Respuesta :
Answer:
(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) [tex]\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[/tex]
[tex]u =\frac{1}{2\times \mu} \times\frac{dP}{dx} \times (y^2 - hy)\\dA = W \times dy[/tex]
The rate of change of velocity is given by;
[tex]\frac{dv}{dt} =\int {u} \, dA[/tex]
[tex]\frac{dv}{dt} =\int\limits^h_0 {\frac{1}{2\times \mu} \times\frac{dP}{dx} \times (y^2 - hy)} \, \times W \times dy[/tex]
[tex]\frac{dv}{dt} =\frac{-1}{12\times \mu} {\times\frac{dP}{dx} \times h^3} \, \times W[/tex]
By substituting, we arrive at
[tex]\frac{dv}{dt} =\frac{-1}{12\times \mu} {\times\frac{dP}{dx} \times h^3} \, \times W[/tex]
[tex]\frac{dv}{dt} =\frac{-1}{12\times 0.031946} {\times-81060 \times 0.0036^3} \, \times 0.55 = 5.43 \times 10^{-3}[/tex]
The volume flow rate is
[tex]\frac{dv}{dt} =5.43 \times 10^{-3} \, m^3/s[/tex]
(b) The velocity, V is given by
[tex]V = \frac{du}{dt}[/tex]
[tex]V =\frac{\frac{dv}{dt } }{W \times h}[/tex], we plug in the values and we have;
[tex]V =\frac{5.43\times10^{-3} }{0.55 \times 0.0036} = 2.7 \, m/s[/tex]
Reynolds's number is given by;
[tex]Re = \frac{\rho \times v \times h}{\mu}[/tex]
Where, ρ for the engine oil is 0.8130 g/cm³ = 813 kg/m³
[tex]Re = \frac{813 \times 2.7 \times 0.0036}{0.031946 } = 251.1[/tex]
As the Reynolds's number value is less than 2040, the flow is laminar flow.
