In this post, I will present the analysis of draining a tank from below containing a power-law fluid. This situation is presented as Exercise 6.31 in the book "Chemical Engineering Fluid Mechanics" by Ronald Darby.
This problem has been adapted to give some more tractable data.
1 The situation
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- Assuming the oil to be Newtonian and neglecting the friction loss from the tank to the tube, answer the following: how fast will the tank drain?
- If the oil is well represented by a power law constitutive equation with an index of 0.4 and an apparent viscosity of $\eta=80 cP$ at a shear rate of $\gamma^\cdot =1 s^{-1}$, answer the following: how fast will the tank drain for this case?
1.1 Important data
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To perform the calculations, additional data needs to be specified. These are,
- fluid density: $54.50 lb_m/ft^3$,
- acceleration due to gravity: $32.17 ft/s^2$,
- pipe roughness: $15.00\times10^-4 mm$,
- pipe material: copper,
- schedule: K.
2 The solution
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The analysis of the situation requires exploring the results if the fluid were Newtonian and if it were represented as power-law. For reference, the proper formula can be found in the post:
Post: Hydraulic equations for non-Newtonian fluids
The situation should be analyzed with the equation of mechanical energy balance,
$\dfrac{P_1}{\gamma}+z_1-h_L+\dfrac{V_1^2}{2g}=\dfrac{P_2}{\gamma}+z_2+\dfrac{V_2^2}{2g}$ Eq. (01)
Before continuing with this, some assumptions are to be made,
- Suppose the tank diameter is so large that the fluid velocity in the tank is zero,
- The discharge pipe discharges at atmosphere, so that $P_2=0$,
- Discard the head loss at the tank exit.
2.1 Some simplifications
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We should notice that the head loss $h_L$ is composed as,
$h_L=h_L^{tank}+h_L^{pipe}$ Eq. (02)
$h_L=f_F^{tank}\dfrac{H}{D^{tank}}\dfrac{V_1^2}{2g}+f_F^{pipe}\dfrac{L}{D^{pipe}}\dfrac{V_2^2}{2g}$ Eq. (03)
Since, the tank diameter is too large then $V_1=0$ and that $P_2=0$, Eqs. (02-03) reduce to,
$\dfrac{P_1}{\gamma}+z_1-h_L=z_2+\dfrac{V_2^2}{2g}$ Eq. (05)
$h_L=f_F^{pipe}\dfrac{L}{D^{pipe}}\dfrac{V_2^2}{2g}$ Eq. (06)
One further simplification to Eq. (05) can be made if we consider that the reference ground is at the fluid surface. Then, $z_1=0$. Also, substitution of $h_L$ given in Eq. (06) into Eq. (05) results in,
$\dfrac{P_1}{\gamma} - f_F^{pipe}\dfrac{L}{D^{pipe}}\dfrac{V_2^2}{2g}=z_2+\dfrac{V_2^2}{2g}$ Eq. (07)
Equation (08) is the one from which the velocity is to be estimated,
$V_2=\sqrt{\dfrac{2g\left( \dfrac{P_1}{\gamma}-z_2 \right)}{1+f_F\dfrac{L}{D}}}$ Eq. (08)
As you can see, $V_2$ from Eq. (08) is independent of the nature of the fluid. This is, Eq. (08) is valid for Newtonian and power-law fluids. However, $f_F$ is not, and an adequate formula and needs to be considered.
Therefore, for any fluid at hand, the solution process is as follows,
- A fluid velocity $V_2$ is guessed,
- The Reynolds number $N_{Re}$ is estimated,
- The friction is estimated from the formula for either the laminar or the turbulent regime,
- Once $f_F$ is known, the fluid velocity $V_2$ is estimated from Eq. (08),
- An error, between the guessed $V_2$ and that found in point 4, needs to be calculated.
- If the error is not too small, say larger than $\times 10^{-8}$, the process should be repeated, taking the $V_2$ found in point 4 as the new guess.
The above procedure can be implemented either in Excel or any other software tool.
2.2 The case of the oil behaving as a Newtonian fluid
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As you can see, Eq. (08) is to be used for the estimation of the velocity $V_2$ and must be solved by using a numerical approach. The friction factor needs to be estimated using the Colebrook equation.
The data, then, needs to be used in the same unit system. In this case, the British units system is to be used. Then, the data is as follows,
- pressure $P_1=288.00 lb_f/ft^2$
- density $\rho=1.69 slug/ft^3$
- dynamic viscosity $\mu=1.67\times 10^{-3} slug/ft\,s$ (for the Newtonian fluid, the apparent viscosity $\eta$ is taken as the newtonian viscosity $\mu$)
- Pipe roughness $\epsilon=4.92\times 10^{-6}ft$
Using the above data, Eq. (08) can be solved by different numerical techniques. In this case, $V_2$ and $f_F^{pipe}$ shall be estimated using a solver available in Python. Therefore, this tool shall be directly used in Excel from the Anaconda add-in.
Thus, I shall directly post below the code,
#-------------------------------------------------------------------------------# Name: Newtonian V2 and fF calculator
# Purpose: Educative
#
# Author: Ilebrando Pérez Reyes
#
# Created: 28/09/2025
# Copyright: (c) ei-perez 2025
# Licence: <your licence>
#-------------------------------------------------------------------------------
import math as m
# Initial values
V2i = 1.0
fFi = 0.001
#
def IterationsV2fF(V2:float, fF:float)->[]:
P1 = float(288.0)
ga = float(54.37)
Lt = 10.0
Di = 0.16
gr = 32.17
z2 = -12.0
ep = 4.92e-6
mu_eta = 1.67e-3
den = 1.69
for j in range(0,50):
for i in range(0,20):
fFc = 0.25/(m.log10(ep/3.72/Di+2.51*mu_eta/den/V2/Di/m.sqrt(fF)))**2
fF = fFc # Subscript c means for calculation only (dummy)
V2c = m.sqrt(2*gr*(P1/ga - z2)/(1+fF*Lt/Di))
Error_check = abs(V2 - V2c)
if Error_check <= 1e-7:
break
else:
V2 = V2c
return[V2c,fF]
Solution = IterationsV2fF(V2i,fFi)
Solution
to be inserted in the console
Notice that in the console, the code output is selected. From this cell, $V_2$ and $f_F$ can be extracted from cell C24. This can be done from the Anaconda console as follows,
From the above calculations, it follows that: $V_2=17.14$.
2.3 The case of the oil behaving as a power-law fluid
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In this case, attention to the friction factor formulas is required.
Post: Hydraulic equations for non-Newtonian fluids
For the power-law fluid, the critical Reynolds number $N_{Re-c}$ needs to be estimated first. This gives,
Since $N_{Re-c}$ remains constant, a set of 3 conditionals is to be considered:
- For laminar regime $N_{Re}<N_{Re-c}$.
- For transition regime $N_{Re-c}<N_{Re}<4,000.00$.
- For turbulent regime $4,000.00<N_{Re}<10^5$.
for the calculation of the friction factor $f_{FF}$. You should recall that for non-Newtonian fluids, the friction factor is known as Fanning's. This procedure is to be embedded within the main calculation procedure outlined in Section 2.1.
Then, some data is required. The apparent viscosity $\eta=1.67\times 10^{-3} slug/ ft\, s$ is already known. However, the parameter $m$ is estimated from,
$m=\dfrac{\eta}{\dot{\gamma}^{n-1}}$ Eq. (09)
which, after the substitution of the known data, gives
In this case, I have chosen a simple implementation in Excel of the formulas and manually carried out the restriction due to the conditionals. One warning should be considered first: as the friction factor used is that of Fanning, the Darcy formula for head loss is expressed as
$h_L=2f_{FF}\dfrac{L}{D}\dfrac{V^2}{2g}$ Eq. (10)
Therefore, Eq. (08) rewrites as,
$V_2=\sqrt{\dfrac{2g\left( \dfrac{P_1}{\gamma}-z_2 \right)}{1+2f_{FF}\dfrac{L}{D}}}$ Eq. (11)
The implementation of the iterative procedure is summarised in the table below.
Finally, you may download the Excel file in the link below,
This is the end of the post. I hope you find it useful.
Any question? Write in the comments, and I shall try to help.
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- Solving the Colebrook equation
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- Example #03: single stage chemical evaporator
- Example #04: triple effect chemical evaporator
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Ildebrando.
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