To derive the expression for the volume (\( V \)) of liquid passing per second through a pipe when the flow is steady using the method of dimensions, we can analyze the variables involved and their dimensions.
Let's start with the variables given:
- \( n \): coefficient of viscosity of the liquid (dimension: \( ML^{-1}T^{-1} \))
- \( r \): radius of the pipe (dimension: \( L \))
- \( \frac{p}{\lambda} \): pressure gradient (dimension: \( ML^{-1}T^{-2} \))
- \( k \): dimensionless constant
We want to find the relationship between these variables and the resulting volume (\( V \)) of liquid passing per second.
The dimensions of volume (\( V \)) are \( L^3T^{-1} \).
Now, let's analyze the given variables in terms of their dimensions:
\[ [V] = L^3T^{-1} \]
\[ [n] = ML^{-1}T^{-1} \]
\[ [r] = L \]
\[ [\frac{p}{\lambda}] = ML^{-1}T^{-2} \]
To derive the expression, we equate the dimensions on both sides of the equation:
\[ L^3T^{-1} = k \cdot (ML^{-1}T^{-1})^a \cdot L^b \cdot (ML^{-1}T^{-2})^x \]
Now, let's equate the dimensions for each term:
1. For dimensions of length (\( L \)):
\[ 3 = 0 + b \]
So, \( b = 3 \).
2. For dimensions of time (\( T \)):
\[ -1 = -a - 2x \]
So, \( a + 2x = 1 \).
3. For dimensions of mass (\( M \)):
\[ 0 = a + x \]
So, \( a = -x \).
Now, substituting \( a = -x \) into \( a + 2x = 1 \):
\[ (-x) + 2x = 1 \]
\[ x = 1 \]
Substituting \( x = 1 \) back into \( a = -x \):
\[ a = -1 \]
Therefore, the correct expression for the volume (\( V \)) of liquid passing per second through the pipe is:
\[ V = k \cdot n^{-1} \cdot r^3 \cdot \left(\frac{p}{\lambda}\right) \]
where \( k \) is a dimensionless constant.
