Fluid dynamics?

Question

An oil with a specific gravity of 0.78 flows at
a velocity of 0.8 m/s in a 2” schedule 40 pipe. Find both
the volume flow rate and the mass flow rat

Answer 1

Use the table in this link: https://www.engineeringtoolbox.com/ansi-...
(about ¼ of the way down, click the blue ‘mm2’ label at the top of the ‘Transverse Areas’ column)

This gives the internal cross-sectional (transverse) area of 2” schedule 40 pipe as 2168mm², which is 2168x10⁻⁶ m² (as 1mm² = 10 ⁶m²).

The volume flow rate (Rv) is Rv = vA = 0.8 x 2168x10⁻⁶ = 0.0017344 m³/s
The should be rounded to 2 sig. figs. to match the precision of the supplied data:
Volume flow rate, Rv = 0.0017m³/s to 2 sig. figs.

The specific gravity of the oil is 0.78 and the density of water is 1000kg/m³. So the density of the oil is ρ = 0.78 x 1000 = 780kg/m³.

The mass flow rate (Rm) is Rm = ρRv = 780 x 0.0017344 = 1.353kg/s
To 2 sig. figs this is 1.4kg/s.

Answer 2

Look up in pipe table (online) the O.D. in mm and the schedule 40 pipe wall thickness, then:
Subtract twice the wall thickness from the O.D. to get the I.D. of schd 40 pipe. This should be = ID = 52.506 mm
Calculate area of pipe in base SI units = π(ID)²/4 = 2165.25E-6 m²
Volume rate of flow = 0.8(2165.25E-6) = 1732.20E-6 ≈ 1730E-6 m³/s ANS
Mass rate of flow = (0.78)(1000)(1732.20E-6) = 135.11 ≈ 135 kg/s ANS

Answer 3

A ≈ 2170 mm² corresponding to ≈ 0.217 dm^2
volum. flow Qv = A*V = 0.217 dm^2*8 dm/sec = 1.74 dm^3/sec
mass flow Qm = 1.74 dm^3/sec*0.80 kg/dm^3 = 1.39 kg/sec
weight flow Qw =Qm*g = 1.39*9.806 = 13.6 N/sec