Engineering: 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?

that sucks im sorryExplanation:...

Engine oil at T = 60 °C is forced to flow between

michael1295

23.01.2022

(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

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) $\frac{dP}{dx} = \frac{0\, atm -1 \, atm}{1.25 \, m} = \frac{0\, bar-1.01325 \, bar}{1.25 \, m} = 0.8106 \, bar = 81060 \, Pa$

$u =\frac{1}{2\times \mu} \times\frac{dP}{dx} \times (y^2 - hy)\\dA = W \times dy$

The rate of change of velocity is given by;

$\frac{dv}{dt} =\int {u} \, dA$

$\frac{dv}{dt} =\int\limits^h_0 {\frac{1}{2\times \mu} \times\frac{dP}{dx} \times (y^2 - hy)} \, \times W \times dy$

$\frac{dv}{dt} =\frac{-1}{12\times \mu} {\times\frac{dP}{dx} \times h^3} \, \times W$

By substituting, we arrive at

$\frac{dv}{dt} =\frac{-1}{12\times \mu} {\times\frac{dP}{dx} \times h^3} \, \times W$

$\frac{dv}{dt} =\frac{-1}{12\times 0.031946} {\times-81060 \times 0.0036^3} \, \times 0.55 = 5.43 \times 10^{-3}$

The volume flow rate is

$\frac{dv}{dt} =5.43 \times 10^{-3} \, m^3/s$

(b) The velocity, V is given by

$V = \frac{du}{dt}$

$V =\frac{\frac{dv}{dt } }{W \times h}$ , we plug in the values and we have;

$V =\frac{5.43\times10^{-3} }{0.55 \times 0.0036} = 2.7 \, m/s$

Reynolds's number is given by;

$Re = \frac{\rho \times v \times h}{\mu}$

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

$Re = \frac{813 \times 2.7 \times 0.0036}{0.031946 } = 251.1$

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

0,0(0 оценок)