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

ASKED: To determine-

(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:

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

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

$\to V=\frac{4\times 45 f{t}^3/h}{\mathrm{\pi }\times {\left(0.5 \mathrm{in}\times {\displaystyle \frac{1 \mathrm{ft}}{12 \mathrm{in}}}\right)}^2}\times \frac{1 h}{60 min}\times \frac{1 min}{60 s}$

$\Rightarrow \boxed{ V=9.167 ft/s }$

The head lost in the pipe can be expressed as:

$HL=Head over the pipe-Velocity head=H-\frac{V^2}{2g}$

here,

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

$\to HL=9 ft-\frac{\left(9.167 ft/s\right)}{2\times 32.2 ft/s^2}$

$\Rightarrow \boxed{ HL=7.695 ft }$

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

$HL=\frac{f\times L\times V^2}{2\times g\times D}$

here,

$f: friction factor, for laminar flow f=\frac{64}{Reynolds number \left(R_e\right)}$

$Reynolds number \left(R_e\right)=\frac{V\times D}{\nu }$

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

$\to 7.695 ft=\frac{f\times 6 ft\times {\left(9.167 ft/s\right)}^2}{2\times 32.2 ft/s^2\times \left(0.5 in\times {\displaystyle \frac{1 ft}{12 in}}\right)}$

$\Rightarrow f=0.0409$

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

$f=\frac{64}{R_e}=\frac{64}{{\displaystyle \frac{V\times D}{\nu }}}$

$\to \nu =\frac{f\times V\times D}{64}$

$\to \nu =\frac{0.0409\times 9.167 ft/s\times \left(0.5 in\times {\displaystyle \frac{1 ft}{12 in}}\right)}{64}$

$\Rightarrow \boxed{ \nu =0.000244 f{t}^2/s }$