.00329 t+ 0.000007 t 2 )

μ=0.0001702 [ 1+0.00329(15.5556 ℃ )+0.000007(15.5556 ℃ )2 ]

μ=1.791987737× 10−4 P

Because the viscosity is in poises, convert it into English unit

1 slug/ft ∙ s
1 .791987737 ×10−4 P× =0.0857820841 slug/ ft ∙ s
0.002089 P

Therefore, the viscosity of air for a temperature of 60°F is

0.0857820841 slug/ ft ∙ s

Since, ρg=γ

Therefore,

γ
ρ=
g

Substitute the value of specific weight and acceleration due to gravity

0.0765 lb /ft 3
ρ=
32.2 ft /s2
ρ=2.375776398× 10−3 lbm /ft 3

After getting the value of density; we need to convert the unit of viscosity

slug 32.174 lb
0.0857820841 × =2.759952774 lb / ft ∙ s
ft ∙ s 1 slug

μ
Since, v=
ρ

Then, substitute the given value

2.759952774 lb /ft ∙ s
v=
2.375776398×10−3 lb m / ft 3

v=1161.705612 ft 2 / s

Because the answer is in English unit, we need to convert it into metric

2 2
ft 2 (30.48) cm
1161.705612 × 2
=1079259.829 cm 2 /s
s 1 ft

Therefore, the kinematic viscosity of air:

In English

v=1161.705612 ft 2 / s

In metric

v=1079259.829 cm2 / s

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