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# issues from the tank in shown at Q = 45 f3/h. Assume laminar flow conditions and neglect entrance and exit losses. (a) ...

issues from the tank in shown at Q = 45 f3/h. Assume laminar flow conditions and neglect entrance and exit losses. (a) What is the velocity of flow in the pipe. Ans: 9.167 ft/s (b) Determine the head lost in the pipe. Ans: 7.695 feet - Determine the kinematic viscosity of the paint in ft?/s. Ans:0.0002444 Transcribed: Problem 7 - 77 Paint issues from the tank in shown at Q = 45 ft/h. Assume laminar flow conditions and neglect entrance and exit losses. 9 ft L=6 ft, d 2 in (a) What is the yelocity of flow in the pipe. Ans: 9.167 ft/s (b) Determine the head lost in the pipe. Ans: 7.695 feet (c) Determine the kinematic viscosity of the paint in ft2/s. Ans: 0.0002444

(a) Velocity of flow (V)

(b) Head lost in the pipe (HL)

(c) Kinematic viscosity of the plant ($\nu$)

Given:

Head over the pipe (H) = 9 ft

Length of the pipe (L) = 6 ft

Diameter of the pipe (D) = 0.5 in

Flow rate through the pipe (Q) = 45 ft3/h

NOTE: The flow is mentioned to be laminar and minor losses are to be neglected.

The flow rate through the pipe can be expressed in terms of the velocity of flow and cross-sectional area of the pipe as:

$Q=V×\left(\frac{\mathrm{\pi }}{4}×{D}^{2}\right)$

$\to V=\frac{4×Q}{\mathrm{\pi }×{\mathrm{D}}^{2}}$

The head lost in the pipe can be expressed as:

here,

g: acceleration due to earth's gravity = 32.2 ft/s2

Using the Darcy-Weisbach formula, head loss in a pipe can be given as:

$HL=\frac{f×L×{V}^{2}}{2×g×D}$

here,

First let us determine the friction factor using the Darcy-Weisbach formula and the head loss computed above.

$⇒f=0.0409$

Now, using the calculated value of friction factor and the formula for friction factor, we can obtain the kinematic viscosity.

$f=\frac{64}{{R}_{e}}=\frac{64}{\frac{V×D}{\nu }}$

$\to \nu =\frac{f×V×D}{64}$

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