Question 2: What is the maximum h that can be achieved in...

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Q2. The laminar flow of a Newtonian fluid between parallel plates is illustrated in the below figure. The velocity distribution for this flow is given by the equation: h2 aP V = 8p dx The total gap between the plates is 3 cm. The viscosity of the fluid is 0.5 Ns/m^2 and the pressure gradient is -1200 N/m^3. Find the magnitude and direction of the shear stress on the upper plate.

Madhur L.

$the-velocity-distribution-for-laminar-flow-between-parallel-plates-is-given-by-fracuu_max-1-leftfr-2$

The velocity distribution for laminar flow between parallel plates is given by \[\frac{u}{u_{\max }}=1-\left(\frac{2 y}{h}\right)^{2}\] where $h$ is the distance separating the plates and the origin is placed midway between the plates, Consider a flow of water at $15^{\circ} \mathrm{C}$ with maximum speed of $0.05 \mathrm{m} / \mathrm{s}$ and $h=0.1 \mathrm{mm}$. Calculate the force on a $1 \mathrm{m}^{2}$ section of the lower plate and give its direction.

Two fixed, horizontal, parallel plates are spaced 0.4 in. apart. A viscous liquid $\left(\mu=8 \times 10^{-3} 16 \cdot \mathrm{s} / \mathrm{ft}^{2}, 3 G=0.9\right)$ flows between the plates with a mean velocity of 0.5 fls. The flow is laminar. Determine the pressure drop per unit length in the direction of flow. What is the maximum velocity in the channel?

Transcript

00:02 Hi, in this question the velocity distribution is given by v is equal to h square by 8 mu into do p by do x of y minus 2y by h the whole square.

00:26 Now we will differentiate this with respect to y.

00:29 So dv by d y is equal to h square by 8 mu into do p by do x of d by d by y of y minus 2 y by h whole square.

00:53 Now upon differentiating this we will get d v by dy as h square by 8 mu x of dupi by x of 1 minus 4 y by its square...

Question

Question 2: What is the maximum $h$ that can be achieved in laminar flow over a 5m plate, based on data from Table A.3? What material/fluid gives this result?

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