Question

A 100-mm pipe is used to transfer oil from a reservoir to a 100- liter tank. It takes 45 minutes to fill the tank with oil that has viscosity and density of 0.005 Pas and 900 kg/m3, respectively Determine a. The volume flow rate of oil b. Mean velocity in the pipe; c. Reynolds number, and d. The maximum velocity in the pipe

Answer 1

Explanation:

Given that,

Diameter =100 mm

Volume = 100 liter

Time = 45 min

Viscosity = 0.005 Pas

Density = 900 kg/m³

(a). We need to calculate the volume flow rate of oil

Using formula of flow rate

$q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{0.1}{2700}$

$Q=3.7\times10^{-5}\ m^3/s$

The volume flow rate of oil is $3.7\times10^{-5}\ m^3/s$

(b). We need to calculate the mean velocity in the pipe

Using formula of mean velocity

$v=\dfrac{Q}{A}$

$v=\dfrac{Q}{\dfrac{\pi}{4}\times d^2}$

Put the value into the formula

$v=\dfrac{3.7\times10^{-5}}{\dfrac{\pi}{4}\times(100\times10^{-3})^2}$

$v=4.7\times10^{-3}\ m/s$

The mean velocity in the pipe is $4.7\times10^{-3}\ m/s$.

(c). We need to calculate the Reynolds number

Using formula of the Reynolds number

$R_{e}=\dfrac{\rho v d}{\mu}$

Put the value in to the formula

$R_{e}=\dfrac{900\times4.7\times10^{-3}\times100\times10^{-3}}{0.005}$

$R_{e}=84.6$

The Reynolds number is 84.6.

(d). We need to calculate the maximum velocity in the pipe

Using formula of maximum velocity

$V_{max}=2v_{avg}$

Put the value into the formula

$v_{max}=2\times4.7\times10^{-3}$

$v_{max}=9.4\times10^{-3}\ m/s$

The maximum velocity in the pipe is $9.4\times10^{-3}\ m/s$

Hence, This is the required solution