The volume flow rate `Q ("in m"^(3) s^(-1))` of a liquid through pipe having diameter d is related to viscosity of water `'eta'` (unit Pascal. s) and the pressure gradient along the pipe `(dP)/(dx)` [pressure gradient `(dP)/(dx)` is rate of change of pressure per unit length along the pipe], by a formula of the form
`Q=k eta^(a)d^(b)((dP)/(dx))^(c)`
Where K is a dimensionless constant. Find a,b and c.

Answer

To solve the problem, we need to determine the values of \( a \), \( b \), and \( c \) in the equation:

\[ Q = k \eta^a d^b \left( \frac{dP}{dx} \right)^c \]

where:
- \( Q \) is the volume flow rate (in \( m^3/s \)),
- \( \eta \) is the viscosity of water (in \( \text{Pascal} \cdot s \)),
- \( d \) is the diameter of the pipe (in meters),
- \( \frac{dP}{dx} \) is the pressure gradient (in \( \text{Pascal/m} \)),
- \( k \) is a dimensionless constant.

### Step 1: Determine the dimensions of each variable

1. **Volume flow rate \( Q \)**:
\[ [Q] = L^3 T^{-1} \]
(where \( L \) is length and \( T \) is time)

2. **Viscosity \( \eta \)**:
\[ \eta = \text{Pascal} \cdot s = \frac{\text{Newton}}{\text{meter}^2} \cdot \text{s} = \frac{(M L T^{-2})}{L^2} \cdot T = M L^{-1} T^{-1} \]
So,
\[ [\eta] = M L^{-1} T^{-1} \]

3. **Diameter \( d \)**:
\[ [d] = L \]

4. **Pressure gradient \( \frac{dP}{dx} \)**:
\[ \frac{dP}{dx} = \frac{\text{Force}}{\text{Area} \cdot \text{Length}} = \frac{M L T^{-2}}{L^2} \cdot \frac{1}{L} = M L^{-2} T^{-2} \]
So,
\[ \left[ \frac{dP}{dx} \right] = M L^{-2} T^{-2} \]

### Step 2: Substitute dimensions into the equation

Now we substitute the dimensions into the equation:

\[ [L^3 T^{-1}] = [k] [M L^{-1} T^{-1}]^a [L]^b [M L^{-2} T^{-2}]^c \]

Since \( k \) is dimensionless, we can ignore its dimensions. This gives us:

\[ L^3 T^{-1} = [M^a L^{-a} T^{-a}] [L^b] [M^c L^{-2c} T^{-2c}] \]

### Step 3: Combine the dimensions

Combining the dimensions on the right side:

\[ L^3 T^{-1} = M^{a+c} L^{-a+b-2c} T^{-a-2c} \]

### Step 4: Set up equations for each dimension

Now we can equate the powers of \( M \), \( L \), and \( T \):

1. For \( M \):
\[ a + c = 0 \quad \text{(1)} \]

2. For \( L \):
\[ -a + b - 2c = 3 \quad \text{(2)} \]

3. For \( T \):
\[ -a - 2c = -1 \quad \text{(3)} \]

### Step 5: Solve the equations

From equation (1):
\[ c = -a \]

Substituting \( c = -a \) into equation (3):
\[ -a - 2(-a) = -1 \implies -a + 2a = -1 \implies a = -1 \]

Now substituting \( a = -1 \) back into equation (1):
\[ c = -(-1) = 1 \]

Now substituting \( a = -1 \) and \( c = 1 \) into equation (2):
\[ -(-1) + b - 2(1) = 3 \implies 1 + b - 2 = 3 \implies b - 1 = 3 \implies b = 4 \]

### Final Values

Thus, we have:
- \( a = -1 \)
- \( b = 4 \)
- \( c = 1 \)

### Summary of the Solution

The values of \( a \), \( b \), and \( c \) are:
- \( a = -1 \)
- \( b = 4 \)
- \( c = 1 \)