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The velocity profile in a 4-cm-diameter transporting 20 C water is given by v(r)=0.1(1-2500r²)m/s. If the viscosity of the water is 1 mPa s and the density is 997 kg/m³, the shearing stress at the wall is nearest:

A. 1.0 Pa
B. 0.1 Pa
C. 0.01 Pa
D. 0.001 Pa

User Gadolin
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1 Answer

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Final answer:

The shear stress at the wall of the pipe can be calculated by taking the derivative of the velocity profile at the wall and multiplying it by the viscosity of the water. The calculated stress value is nearest to 0.1 Pa.

Step-by-step explanation:

The student is asking about the shear stress that water exerts on the walls of a pipe given a specific velocity profile, the viscosity of the water, and the density of the water. The shear stress at the wall of the pipe (τ) can be found using the formula τ = η * (dv(r)/dr)|r=R, where η is the viscosity of the fluid and dv(r)/dr is the radial derivative of the velocity profile evaluated at the wall (r=R).

To find dv(r)/dr, differentiate the given velocity profile v(r)=0.1(1-2500r²) with respect to r, which yields dv(r)/dr = -0.1 * (5000r). Since shear stress is evaluated at the wall, use the radius R = D/2 = 2 cm = 0.02 m. Thus, the derivative at the wall is dv(r)/dr|r=R = -0.1 * 5000 * 0.02 m/s.

After calculating the derivative, multiply the result by the given viscosity η = 1 mPa s = 1 * 10^-3 Pa s to obtain the shear stress at the wall. After performing the multiplication, you shall find that the shear stress is nearest to option B, 0.1 Pa.

User Fomite
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