(1) To determine the values of velocity (v) and height (h) for points (1), (2), and (3), we will use the Bernoulli’s equation, which states that the total energy at any point along a streamline is constant. We will assume that the flow is steady, incompressible, and has negligible losses due to friction.

At point (1), the height is not given, so we will assume that it is at the same level as point (2) and point (3). Therefore, the height at all three points is h = 0.

Using Bernoulli’s equation between points (1) and (2), we have:

P1/ρg + v1^2/2g + h1 = P2/ρg + v2^2/2g + h2

where P is pressure, ρ is density, g is acceleration due to gravity, v is velocity, and h is height.

Since the pipe between the two fittings is vertical, the velocity at point (1) is zero (v1 = 0). Therefore, we can simplify the equation to:

P1/ρg + h1 = P2/ρg + v2^2/2g + h2

Plugging in the given values, we get:

12 kPa / (1000 kg/m^3 x 9.81 m/s^2) + 0 = 11 kPa / (1000 kg/m^3 x 9.81 m/s^2) + v2^2 / (2 x 9.81 m/s^2) + 10 m / 1000 = 0

Simplifying and solving for v2, we get:

v2 = sqrt((12 – 11) x 1000 / 2) = 7.07 m/s

Using Bernoulli’s equation between points (2) and (3), we have:

P2/ρg + v2^2/2g + h2 = P3/ρg + v3^2/2g + h3

Again, since the pipe between the two fittings is vertical, the velocity at point (3) is zero (v3 = 0). Therefore, we can simplify the equation to:

P2/ρg + v2^2/2g + h2 = P3/ρg

Plugging in the given values, we get:

11 kPa / (1000 kg/m^3 x 9.81 m/s^2) + 7.07^2 / (2 x 9.81) + 0 = 10 kPa / (1000 kg/m^3 x 9.81 m/s^2)

Simplifying and solving for h2, we get:

h2 = (11 – 10) x 1000 / (2 x 9.81) – 7.07^2 / (2 x 9.81) = -6.06 m

Therefore, the values of velocity and height for points (1), (2), and (3) are:

Point (1): v1 = 0 m/s, h1 = 0 m Point (2): v2 = 7.07 m/s, h2 = -6.06 m Point (3): v3 = 0 m/s, h3 = 0 m

(2) To determine the friction head loss between points (1) and (2), we can use the Darcy-Weisbach equation:

hf = f * (L/D) * (v^2/2g)

where hf is the friction head loss, f is the friction factor, L is the length of the pipe, D is the diameter of the pipe, v is the velocity of the water, and g is the acceleration due to gravity.

First, we need to calculate the velocity of the water at point (2):

A2 = π/4 * (0.03 m)^2 = 7.07 x 10^-4 m^2 (cross-sectional area) Q = 5 L/min = 8.33 x 10^-5 m^3/s (volumetric flow rate) v2 = Q/A2 = 1.18 m/s (velocity)

Now we can calculate the friction head loss:

f = (64/Re) Re = (ρ * v * D) / μ = (1000 kg/m^3 * 1.18 m/s * 0.03 m) / (1 x 10^-3 Pa s) = 35,400 f = 0.022 (using Moody chart for Re and ε/D = 1.8 x 10^-4)

hf1-2 = f * (L/D) * (v^2/2g) = 0.022 * (10 m / 0.03 m) * (1.18 m/s)^2 / (2 * 9.81 m/s^2) = 0.67 m

To determine the friction head loss between points (1) and (3), we follow the same procedure:

A3 = π/4 * (0.03 m)^2 = 7.07 x 10^-4 m^2 (cross-sectional area) Q = 5 L/min = 8.33 x 10^-5 m^3/s (volumetric flow rate) v3 = Q/A3 = 1.18 m/s (velocity)

f = (64/Re) Re = (ρ * v * D) / μ = (1000 kg/m^3 * 1.18 m/s * 0.03 m) / (1 x 10^-3 Pa s) = 35,400 f = 0.022 (using Moody chart for Re and ε/D = 1.8 x 10^-4)

hf1-3 = f * (L/D) * (v^2/2g) = 0.022 * (2 m / 0.03 m) * (1.18 m/s)^2 / (2 * 9.81 m/s^2) = 0.046 m

The friction head loss between points (1) and (2) is higher than between points (1) and (3) (0.67 m vs 0.046 m). This is because the pipe between points (1) and (2) is longer and has a smaller diameter, which increases the frictional resistance to flow.

(3) To calculate the friction loss factors (K) of the two fittings in the pipe system, we can use the Darcy-Weisbach equation:

ΔP = f (L/D) (v^2/2g)

where ΔP is the pressure drop, L is the length of the pipe, D is the diameter of the pipe, v is the velocity of the fluid, g is the acceleration due to gravity, and f is the friction factor.

For each fitting, we can rearrange the Darcy-Weisbach equation to solve for the friction factor:

f = (ΔP / (L/D) (v^2/2g))

Assuming fully developed, turbulent flow, we can use the Colebrook equation to determine the friction factor:

(1/√f) = -2.0 log10[(ε/D)/3.7 + (2.51/Re√f)]

where ε is the roughness of the pipe, and Re is the Reynolds number.

For fitting 1 (the pipe contraction), we have:

ΔP = 1 kPa L/D = 2/(30/1000) = 66.67 v = Q/A = (5/60)/(π*(50/1000)^2/4) = 1.273 m/s g = 9.81 m/s^2 ε = 1.8 x 10^-4 m Re = (ρvD/μ) = (10001.27350/1000)/(1*10^-3) = 636.5

Using the Colebrook equation, we can solve for f:

(1/√f) = -2.0 log10[(ε/D)/3.7 + (2.51/Re√f)] (1/√f) = -2.0 log10[(1.8 x 10^-4/50)/3.7 + (2.51/(636.5√f))] f = 0.0273

Therefore, the friction loss factor (K) for fitting 1 is:

K = f(D/2) = 0.0273(50/1000)/2 = 6.83 x 10^-4

For fitting 2 (the pipe expansion), we have:

ΔP = 1 kPa L/D = 2/(30/1000) = 66.67 v = Q/A = (5/60)/(π*(30/1000)^2/4) = 3.98 m/s g = 9.81 m/s^2 ε = 1.8 x 10^-4 m Re = (ρvD/μ) = (10003.9830/1000)/(1*10^-3) = 1194

Using the Colebrook equation, we can solve for f:

(1/√f) = -2.0 log10[(ε/D)/3.7 + (2.51/Re√f)] (1/√f) = -2.0 log10[(1.8 x 10^-4/30)/3.7 + (2.51/(1194√f))] f = 0.0121

Therefore, the friction loss factor (K) for fitting 2 is:

K = f(D/2) = 0.0121(30/1000)/2 = 1.82 x 10^-4

So the friction loss factor for fitting 1 is larger than that for fitting 2.