applied hydraulics

Question

In the flow of a fluid there is a galvanized steel duct (e = 0.180 mm; D = 100 mm); Q = 20 L/s; ν = 3604 x 10-4 m2/s; γ = 1,324 kgf/m3. The question is: what is the head loss in fluid m.c. and the equivalent value in m.c.a.

Answer 1

For a section of pipe, the loss in pressure is given by ΔP = (1/2)*f*(L/D)*ρ*(V^2) where f is the loss coefficient, L is the pipe length, D is the pipe diameter, ρ is the fluid density, and v is the fluid velocity.

Here, the weight density 'γ' is given in units of kgf/m^3, which means that its mass density 'ρ' is the same number (ρ=1324 kg/m^3)

The velocity is related to the flow rate by Q=V*A, where A is the cross sectional area. In the case of a circular duct, A=(π/4)*D^2, where D is the diameter.

This gives Q=V*(π/4)*D^2 => V=4Q/(πD^2). Converting the flow rate to 0.02 m^3/s and the diameter to 0.1 m gives V=2.55 m/s

To calculate the friction factor, we first need the Reynolds Number of the flow: Re = ρVD/μ = VD/ν, where 'ν' is the kinematic viscosity. In this case, assuming ν=3.604*(10^-4) m^2/s, this means that Re=708. This is in the region of laminar flow, meaning that the friction factor is 64/Re, so f=64/708 = 0.09.

You now have every component you need to calculate the pressure loss, except for the length of the pipe. By simply not including the length of the pipe, you can calculate the head loss per meter of pipe.

To convert pressure loss to head (m.c.), divide the answer by the acceleration due to gravity (9.81 m/s^2), and the fluid density (1324 kg/m^3). To convert pressure loss to head loss of water (m.c.a.), divide the answer by the acceleration due to gravity and the density of water (1000 kg/m^3).