Water flows at a rate of 0.035 m3/s in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pressure at the centerline is measured to be 480 kPa and 445 kPa before and after the reducer, respectively, determine the head loss in the reducer. Express your answer in m.

To solve this problem, we can use the Bernoulli's equation, which states that the total energy at any point in a fluid flow system is constant. The Bernoulli's equation can be written as:

P1/ρg + v1^2/2g + z1 = P2/ρg + v2^2/2g + z2 + hL

Where: P1 and P2 are the pressures at points 1 and 2, respectively. ρ is the density of the fluid. g is the acceleration due to gravity. v1 and v2 are the velocities at points 1 and 2, respectively. z1 and z2 are the elevations at points 1 and 2, respectively. hL is the head loss.

In this case, we can assume that the elevation at both points is the same, so z1 = z2.

Given: P1 = 480 kPa = 480,000 Pa P2 = 445 kPa = 445,000 Pa Q = 0.035 m^3/s (flow rate) d1 = 15 cm = 0.15 m (diameter at point 1) d2 = 8 cm = 0.08 m (diameter at point 2) ρ = 1000 kg/m^3 (density of water) g = 9.81 m/s^2 (acceleration due to gravity)

First, let's calculate the velocities at points 1 and 2 using the flow rate:

v1 = Q/A1 v2 = Q/A2

Where: A1 = π(d1/2)^2 (cross-sectional area at point 1) A2 = π(d2/2)^2 (cross-sectional area at point 2)

Substituting the values:

A1 = π(0.15/2)^2 = 0.01767 m^2 A2 = π(0.08/2)^2 = 0.00503 m^2

v1 = 0.035/0.01767 = 1.98 m/s v2 = 0.035/0.00503 = 6.96 m/s

Now, let's substitute the values into the Bernoulli's equation:

480,000/(10009.81) + (1.98)^2/(29.81) = 445,000/(10009.81) + (6.96)^2/(29.81) + hL

Solving the equation for hL gives:

hL = (480,000 - 445,000)/(10009.81) + (1.98)^2/(29.81) - (6.96)^2/(2*9.81)

hL = 35,000/9810 + 0.20 - 2.47

hL = 3.57 - 2.27

hL = 1.3 m

Therefore, the head loss in the reducer is 1.3 m.