Respuesta :
Answer:
Fluid viscosity, [tex]\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]
Explanation:
Container depth, D = 12 cm = 0.12 m
Tube length, l = 30 cm = 0.3 m
Specific Gravity, [tex]\rho[/tex] = 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, [tex]Q = A_2 v_2[/tex]
[tex]Area, A_2 = \pi d^{2} /4\\A_2 = \pi/4 * 0.002^2\\A_2 = 3.14159 * 10^{-6} m^2[/tex]
[tex]v_{2} = Q/A_{2} \\v_{2} =\frac{1.9 * 10^{-6}}{3.14 * 10^{-6}} \\v_{2} = 0.605 m/s[/tex]
Applying the Bernoulli (energy flow) equation between Point 1 and point 2 to calculate the head loss:
[tex]\frac{p_{1} }{\rho g} + \frac{v_{1}^2 }{2 g} + z_1 = \frac{p_{2} }{\rho g} + \frac{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\\\frac{p_{atm} }{\rho g} + \frac{0^2 }{2 g} + 0.42= \frac{p_{atm} }{\rho g} + \frac{0.605^2 }{2 *9.8} +0 + h_f\\h_f = 0.401 m[/tex]
For laminar flow, the head loss is given by the formula:
[tex]h_f = \frac{128 Q \mu l}{\pi \rho g d^4} \\\\0.401 = \frac{128 * 1.9 * 10^{-6} * 0.3 \mu}{\pi *0.95* 9.8* 0.002^4}\\\\\\\mu = \frac{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}[/tex]
Answer:
0.00257 kg / m.s
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.
[tex]Q = A*V_2[/tex]
Where,
A: The cross sectional area of the tube
- The cross sectional area of the tube ( A ) is expressed as:
[tex]A = \pi \frac{D^2}{4} \\\\A = \pi \frac{0.002^2}{4} \\\\A = 3.14159 * 10^-^6 m^2[/tex]
- The velocity at the exit can be determined from the flow rate equation:
[tex]V_2 = \frac{Q}{A} \\\\V_2 = \frac{1.9*10^-^6}{3.14159*10^-^6} \\\\V_2 = 0.605 \frac{m}{s}[/tex]
- We will apply the energy balance ( head ) between the points of top-surface ( free surface ) and the exit of the vertical tube.
[tex]\frac{P_1}{p*g} + \frac{V^2_1}{2*g} + z_t_o_p = \frac{P_2}{p*g} + \frac{V^2_2}{2*g} + z_d_a_t_u_m + h_L[/tex]
- 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:
[tex]h_L = \frac{32*u*L*V}{S.G*p*g*D^2}[/tex]
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:
[tex]h + L = \frac{V_2^2}{2*g} + \frac{32*u*L*V_2}{S.G*p*g*D^2} \\\\0.42 = \frac{0.605^2}{2*9.81} + \frac{32*u*(0.3)*(0.605)}{0.95*998*9.81*0.002^2} \\\\u = 0.00257 \frac{kg}{m.s}[/tex]
- Lets check the validity of the Laminar Flow assumption to calculate the major losses:
[tex]Re = \frac{S.G*p*V_2*D}{u} \\\\Re = \frac{0.95*998*0.605*0.002}{0.00257} \\\\Re = 446 < 2100[/tex]( Laminar Flow )
