**Problem Statement:**
A pipeline with a diameter of 0.8 m and a length of 3 km is given. To increase the discharge, another pipe with the same diameter is introduced parallel to the first pipe in the second half of its length. The Darcy-Weisbach friction factor is given as 0.04 and the head at the inlet is 40 m. Find the increase in discharge.

**Solution:**

To find the increase in discharge, we need to calculate the difference in discharge between the initial pipeline and the augmented pipeline.

**Step 1: Calculate the initial discharge in the pipeline (Q1)**

The discharge in a pipeline can be calculated using the Darcy-Weisbach equation:

Q = (π/4) x D^2 x v

Where:
Q = Discharge (m^3/s)
D = Diameter of the pipeline (m)
v = Velocity of flow (m/s)

Given:
D = 0.8 m

To find the velocity of flow, we can use the equation of continuity:

A1 x v1 = A2 x v2

Where:
A = Cross-sectional area of the pipeline (m^2)
v = Velocity of flow (m/s)

Since the pipeline is circular, we can use the formula for the cross-sectional area of a circle:

A = (π/4) x D^2

For the initial pipeline:
A1 = (π/4) x (0.8)^2
A1 = 0.502 m^2

Using the equation of continuity, we can rearrange it to find v1:

v1 = (A2 x v2) / A1

Since the diameter of the pipeline is the same throughout, A1 = A2. Therefore, we can simplify the equation to:

v1 = v2

Now, we can substitute the values into the discharge equation:

Q1 = (π/4) x (0.8)^2 x v1

**Step 2: Calculate the augmented discharge in the pipeline (Q2)**

Since another pipe is introduced parallel to the first pipe in the second half of its length, the total cross-sectional area of the flow will be doubled. Therefore, the cross-sectional area of the augmented pipeline will be:

A2 = 2 x A1
A2 = 2 x 0.502
A2 = 1.004 m^2

Using the equation of continuity, we can calculate the velocity of flow in the augmented pipeline:

v2 = (A1 x v1) / A2

Now, we can substitute the values into the discharge equation:

Q2 = (π/4) x (0.8)^2 x v2

**Step 3: Calculate the increase in discharge**

The increase in discharge can be calculated by subtracting the initial discharge from the augmented discharge:

ΔQ = Q2 - Q1

Substituting the values, we get:

ΔQ = (π/4) x (0.8)^2 x v2 - (π/4) x (0.8)^2 x v1

Simplifying the equation, we get:

ΔQ = (π/4) x (0.8)^2 x (v2 - v1)

**Step 4: Calculate the velocity difference (v2 - v1)**

To calculate