Question 6.2: For a steady, fully developed laminar flow through a duct, t...
For a steady, fully developed laminar flow through a duct, the pressure drop per unit length of the duct \Delta p / l is constant in the direction of flow and depends on the average flow velocity V, the hydraulic diameter of the duct D_{h} , the density \mathcal{\rho} , and the viscosity \mu , of the fluid. Find out the pertinent dimensionless groups governing the problem by the use of Buckingham’s \pi theorem.
The variables involved in the problem are
\frac{\Delta p}{l}, V, D_{h}, \rho, \muHence, m=5.
The fundamental dimensions in which these five variables can be expressed are M (mass),L (length) and T time . Therefore, n = 3. According to Pi theorem, the number of independent \pi terms is (5–3) = 2, and the problem can be expressed as
f\left(\pi_{1} \pi_{2}\right)=0 (6.10)
In determining \pi_{1} and \pi_{2} , the number of repeating variables that can be taken is 3. The term \Delta p / l being the dependent variable should not be taken as the repeating one.
Therefore, choices are left with V, D_{h}, \rho and \mu . Incidentally,any combination of three out of these four quantities involves all the fundamental dimensions M, L and T.
Hence any one of the following four possible sets of repeating variables can be used:
V, D_{h}, \rhoV, D_{h}, \mu
D_{h}, r, \mu
V, r, \mu
Let us first use the set V, D_{h} and \rho . Then the \pi terms can be written as
\pi_{1}=V^{a} D_{h}^{b} \rho^{c} \Delta p / l (6.11)
\pi_{2}=V^{a} D_{h}^{b} \rho^{c} \mu (6.12)
Expressing the Eqs (6.11) and (6.12) in terms of the fundamental dimensions of the variables, we get
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c} \mathrm{ML}^{-2} \mathrm{~T}^{-2} (6.13)
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c} \mathrm{ML}^{-1} \mathrm{~T}^{-1} (6.14)
Equating the exponents of M, L and T on both sides of Eq. (6.13) we have,
c + 1 = 0
a + b – 3c – 2 = 0
– a – 2 = 0
which give a=-2, b=1 \quad \text { and } \quad c=-1
\therefore p_{1}=\frac{\Delta p D_{h}}{l \rho V^{2}}
Similarly from Eq. (6.14)
c + 1 = 0
a + b – 3c – 1 = 0
– a – 1 = 0
which give a = –1, b = –1, and c = –1
Therefore, \pi_{2}=\frac{\mu}{V D_{h} \rho}
Hence, Eq. (6.10) can be written as
F\left(\frac{\Delta p D_{h}}{l \rho V^{2}}, \frac{\mu}{V D_{h} \rho}\right)=0 (6.15)
The term \pi_{2} is the reciprocal of Reynolds number, Re, as defined earlier.Equation (6.15) can also be expressed as
f\left(\frac{\Delta p D_{h}}{l \rho V^{2}}, \frac{V D_{h} \rho}{\mu}\right)=0 (6.16)
or \frac{\Delta p D_{h}}{l \rho V^{2}}=\phi(\mathrm{Re}) (6.17)
The term \pi_{1} \text {,i.e., } \frac{\Delta p D_{h}}{l \rho V^{2}} is known as the friction factor in relation to a fully developed flow through a closed duct.
Let us now choose V, D_{h} and \mu as the repeating variables.
Then
\pi_{1}=V^{a} D_{h}^{b} \mu^{c}(\Delta p / l) (6.18)
\pi_{2}=V^{a} D_{h}^{b} \mu^{c} \rho (6.19)
Expressing the right-hand side of Eq. (6.18) in terms of fundamental dimensions, we have
\mathbf{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a} \mathrm{~L}^{b}\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)^{c} \mathrm{ML}^{-2} \mathrm{~T}^{-2}Equating the exponents of M, L and T from above,
c + 1 = 0
a + b – c – 2 = 0
– a – c – 2 = 0
Finally, a=-1, b=2, c=-1
Therefore, \pi_{1}=\frac{\Delta p}{l} \frac{D_{h}^{2}}{V \mu}
Similarly, equating the exponents of fundamental dimensions of the variables on both sides of Eq. (6.19) we get
a=1, b=1, c=-1Therefore, \pi_{2}=\frac{\rho V D_{h}}{\mu}
Hence, the same problem which was defined by Eq. (6.15) can also be defined by the equation
f\left(\frac{\Delta p}{l} \frac{D_{h}^{2}}{V \mu}, \frac{\rho V D_{h}}{\mu}\right)=0 (6.20)
Though the Eqs (6.15) and (6.20) are not identical, but they are interdependent.
Now if we write the two sets of \pi terms obtained straightforward from the application of \pi theorem as
|
\pi_{1} |
\pi_{2} |
|
|
Set 1: |
\frac{\Delta p D_{h}}{l \rho V^{2}}, |
\frac{\mu}{\rho V D_{h}} |
|
Set 2: |
\frac{\Delta p D_{h}^{2}}{l V \mu}, |
\frac{\rho V D_{h}}{\mu} |
We observe that
\left(1 / \pi_{2}\right) of Set 2 =\left(\pi_{2}\right) of Set 1
And \left(\pi_{1} / \pi_{2}\right) of Set 2 =\left(\pi_{1}\right) of Set 1
Therefore, it can be concluded that, from one set of \pi terms, one can obtain the other set by some combination of the \pi terms of the existing set. It is justified both mathematically and physically that the functional relationship of \pi terms representing a problem in the form
f\left(\pi_{1}, \pi_{2}, \ldots \pi_{r}\right)=0is equivalent to any implicit functional relationship between other \pi terms obtained from any arbitrary mathematical combination of \pi terms of the existing set, provided the total number of independent p terms remains the same. For example, Eq. (6.20) and Eq. (6.15) can be defined in terms of \pi parameters of the Set 2 as f\left(\pi_{1}, \pi_{2}\right)=0 \text { and } F\left(\pi_{1} /\right. \left.\pi_{2}, 1 / \pi_{2}\right)=0 , respectively.
Table 6.3 shows different mutually interdependent sets of \pi terms obtained from all possible combinations of the repeating variables of Example 6.2. Though the different sets of \pi terms as shown in Column 2 of Table 6.3 are mathematically meaningful, many of them lack physical significance. The physically meaningful parameters of the problem are \Delta p D_{h} / l \rho V^{2} are \rho V D_{h} / \mu and are known as friction factor and Reynolds number, respectively. Therefore while selecting the repeating variables, for a fluid flow problem, it is desirable to choose one variable with geometric characteristics, another variable with flow characteristics and yet another variable with fluid properties. This ensures that the dimensionless parameters obtained will be the meaningful ones with respect to their physical interpretations.
Table 6.3 Different sets of \pi terms resulting from different combinations of repeating variables of a pipe flow problem
|
Repeating variables |
Set of \pi terms |
Functional relation |
|
| \pi_{l} |
\pi_{2} |
||
|
V, D_{h}, \rho |
\frac{\Delta p D_{h}}{l \rho V^{2}} | \frac{\mu}{\rho V D_{h}} | F\left(\frac{\Delta p D_{h}}{l \rho V^{2}}, \frac{\mu}{\rho V D_{h}}\right)=0 |
| V, D_{h}, \mu | \frac{\Delta p D_{h}^{2}}{l V \mu} | \frac{\rho V D_{h}}{\mu} |
f\left(\frac{\Delta p D_{h}^{2}}{I V \mu}, \frac{\rho V D_{h}}{\mu}\right)=0 |
|
D_{h}, \rho, \mu |
\frac{\Delta p D_{h}^{3} \rho}{I \mu^{2}} | \frac{\rho V D_{h}}{\mu} |
\phi\left(\frac{\Delta p D_{h}^{3} \rho}{l \mu^{2}}, \frac{\rho V D_{h}}{\mu}\right)=0 |
|
V, \rho, m |
\frac{\Delta p \mu}{l V^{3} \rho^{2}} | \frac{\rho V D_{h}}{\mu} |
\psi\left(\frac{\Delta p \mu}{l V^{3} \rho^{2}}, \frac{\rho V D_{h}}{\mu}\right)=0 |
The above discussion on Buckingham’s p theorem can be summarised as follows:
The above discussion on Buckingham’s p theorem can be summarised as follows:
(i) List the m physical quantities involved in a particular problem. Note the number n, of the fundamental dimensions to express the m quantities. There will be (m–n) \pi terms.
(ii) Select n of the m quantities, excluding any dependent variable, none dimensionless and no two having the same dimensions. All fundamental dimensions must be included collectively in the quantities chosen.
(iii) The first \pi term can be expressed as the product of the chosen quantities each raised to an unknown exponent and one other quantity.
(iv) Retain the quantities chosen in (ii) as repeating variables and then choose one of the remaining variables to establish the next \pi term in a similar manner as described in (iii). Repeat this procedure for the successive \pi terms.
(v) For each \pi term, solve for the unknown exponents by dimensional analysis
(vi) If a quantity out of m physical variables is dimensionless, it is a \pi term.
(vii) If any two physical quantities have the same dimensions, their ratio will be one of the \pi terms
(viii) Any \pi term may be replaced by the term, raised to an exponent. For example, \pi_{3} may be replaced by \pi_{3}^{2} or \pi_{2} by \sqrt{\pi_{2}}
(ix) Any \pi term may be replaced by multiplying it by a numerical constant. For example, \pi_{1} may be replaced by 3 \pi_{1}