Chapter 13, Problem 13.4P

Interpretation Introduction

Interpretation:

The power required to pump the oil in the steel pipe is to be calculated.

Concept Introduction:

The formula used to calculate the friction factor $\left(f_f\right)$ for $4\times {10}^4\le \mathrm{Re}\le {10}^8$ and $0\le e/D\le 0.05$ is:

  $f_f=\frac{1}{{\left[-3.6{\log }_{10}\left[\frac{6.9}{\mathrm{Re}}+{\left(\frac{1}{3.7}\frac{e}{D}\right)}^{10/9}\right]\right]}^2}$ ........... (1)

Here,

  $\begin{array}{c}\mathrm{Re}=\text{Reynolds number}\end{array}$

  $\begin{array}{c}\frac{e}{D}=\text{Relative roughness}\end{array}$

The formula for head loss due to pipe fittings to be used is:

  $h_L=2f_f\frac{L}{D}\frac{v^2}{g}$ .....(2)

Here,

  $\begin{array}{c}{h}_L=\text{Head loss}\end{array}$

  $\begin{array}{c}L=\text{Pipe length}\end{array}$

  $\begin{array}{c}D=\text{Diameter of the pipe}\end{array}$

  $\begin{array}{c}v=\text{Velocity of the fluid in the pipe}\end{array}$

  $\begin{array}{c}g=\text{Acceleration due to gravity}\end{array}$

The general energy balance equation according to the first law of thermodynamics written for a control volume is:

  $\frac{\delta Q}{dt}-\frac{\delta W}{dt}={\displaystyle {\iint }_{C.S}\left(e+\frac{P}{\rho }\right)\rho \left(v\cdot n\right)dA}+\frac{\partial }{\partial t}{\displaystyle {\iiint }_{C.V}e\rho dV}+\frac{\delta W_{\mu }}{dt}$.... (3)

Here, C.S is the control surface over which the integral dA is taken, C.V is the control volume over which the integral dV is taken, $\rho$ is the density of the fluid, $v$ is the velocity vector, $n$ is the direction of the vector $v$ , $\delta Q/dt$ is the rate of heat added (positive) or removed (negative) from the system, $\delta W/dt$ is the rate of work done by (positive) or on (negative) the system, and $\delta W_{\mu }/dt$ is the rate of work done to overcome the viscous effect at the control surface. The product $v\cdot n$ is scalar defined as:

  $v\cdot n=\left|v\right|\left|n\right|\cos \theta$