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Engine oil (η = 0.20 Pa⋅s) passes through a fine 1.90-mm-diameter tube that is 9.40cm long.
What pressure difference is needed to maintain a flow rate of 6.4mL/min ?,

Respuesta :

meerkat18
In this problem which involves pressure difference  and flow  rate, you can use the formula:

Q = πPr^4/(8lη) 

Where:
Q = Flow in Liters/second 
η= Viscosity in Pa.s 
P = Pressure in pascals 
r = Radius of the tube in meters 
l = Length of the tube in meters 
The given needed to be converted to the right units for the formula.

Q = 6.4 mL/minute or (0.0064 /60) Liters /second 
η= 0.2 Pa.s 
r = 0.95 mm or 0.00095 m
l = 9.40 cm or 0.094 m

(0.0064/60) = π P (0.00095) ^4 /(8 x 0.094 x 0.2) 

The pressure difference can be found by solving for P in the equation.
Nyctalus

Here we have to calculate the pressure difference between two ends of the tube.

The diametre of tube [d] =1.90 mm

we know that  1 mm=[tex]10^{-3}[/tex]

The radius of tube [r]=0.95 mm i.e 0.00095 m

The length of the tube [l]=9.40 cm i.e 0.0940 m

The coefficient of visocity of the oil [tex][\eta][/tex] =0.20 pa.s

The rate of flow of oil [tex][\frac{dv}{dt} ]=6.4mL/min[/tex]

we know that 1 mL=[tex]10^{-3} L=10^{-6}m^{3}[/tex]  [1L=[tex]10^{-3} m^{3} ][/tex]

     Hence 6.4mL/m^3= [tex]1.067*10^{-7}[/tex] m^3/s

we know that [tex]\frac{dv}{dt} =\frac{\pi pr^4}{8\eta l}[/tex]

Hence [tex]p=\frac{dv}{dt} *8\eta l*\frac{1}{\pi r^4}[/tex]

        ⇒ [tex]p=1.067*10^{-7} *8*0.20*0.0940 *\frac{1}{ 3.14*[0.00095]^4}[/tex] pa

        ⇒ P=6.274630973*[tex]10^{3}[/tex] pa