Question

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 ?,

Answer 1

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=
`10^(-3)`

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
`[\eta]` =0.20 pa.s

The rate of flow of oil
`[(dv)/(dt) ]=6.4mL/min`

we know that 1 mL=
`10^(-3) L=10^(-6)m^(3)` [1L=
`10^(-3) m^(3) ]`

Hence 6.4mL/m^3=
`1.067*10^(-7)` m^3/s

we know that
`(dv)/(dt) =(\pi pr^4)/(8\eta l)`

Hence
`p=(dv)/(dt) *8\eta l*(1)/(\pi r^4)`

⇒
`p=1.067*10^(-7) *8*0.20*0.0940 *(1)/( 3.14*[0.00095]^4)` pa

⇒ P=6.274630973*
`10^(3)` pa

Answer 2

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.