In the figure below, methane (molecular weight = 16 g/mol) is flowing through a pipeline 40 km long to receiving station A. At a point 5 km before A, a branch leads off from the main pipeline and runs 3 km to receiving station B. For all pipes, the internal diameter is 15 cm and the Fanning friction factor is fF = 0.0025. At the entry to the pipeline, the pressure is 1500 kPa and the mass flow rate is 1.25 kg/s. The pressure at station A is 500 kPa. Assume that the flow is isothermal (T = 25°C). Ignore energy losses at the entrance and exit of the pipeline and at the branch point, and assume kinetic energy is negligible. A. [5 marks] Determine the pressure at the point where the pipe branches. B. [6 marks] Determine the mass flow rate of gas entering receiving station B. C. [6 marks] Determine the velocity of the gas entering receiving station B.

To solve this problem, we will use the Bernoulli equation and the Darcy-Weisbach equation for fluid flow in pipes.

A. Determine the pressure at the point where the pipe branches.

First, we need to find the pressure drop from the start of the pipeline to the branch point. We can use the Darcy-Weisbach equation for this:

ΔP = fF * (L/D) * (ρv²/2)

where: fF = Fanning friction factor = 0.0025 L = length of the pipe = 35 km = 35000 m (from the start to the branch point) D = diameter of the pipe = 15 cm = 0.15 m ρ = density of the gas. We can find this from the ideal gas law, ρ = P/(RT), where P = pressure = 1500 kPa = 1500000 Pa, R = gas constant = 8.314 J/(mol.K), and T = temperature = 25°C = 298.15 K. For methane, the molar mass M = 16 g/mol = 0.016 kg/mol, so R = R'/M = 8.314/0.016 = 519.625 m²/s².K. Therefore, ρ = 1500000/(519.625298.15) = 9.68 kg/m³. v = velocity of the gas. We can find this from the mass flow rate and the cross-sectional area of the pipe, v = m'/(ρA), where m' = mass flow rate = 1.25 kg/s, and A = πD²/4 = 3.14160.15²/4 = 0.0177 m². Therefore, v = 1.25/(9.68*0.0177) = 7.5 m/s.

Substituting these values into the Darcy-Weisbach equation, we get:

ΔP = 0.0025 * (35000/0.15) * (9.68*7.5²/2) = 328125 Pa = 328.125 kPa

Therefore, the pressure at the branch point is P = P0 - ΔP = 1500 - 328.125 = 1171.875 kPa.

B. Determine the mass flow rate of gas entering receiving station B.

The mass flow rate at the branch point is the same as at the start of the pipeline, m' = 1.25 kg/s, because there are no losses. Therefore, the mass flow rate entering station B is also 1.25 kg/s.

C. Determine the velocity of the gas entering receiving station B.

The velocity of the gas at the branch point is the same as at the start of the pipeline, v = 7.5 m/s, because the diameter of the pipe is the same and the flow is incompressible. Therefore, the velocity of the gas entering station B is also 7.5 m/s.