Question

Engine oil at T = 60 °C is forced to flow between two very large, stationary, parallel flat plates separated by a thin gap height h = 3.60 mm. The plate dimensions are L = 1.25 m and W = 0.550 m. The outlet pressure is atmospheric, and the inlet pressure is 1 atm gage pressure.

Estimate the volume flow rate of oil. Also calculate the Reynolds number of the oil flow, based on gap height h and average velocity V.
Is the flow laminar or turbulent?

Answer 1

(a) The volume flow rate is 5.43 × 10⁻³ m³/s

(b) The Reynolds's number is 251.1

The flow is laminar as the Reynolds's number is below the critical value of 2040

Step-by-step explanation:

Here, we have

The engine oil temperature = 60 °C

Plate gap = 3.60 m

Plate dimensions =

Length, L = 1.25 m

Width, W = 0.550 m

Inlet pressure = 1 atm

Viscosity of engine oil at SAE 10W - 60

μ = 31.946 mPa.s = 0.031946 kg/m·s

(a)
`(dP)/(dx) = (0\, atm -1 \, atm)/(1.25 \, m) = (0\, bar-1.01325 \, bar)/(1.25 \, m) = 0.8106 \, bar = 81060 \, Pa`

`u =(1)/(2* \mu) *(dP)/(dx) * (y^2 - hy)\\dA = W * dy`

The rate of change of velocity is given by;

`(dv)/(dt) =\int {u} \, dA`

`(dv)/(dt) =\int\limits^h_0 {(1)/(2* \mu) *(dP)/(dx) * (y^2 - hy)} \, * W * dy`

`(dv)/(dt) =(-1)/(12* \mu) {*(dP)/(dx) * h^3} \, * W`

By substituting, we arrive at

`(dv)/(dt) =(-1)/(12* \mu) {*(dP)/(dx) * h^3} \, * W`

`(dv)/(dt) =(-1)/(12* 0.031946) {*-81060 * 0.0036^3} \, * 0.55 = 5.43 * 10^(-3)`

The volume flow rate is

`(dv)/(dt) =5.43 * 10^(-3) \, m^3/s`

(b) The velocity, V is given by

`V = (du)/(dt)`

`V =((dv)/(dt ) )/(W * h)`, we plug in the values and we have;

`V =(5.43*10^(-3) )/(0.55 * 0.0036) = 2.7 \, m/s`

Reynolds's number is given by;

`Re = (\rho * v * h)/(\mu)`

Where, ρ for the engine oil is 0.8130 g/cm³ = 813 kg/m³

`Re = (813 * 2.7 * 0.0036)/(0.031946 ) = 251.1`

As the Reynolds's number value is less than 2040, the flow is laminar flow.