Suppose that the fluid being sheared in Fig. 1.8 is SAE 30 oil at 20°C. Compute the shear stress in the oil if V = 3 m/s and h = 2 cm.

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Accepted Answer

• System sketch: This is shown earlier in Fig. 1.8.
• Assumptions: Linear velocity profile, laminar newtonian fluid, no slip at either plate surface.

• Approach: The analysis of Fig. 1.8 leads to Eq. (1.26) for laminar flow.

[latex]u = V \frac{y}{h}[/latex] (1.26)
• Property values: From Table 1.4 for SAE 30 oil, the oil viscosity µ = 0.29 kg/(m-s).

Table 1.4 Viscosity and Kinematic Viscosity of Eight Fluids at 1 atm and 20°C
Fluid	µ, kg/(m·s)[latex]^†[/latex]	Ratio µ/µ([latex]H_2[/latex])	ρ, kg/[latex]m^3[/latex]	v, [latex]m^2[/latex]/[latex]s^†[/latex]	Ratio v/v(Hg)
Hydrogen	9.0 E-6	1.0	0.084	1.05 E-4	910
Air	1.8 E-5	2.1	1.20	1.50 E-5	130
Gasoline	2.9 E-4	33	680	4.22 E-7	3.7
Water	1.0 E-3	114	998	1.01 E-6	8.7
Ethyl alcohol	1.2 E-3	135	789	1.52 E-6	13
Mercury	1.5 E-3	170	13,550	1.16 E-7	1.0
SAE 30 oil	0.29	33,000	891	3.25 E-4	2,850
Glycerin	1.5	170,000	1,260	1.18 E-3	10,300

[latex]^†[/latex]1 kg/(m·s) = 0.0209 slug/(ft·s); 1 [latex]m^2[/latex]/s = 10.76 [latex]ft^2[/latex]/s.
• Solution steps: In Eq. (1.26), the only unknown is the fluid shear stress:

[latex] au = µ\frac{V}{h} = \left(0.29\frac{kg}{m \cdot s} ight) \frac{(3 m/s)}{(0.02 m)} = 43.5 \frac{kg \cdot m/s^2}{m^2} = 43.5\frac{N}{m^2} \approx 44 Pa[/latex] Ans.

• Comments: Note the unit identities, 1 kg-m/[latex]s^2[/latex] ≡ 1 N and 1 N/[latex]m^2[/latex] ≡ 1 Pa. Although oil is very viscous, this shear stress is modest, about 2400 times less than atmospheric pressure. Viscous stresses in gases and thin (watery) liquids are even smaller.

Question 1.7: Suppose that the fluid being sheared in Fig. 1.8 is SAE 30 o...

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