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Dynamic viscosity Formula is used as the tangential force required to move one horizontal plane of a fluid with respect to another. Viscosity is helpful to describe how thick a material is and its range of fluidity.
- Dynamic viscosity formula is an important property of fluid materials.
- It is useful in understanding the behaviour of the fluid.
- The viscosity decreases with an increase in temperature.
- It will show movement when it comes in contact with solid boundaries.
- Dynamic Viscosity Formula deforms in case of mechanical oscillation.
- It depends upon the temperature, frequency and time.
- They exhibit the property of bulky materials in the case of absolute physical sense.
- The device used to measure dynamic viscosity is a capillary viscometer.
- The dynamic viscosity of water is calculated to be 11.4 × 10-3 poise.
Read More: Terminal Velocity
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Key Terms: Dynamic Viscosity Formula, Dynamic Viscosity, Viscosity, Tangential Force, Fluid, Intermolecular Friction, Plane, Shear Stress, Kinetic Viscosity
What is Dynamic Viscosity?
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The dynamic viscosity of a fluid is the measure of the resistance to its progressive deformation by fraction or shear. Shear stress in the fluid is possible due to the intermolecular friction exerted when layers of fluids attempt to flow over each other.
- The dynamic viscosity formula for the fluid will define its internal resistance to flow due to shear force.
- It is a type of tangential force that acts when one horizontal plane moves with another.
- Viscosity acts as an important property of fluid.
- Dynamic viscosity is the force required to overcome its internal molecular friction.
- It analyses fluid behaviour and fluid movement near solid limits.
The dynamic viscosity formula can be expressed as the tangential force per unit area required to move the fluid in a horizontal plane relative to another plane. In this, the particles move with a unit value velocity while the fluid molecules maintain a distance unitary of each other.
Dynamic Viscosity
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Viscosity Detailed Video Explanation:
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Formula for Dynamic Viscosity
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The formula for the dynamic viscosity is given by,
Dynamic viscosity = shear stress / change in shear rate
In the form of the equation we can write it as,
η = T/γ
- Where,
- η → Dynamic viscosity
- τ → Shearing stress
- γ → Shear rate
Dynamic Viscosity Formula
Read More: Mechanical Properties of Fluids
Unit of Dynamic Viscosity
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A rotary viscometer is a popular instrument, used to measure dynamic viscosity. These instruments will spin a probe in the liquid sample. Viscosity is determined by measuring force, which is the torque required to rotate the probe.
- The SI actual unit of dynamic viscosity (μ) is Pascal-second (Pa s), which also can be represented as 1 kg m-1 s-1.
The dynamic viscosity of various materials are tabulated below.
| Liquid | Absolute Viscosity (N s/m2, Pa s) |
|---|---|
| Air | 1.983 x 10-5 |
| Water | 10-3 |
| Olive Oil | 10-1 |
| Glycerol | 100 |
| Liquid honey | 101 |
| Golden syrup | 102 |
| Glass | 1040 |
Read More: Poisson's Ratio
Factors affecting Dynamic Viscosity
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The factors affecting dynamic viscosity are as follows:
- The dynamic viscosity of liquid decreases with increase in temperature.
- Similary dynamic viscosity of gases increase with increase in temperature.
Read More: Barometer
Differences Between Dynamic Viscosity and Kinetic Viscosity
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Here are the basic differences between Dynamic Viscosity and Kinetic Viscosity-
| Dynamic Viscosity | Kinetic Viscosity |
|---|---|
| Dynamic Viscosity can also be known as Absolute Viscosity. | Kinetic Viscosity is also know Absolute Viscosity. |
| The symbol of Dynamic Viscosity is µ, and the unit of it is Ns/m2 | The symbol of Kinetic Viscosity is v, and the unit of it is m2/s |
| It is represented as the ratio between the sheer strees and sheer strain | Kinetic Viscosity can be represented as the ratio between dynamic viscosity and density. |
| Dynamic Viscosity is the representation of the viscous force of the fluid. | Kinetic Viscosity is the representation of the inertia along the viscous force. |
Read More: Bulk Modulus
Things to Remember
- Dynamic viscosity formula represents ratio between sheer stress and sheer strain.
- It is defined as the tangential force required to move one horizontal plane of a fluid with respect to another.
- The dynamic viscosity formula is given by η = T/γ
- Rotary viscometer is a popular instrument that is used to measure dynamic viscosity.
- The SI actual unit of dynamic viscosity (μ) is Pascal-second (Pa s).
- It can be represented as 1 kg m-1 s-1.
Read More:
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Sample Questions
Ques. We have a fluid with a shear rate of 0.5 s-1 and shearing stress of 0.76 N/m2. According to its dynamic viscosity, to which one of these fluids corresponds? (3 marks)
Water: 1 Pa*s
Air: 0.018 Pa*s
Mercury: 1.526 Pa*s
Ans. First, calculate the dynamic viscosity using the formula :
η = Tγ
where,
τ = 0.76 N/m2 and
γ = 0.5 s-1
Now, η = τγ
η = (0.76 N/m2) / (0.5 (1/s))
η = 1.52 (N*s) / m2
η= 1.52 Pa*s
The fluid is mercury.
Ques. What is the pressure necessary to move a plane of fluid with a shear rate of 0.35 s-1 and a dynamic viscosity of 0.018 Pa*s? (3 marks)
Ans. From the formula of dynamic viscosity, we can find the shear stress.
η = Tγ
and then substituting the values,
(0.018 Pa*s) = T* (0.35 s-1)
τ = (0.018 Pa*s)*(0.35 s-1)
τ= 0.0063 Pa
Now,
τ = 0.0063
Thus, Pa = 0.0063 N/m2
Therefore, The Pressure required is Pa = 0.0063 N/m2
Ques. A fluid with a shear rate of 0.5 per second, and the shearing stress 0.76 N per m2. According to its dynamic viscosity, to which one of these fluids corresponds? (3 marks)
Water with dynamic viscosity 1 Pas
Air with dynamic viscosity 0.018 Pas
Mercury with dynamic viscosity 1.526 Pas
Ans. First calculate the dynamic viscosity using the following formula, using the given values,
T= 0.76 N per m2 and
γ = 0.5 per second
η = Tγ
η = ( 0.76 N per m2) * ( 0.5 per second)
η = 0.760.5
η = 1.52 Pas
Therefore it is clear that Mercury fluid will correspond with this fluid.
Ques. The flow rate of water from a tap of diameter 1.25 cm is 0.48 L/min. The coefficient of viscosity of water is 10-3 Pa s. After some time the flow rate is increased to 3 L/min. Characterise the flow for both the flow rates. (5 marks)
Ans. Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d2/4
v = 4 Q / d2 π
We then estimate the Reynolds number to be
Re = 4 ρ Q / π d η
Re = 4 ×103 kg m-3 ×Q/(3.14 ×1.25 ×10-2 m ×10-3 Pa s)
= 1.019 × 108 m-3 s Q
Since initially
Q = 0.48 L / min = 8 cm3/ s = 8 × 10-6 m3 s-1,
we obtain,
Re = 815
Since this is below 1000, the flow is steady.
After some time when
Q = 3 L / min = 50 cm3 / s
Q = 5 × 10-5 m3/s-1
we obtain,
Re = 5095
The flow will be turbulent.
Ques. Calculate the dynamic viscosity of oil, which is used for lubrication between a square plate of size 0.8m × 0.8m & an inclined plane with an angle of inclination of 30°.
The weight of the square plate is 300N & it slides at a uniform velocity of 0.3m/s.
The thickness of the oil firm is 1.5 mm. (3 marks)
Ans. M = ?
A = 0.8 × 0.8 = 0.64 m2
W = 300N
U = 0.3 m/s
t = 1.5 mm or 1.5 × 10-3 m
Therefore, the force acting along with the plane = Component of weight parallel to the plane.
Therefore, F = 300 sin 30°
Therefore , F = 150N
Therefore, Shear stress is given by:
τ = F/A
= 150/0.64
Therefore, Shear stress is 150/0.64 N/m2
Also, we know that,
τ = µ du/dy
150/0.64 = µ 0.3/1.5×10-3
µ = 150× 1.5 × 10-3/ 0.64× 0.3
= 1.17 NS/m2
Therefore, the dynamic viscosity is 1.17 NS/m2
Ques. The dynamic viscosity of the oil used for lubrication between a shaft and sleeve is 6 poise. The shaft is of diameter 0.4 m and rotates at 190 Rpm. Calculate the power lost in the bearing for a sleeve length of 90 mm. The thickness of the oil firm is 1.5 mm. (5 marks)
Ans. Given,
M = 6 poise
= 6/10 NS/m2
= 0.6 NS/m2
D = 0.4 m
N = 190 Rpm
L = 90 mm = 90 × 10-3 m
t = 1.5 mm = 1.5 × 10-3 m
Velocity = πDN/60
= π×0.4×190/60
= 3.98 m/s
τ = µ du/dy
= 0.6 × 3.98/1.5×10-3
= 1592 N/m2
Now,
Force = τ × C. A
= 1592 × π × 0.4 × 90 × 10-3
= 1592 × 3.14159 × 0.4 × 90 × 10-3
= 180.05
Now,
Power = force × velocity
= 180.05 × 3.98
= 716.48 watt
Therefore, the power is 716.48 watt.
Ques. What is Archimedes Principle? (2 marks)
Ans. When a body is partially or completely immersed in a liquid, it loses some of its weight. The loss in weight of the body in the liquid is equal to the weight of the liquid displaced by the immersed part of the body. The upward force exerted by the liquid displaced when a body is immersed is called buoyancy. Due to this, there is an apparent loss in the weight experienced by the body.
Ques. Calculate the density of the fluid having an absolute viscosity of 0.89 N s per square m and kinematic viscosity of 2 m2s-1 (2 marks)
Ans.Known parameters are,
Absolute viscosity μ=0.89 N s per square
Kinematic viscosity,v= 2m2s-1
The density is: v=μ rho
i.e. ρ = vμ = 0.892
= 0.445 kg per cubic m
Therefore, the density of the fluid is 0.445 kg per cubic m.
Ques. Define the Energy of a liquid. (2 marks)
Ans. A liquid can possess three types of energies: (i) kinetic energy, (ii) potential energy and (iii) pressure energy
The energy possessed by a liquid due to its motion is called kinetic energy i.e., 1/2mv2.
The potential energy of a liquid of mass m at a height h is given by P.E. = mgh
The energy possessed by a liquid by virtue of its pressure is called pressure energy. Pressure energy of liquid in volume dV = PdV
Pressure energy per unit mass of the liquid
\(\frac{PdV}{\rho dV} = \frac{P}{\rho}(m = \rho d V)\)
Ques. The flow rate of water from a tap of diameter 1.25 cm is 0.48 L/min. The coefficient of viscosity of water is 10-3 Pa s. After sometime the flow rate is increased to 3 L/min. Characterise the flow for both the flow rates. (5 marks)
Ans. Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d2 / 4
v = 4 Q / d2 π
We then estimate the Reynolds number to be
Re = 4 ρ Q / π d η
= 4 ×103 kg m-3 ×Q/(3.14 ×1.25×10-2 m ×10-3 Pa s = 1.019 × 108 m-3 s Q
Since initially,
Q = 0.48 L / min = 8 cm3 / s = 8 × 10-6m3/s-1,
we obtain,
Re = 815
Since this is below 1000, the flow is steady.
After sometime when
Q = 3 L / min = 50 cm3 / s = 5 × 10-5 m3s-1,
we obtain,
Re = 5095
The flow will be turbulent.
Ques. What is the pressure necessary to move a plane of fluid with a shear rate of 0.35 s-1 and a dynamic viscosity of 0.017 Pa*s? (3 marks)
Ans. From the formula of dynamic viscosity, we can find the shear stress.
η = Tγ
and then substituting the values,
(0.017 Pa*s) = T* (0.35 s-1)
τ = (0.017 Pa*s)*(0.35 s-1)
τ= 0.00595 Pa
Now,
τ = 0.00595
Thus, Pa = 0.00595 N/m2
Therefore, The Pressure required is Pa = 0.00595 N/m2
Ques. What is the dynamic viscosity of various materials? (5 marks)
Ans. The dynamic viscosity of various materials are tabulated below.
| Liquid | Absolute Viscosity (N s/m2, Pa s) |
|---|---|
| Air | 1.983 x 10-5 |
| Water | 10-3 |
| Olive Oil | 10-1 |
| Glycerol | 100 |
| Liquid honey | 101 |
| Golden syrup | 102 |
| Glass | 1040 |
Ques. What is the difference between dynamic viscosity and kinetic viscosity? (5 marks)
Ans. The difference between dynamic viscosity and kinetic viscosity are as follows:
| Dynamic Viscosity | Kinetic Viscosity |
|---|---|
| Dynamic Viscosity can also be known as Absolute Viscosity. | Kinetic Viscosity is also know Absolute Viscosity. |
| The symbol of Dynamic Viscosity is µ, and the unit of it is Ns/m2 | The symbol of Kinetic Viscosity is v, and the unit of it is m2/s |
| It is represented as the ratio between the sheer strees and sheer strain | Kinetic Viscosity can be represented as the ratio between dynamic viscosity and density. |
| Dynamic Viscosity is the representation of the viscous force of the fluid. | Kinetic Viscosity is the representation of the inertia along the viscous force. |
Ques. The flow rate of water from a tap of diameter 2.25 cm is 0.48 L/min. The coefficient of viscosity of water is 10-3 Pa s. After sometime the flow rate is increased to 2 L/min. Characterise the flow for both the flow rates. (5 marks)
Ans. Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d2 / 4
v = 4 Q / d2 π
We then estimate the Reynolds number to be
Re = 4 ρ Q / π d η
= 4 ×103 kg m-3 ×Q/(3.14 ×2.25×10-2 m ×10-3 Pa s = 0.566 × 108 m-3 s Q
Since initially,
Q = 0.48 L / min = 8 cm3 / s = 8 × 10-6m3/s-1,
we obtain,
Re = 815
Since this is below 1000, the flow is steady.
After sometime when
Q = 2 L / min = 3.33 × 10-5 m3s-1,
we obtain,
Re = 5095
The flow will be turbulent.
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