Question

To determine the viscosity of a liquid of specific gravity 0.95, you fill to a depth of 12 cm a large container that drains through a 30 cm long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of draining is found to be 1.9 cm3 /s. What is the fluid viscosity (assume laminar flow)

Answer 1

0.00257 kg / m.s

Step-by-step explanation:

Given:-

- The specific gravity of a liquid, S.G = 0.95

- The depth of fluid in free container, h = 12 cm

- The length of the vertical tube , L = 30 cm

- The diameter of the tube, D = 2 mm

- The flow-rate of the fluid out of the tube into atmosphere, Q = 1.9 cm^3 / s

Find:-

To determine the viscosity of a liquid

Solution:-

- We will consider the exit point of the fluid through the vertical tube of length ( L ), where the flow rate is measured to be Q = 1.9 cm^3 / s

- The exit velocity ( V2 ) is determined from the relation between flow rate ( Q ) and the velocity at that point.

`Q = A*V_2`

Where,

A: The cross sectional area of the tube

- The cross sectional area of the tube ( A ) is expressed as:

`A = \pi (D^2)/(4) \\\\A = \pi (0.002^2)/(4) \\\\A = 3.14159 * 10^-^6 m^2`

- The velocity at the exit can be determined from the flow rate equation:

`V_2 = (Q)/(A) \\\\V_2 = (1.9*10^-^6)/(3.14159*10^-^6) \\\\V_2 = 0.605 (m)/(s)`

- We will apply the energy balance ( head ) between the points of top-surface ( free surface ) and the exit of the vertical tube.

`(P_1)/(p*g) + (V^2_1)/(2*g) + z_t_o_p = (P_2)/(p*g) + (V^2_2)/(2*g) + z_d_a_t_u_m + h_L`

- The free surface conditions apply at atmospheric pressure and still ( V1 = 0 ). Similarly, the exit of the fluid is also to atmospheric pressure. Where, z_top is the total change in elevation from free surface to exit of vertical tube.

- The major head losses in a circular pipe are accounted using Poiessel Law:

`h_L = (32*u*L*V)/(S.G*p*g*D^2)`

Where,

μ: The dynamic viscosity of fluid

L: the length of tube

V: the average velocity of fluid in tube

ρ: The density of water

- The average velocity of the fluid in the tube remains the same as the exit velocity ( V2 ) because the cross sectional area ( A ) of the tube remains constant throughout the tube. Hence, the velocity also remains constant.

- The energy balance becomes:

`h + L = (V_2^2)/(2*g) + (32*u*L*V_2)/(S.G*p*g*D^2) \\\\0.42 = (0.605^2)/(2*9.81) + (32*u*(0.3)*(0.605))/(0.95*998*9.81*0.002^2) \\\\u = 0.00257 (kg)/(m.s)`

- Lets check the validity of the Laminar Flow assumption to calculate the major losses:

`Re = (S.G*p*V_2*D)/(u) \\\\Re = (0.95*998*0.605*0.002)/(0.00257) \\\\Re = 446 < 2100`( Laminar Flow )

Answer 2

Fluid viscosity,
`\mu = 2.57 * 10^(-3) kgm^(-1)s^(-1)`

Step-by-step explanation:

Container depth, D = 12 cm = 0.12 m

Tube length, l = 30 cm = 0.3 m

Specific Gravity,
`\rho` = 0.95

Tube diameter, d = 2 mm = 0.002 m

Rate of flow, Q = 1.9 cm³/s = 1.9 * 10⁻⁶ m³/s

Calculate the velocity at point 2 ( check the diagram attached)

Rate of flow at section 2,
`Q = A_2 v_2`

`Area, A_2 = \pi d^(2) /4\\A_2 = \pi/4 * 0.002^2\\A_2 = 3.14159 * 10^(-6) m^2`

`v_(2) = Q/A_(2) \\v_(2) =(1.9 * 10^(-6))/(3.14 * 10^(-6)) \\v_(2) = 0.605 m/s`

Applying the Bernoulli (energy flow) equation between Point 1 and point 2 to calculate the head loss:

`(p_(1) )/(\rho g) + (v_(1)^2 )/(2 g) + z_1 = (p_(2) )/(\rho g) + (v_(2)^2 )/(2 g) + z_2 + h_f\\ z_1 = L + l = 0.12 + 0.3\\z_1 = 0.42\\p_1 = p_(atm)\\v_1 = 0\\z_2 = 0\\(p_(atm) )/(\rho g) + (0^2 )/(2 g) + 0.42= (p_(atm) )/(\rho g) + (0.605^2 )/(2 *9.8) +0 + h_f\\h_f = 0.401 m`

For laminar flow, the head loss is given by the formula:

`h_f = (128 Q \mu l)/(\pi \rho g d^4) \\\\0.401 = (128 * 1.9 * 10^(-6) * 0.3 \mu)/(\pi *0.95* 9.8* 0.002^4)\\\\\\\mu = (0.401 * \pi *0.95* 9.8* 0.002^4)/(128 * 1.9 * 10^(-6) * 0.3) \\\\\mu = 2.57 * 10^(-3) kgm^(-1)s^(-1)`