The pressure drop, \(\Delta p,\) along a straight pipe of diameter \(D\) has been experimentally studied, and it is observed that for laminar flow of a given fluid and pipe, the pressure drop varies directly with the distance, \(\ell\), between pressure taps. Assume that \(\Delta p\) is a function of \(D\) and \(\ell\), the velocity, \(V\), and the fluid viscosity, \(\mu\). Use dimensional analysis to deduce how the pressure drop varies with pipe diameter.

The known variables are Pressure drop, \(\Delta p\), Diameter of pipe, \(D\), Distance between pressure taps, \(\ell\), Velocity, \(V\), and Fluid viscosity, \(\mu\).

The dimensions of each variable are as follows: - Pressure drop, \(\Delta p\), is a force per unit area, which in terms of primary dimensions is Mass (M) Length (L) Time (T): \([M^1L^{-1}T^{-2}]\)- Diameter, \(D\), and Distance between pressure taps, \(\ell\), are both Length (L): \([L^1]\)- Velocity, \(V\), is Length (L) Time (T): \([L^1T^{-1}]\)- Viscosity, \(\mu\), is Mass (M) per Length (T) Time (L): \([ML^{-1}T^{-1}]\)

According to the Buckingham's Pi Theorem, we if have n-variables and m-primary dimensions (M, L & T), there will be n-m (5-3=2) dimensionless variables (often referred as Pi terms). Now, Let's perform dimensional analysis, we assume the relationship between variables as: \(\Delta p = C D^a \ell ^b V^c \mu^d\) \(where, C, a, b, c, and d are constants to be determined\) Equating the dimensions on both sides, we have: \( [M^1L^{-1}T^{-2}] = [L]^a . [L]^b .[L^1T^{-1}]^c . [ML^{-1}T^{-1}]^d As we know the pressure drop, \(\Delta p\), varies directly with the distance, \(\ell\), b is found to be 1. Therefore, we can write the relationship as, \(\Delta p = C D^a \ell V^c \mu^d\) \(or, \) \( [M^1L^{-1}T^{-2}] = [L]^a . [L] . [L^1T^{-1}]^c . [ML^{-1}T^{-1}]^d)

To find the remaining constants, we equate the powers of primary dimensions on the left-hand side and right-hand side. Equating powers of M gives d = 1. Equating powers of L gives a + c = -1. Equating powers of T gives -c - d = -2. Solving these equations, we get a = -1, c = 0 and d = 1.

Substituting the values of a, c, d in the equation, we get \(\Delta p = C D^{-1}\ell V^0 \mu^1\). Thus, the pressure drop varies inversely with the pipe diameter for a given fluid and pipe under laminar flow.