Sure! Let's the problem **step step** using **Buckingham Theorem**--- ## **Step 1: List All Variables and Their Dimensions** Given variables: - \( Q \): Flowrate (unknown if mass or volumetric; assume **volumetric**: \( [L^3 T^{-1}] \)) - \( \rho_a \): Density of ambient air \([M L^{-3}]\) - \( \rho_g \): Density of gas in stack \([M L^{-3}]\) - \( g \): Acceleration due to gravity \([L T^{-2}]\) - \( h \): Stack height \([L]\) - \( d \): Stack diameter \([L]\) --- ## **Step 2: Express the Functional Relationship** \[ Q = f(\rho_a, \rho_g, g, h, d) \] --- ## **Step 3: Count Number of Variables and Fundamental Dimensions** - **Variables (\(n\))**: 6 (\(Q, \rho_a, \rho_g, g, h, d\)) - **Fundamental dimensions (\(k\))**: 3 (Mass \(M\), Length \(L\), Time \(T\)) --- ## **Step 4: Number of Pi Terms** \[ \text{Number of } \pi \text{ terms} = n - k = 6 - 3 = 3 \] --- ## **Step 5: Choose Repeating Variables** We are told to use **\(\rho_a\), \(d\), and \(g\)** as repeating variables. --- ## **Step 6: Form Pi Terms** Each Pi term is dimensionless. ### **General Form** \[ \pi = Q^a \, \rho_a^b \, d^c \, g^d \, X^e \] where \(X\) is the non-repeating variable in that Pi term. Let's form the Pi terms for each remaining variable (\(Q, \rho_g, h\)): --- ### **Pi Term 1: Involving Q** \[ \pi_1 = Q \cdot \rho_a^a d^b g^c \] Dimensions: - \( Q: [L^3 T^{-1}] \) - \( \rho_a: [M L^{-3}] \) - \( d: [L] \) - \( g: [L T^{-2}] \) Combine: \[ [L^3 T^{-1}] [M L^{-3}]^a [L]^b [L T^{-2}]^c \] \[ = M^a \, L^{3 - 3a + b + c} \, T^{-1 - 2c} \] Set exponents to zero for dimensionless: \[ \begin{cases} M: & a = \\ L: & 3 - 3a + b + c = \\ T: & -1 - 2c = \end{cases} \] Solve: - \(a = \) - \( -1 - 2c = \implies c = -.5 \) - \( 3 + b - .5 = \implies b = -2.5 \) So, \[ \pi_1 = \dfrac{Q}{d^{2.5} g^{.5}} \] Or, more commonly, rationalize the exponents: \[ \boxed{ \pi_1 = \frac{Q}{d^{5/2} g^{1/2}} } \] Alternatively, using \(\rho_a\) with exponent (so it does not appear). --- ### **Pi Term 2: Involving \(\rho_g\)** \[ \pi_2 = \rho_g \cdot \rho_a^a d^b g^c \] Dimensions: - \( \rho_g: [M L^{-3}] \) \[ [M L^{-3}] [M L^{-3}]^a [L]^b [L T^{-2}]^c = M^{1+a} L^{-3 + b + c - 3a} T^{-2c} \] Set exponents to zero: \[ \begin{cases} M: & 1+a = \implies a = -1 \\ L: & -3 - 3a + b + c = \implies -3 + 3 + b + c = \implies b + c = \\ T: & -2c = \implies c = \end{cases} \] So, \(a = -1\), \(c = \), \(b = \): \[ \pi_2 = \frac{\rho_g}{\rho_a} \] --- ### **Pi Term 3: Involving \(h\)** \[ \pi_3 = h \cdot \rho_a^a d^b g^c \] Dimensions: - \( h: [L] \) \[ [L] [M L^{-3}]^a [L]^b [L T^{-2}]^c = M^a L^{1-3a+b+c} T^{-2c} \] Set exponents to zero: \[ \begin{cases} M: & a = \\ L: & 1 + b + c = \implies b + c = -1 \\ T: & -2c = \implies c = \end{cases} \] So, \(a=\), \(c=\), \(b=-1\): \[ \pi_3 = \frac{h}{d} \] --- ## **Step 7: Final Set of Pi Terms** \[ \boxed{ \begin{align*} \pi_1 &= \frac{Q}{d^{5/2} g^{1/2}} \\ \pi_2 &= \frac{\rho_g}{\rho_a} \\ \pi_3 &= \frac{h}{d} \end{align*} } \] ## **Summary/Final Answer** The set of dimensionless Pi terms describing the problem, using \(\rho_a\), \(d\), and \(g\) as repeating variables, is: \[ \boxed{ \left\{ \frac{Q}{d^{5/2} g^{1/2}}, \quad \frac{\rho_g}{\rho_a}, \quad \frac{h}{d} \right\} } \] These terms can be used to write the functional relationship in dimensionless form: \[ f\left( \frac{Q}{d^{5/2} g^{1/2}}, \frac{\rho_g}{\rho_a}, \frac{h}{d} \right) = \]