Question

Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the channel is 26 kPa/m, and the distance between the plates is 4 mm. The flow is laminar. First, determine the volume rate of flow (per meter of width).

Answer 1

Volume Rate = 3.65 x 10^(-4) m²/s

Step-by-step explanation:

This is a laminar flow and the formula for the volume rate of flow is given as;

q = ((2h³)/3μ) (ΔP/L)

Where;

h is hydraulic depth

μ is viscosity of oil (SAE 30) at 15.6°

(ΔP/L) is the pressure drop per unit length.

Now, distance between plates is 4mm and h = d/2 = 4/2 = 2mm or 0.002 m

μ is traced out from the graph i attached below and and it's approximately 0.38 Pa.s

(ΔP/L) = 26 KPa/m = 26000 Pa/m

So q = ((2 x 0.002³)/(3 x 0.38))(26,000) = 3.65 x 10^(-4) m²/s

Answer 2

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

Step-by-step explanation:

For laminar flow between two parallel horizontal plates, the volumetric flow per metre of width is given as

q = (2h³/3μ) (ΔP/L)

h = hydraulic depth = 4mm/2 = 2mm = 0.002 m

μ = viscosity of oil (SAE 30) at 15.6°C = 0.38 Pa.s

(ΔP/L) = 26 KPa/m = 26000 Pa/m

q = (2h³/3μ) (ΔP/L)

q = (26000) × (2(0.002³)/(3×0.38))

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s