Problem 5 A large plate is pulled at a constant speed u = 4...

Problem 5 A large plate is pulled at a constant speed $u = 4 m/s$ over a fixed plate on a film of engine oil (thickness $h = 5mm$). $h = 5mm$ $U = 4 m/s$ Assuming the no-slip boundary conditions for both the top and bottom interfaces, determine the velocity profile $u(y)$ along the $y$ direction in the oil film if the velocity distribution is known as linear. Then determine the shear stress $\tau$ on the upper plate and its direction (positive or negative). (Hint: The shear stress $\tau = \frac{du}{dy}$ from the lecture notes.)

Transcript

00:01 Hello students, here a large plate is pulled at a constant speed of 4 m per second and the height of the plate is 5 mm and the temperature here at which it is fixed is 20 degrees celsius.

00:14 From that the linear velocity of the oil, this is a constant value that is given as 0 .8374 pascal second.

00:24 In this case, we have to determine the shear stress developed on the upper plate and for that we have to find that the parabolic velocity and the shear stress and again if we consider that if it is linear then we have to find the shear stress again.

00:47 That is first thing is we have to find the shear stress assuming that the velocity is parabolic and another thing is again we have to find the shear stress assuming that the linear velocity is linear.

01:06 So, two cases we have to find and so considering the velocity profile, considering velocity profile the parabolic velocity is given by profile is given by u is equal to u into h square by the h square and let it be equation 1 and differentiating this with respect to h that is du by dh is equal to 2 u h by h square and let it be equation 2.

01:55 Now, let's write the shear stress expression.

01:59 So, the shear stress which is given by tau.

02:07 So, tau is equal to mu into du by dh where mu is the linear viscosity and du by dh is the velocity profile and now here if we substitute the du by dh mu into 2 u h by h square here h and h are equal that is equal to 5 millimeter and plucking the values tau is equal to 0 .8374 which is the linear velocity linear viscosity and 2 here u is 4 meter per second and h is the height which is 0 .005 divided by 0 .005 the whole square and we get the shear stress which is equal to 1339 .84 newton per meter square...

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