Unit of Viscosity S.I. unit: N.s.m2or Pa.s Other common units: Poise (dyne.s.cm2) or cP (centipoise) Prove this equation: 1 Pa.s= 10 Poise = 103 cP
The Correct Answer and Explanation is :
To prove the equation ( 1 \, \text{Pa} \cdot \text{s} = 10 \, \text{Poise} = 10^3 \, \text{cP} ), we need to understand the relationship between these units of viscosity.
SI Unit of Viscosity: Pascal-second (Pa·s)
In the International System of Units (SI), the unit of viscosity is the Pascal-second (Pa·s), which is defined as the force of one Pascal (Pa) per square meter applied over a one-second period. The Pascal is a unit of pressure, and since viscosity involves force per unit area (dynamic viscosity), the SI unit is represented as:
[
\text{Pa} = \frac{\text{N}}{\text{m}^2}
]
where N is Newton, which is the unit of force. Thus, we can write:
[
\text{Pa} \cdot \text{s} = \frac{\text{N} \cdot \text{s}}{\text{m}^2}
]
Now, let’s examine the relationship between Pa·s and the cgs (centimeter-gram-second) system units, which use Poise and centipoise.
Poise and Centipoise
The Poise (P) is a cgs unit of dynamic viscosity, and it is defined as the force of one dyne per square centimeter per velocity gradient of one centimeter per second. The unit of Poise can be expressed as:
[
1 \, \text{Poise} = \frac{1 \, \text{dyne} \cdot \text{s}}{\text{cm}^2}
]
And, the dyne is defined as ( 1 \, \text{dyne} = 10^{-5} \, \text{N} ), and ( 1 \, \text{cm} = 10^{-2} \, \text{m} ). Therefore, we can convert Poise to SI units:
[
1 \, \text{Poise} = \frac{1 \cdot 10^{-5} \, \text{N} \cdot \text{s}}{(10^{-2} \, \text{m})^2} = 10^{-3} \, \text{Pa} \cdot \text{s}
]
Thus, we have:
[
1 \, \text{Pa} \cdot \text{s} = 10 \, \text{Poise}
]
Conversion to Centipoise (cP)
The centipoise (cP) is simply ( 1 \, \text{Poise} = 100 \, \text{cP} ). Therefore, we have:
[
1 \, \text{Pa} \cdot \text{s} = 10^3 \, \text{cP}
]
Conclusion
Thus, we have shown the relationship:
[
1 \, \text{Pa} \cdot \text{s} = 10 \, \text{Poise} = 10^3 \, \text{cP}
]
This completes the proof.