If the viscosity of water at 70 °C is 0.00402 poise and its specific gravity is 0.978 determine its absolute viscosity in Pa-s and its kinematic viscosity in m²/s and in stokes.
The correct answer and explanation is :
To solve this problem, we need to convert the given viscosity values from one unit system to another and understand their relationships.
Given:
- Viscosity (dynamic) in Poise:
$\mu = 0.00402 \, \text{P}$ - Specific gravity (SG):
$SG = 0.978$
1. Convert Dynamic (Absolute) Viscosity to SI Units (Pa·s):
1 Poise = 0.1 Pa·s
$$
\mu = 0.00402 \, \text{P} \times 0.1 = 0.000402 \, \text{Pa·s}
$$
2. Find the Density of Water at 70 °C:
Specific gravity is the ratio of the fluid’s density to water at 4 °C (which is 1000 kg/m³):
$$
\rho = SG \times 1000 = 0.978 \times 1000 = 978 \, \text{kg/m}^3
$$
3. Calculate Kinematic Viscosity (ν):
Kinematic viscosity is the ratio of dynamic viscosity to density:
$$
\nu = \frac{\mu}{\rho} = \frac{0.000402}{978} \approx 4.11 \times 10^{-7} \, \text{m}^2/\text{s}
$$
4. Convert Kinematic Viscosity to Stokes:
1 Stokes = $1 \times 10^{-4} \, \text{m}^2/\text{s}$
$$
\nu = \frac{4.11 \times 10^{-7}}{1 \times 10^{-4}} = 0.00411 \, \text{St}
$$
Final Answers:
- Absolute viscosity: $0.000402 \, \text{Pa·s}$
- Kinematic viscosity:
- $4.11 \times 10^{-7} \, \text{m}^2/\text{s}$
- $0.00411 \, \text{St}$
Explanation (300 Words):
Viscosity is a key property in fluid mechanics that describes a fluid’s resistance to flow. It comes in two forms: dynamic (absolute) viscosity, which measures the internal resistance of a fluid to shear forces, and kinematic viscosity, which considers the effect of fluid density.
The unit Poise (P) is part of the CGS system, commonly used in older texts and industry, but SI units (Pa·s for dynamic and m²/s for kinematic) are more standard today. To convert from Poise to Pascal-seconds, we use the relation: $1 \, \text{P} = 0.1 \, \text{Pa·s}$. Thus, the given dynamic viscosity of 0.00402 P equals $0.000402 \, \text{Pa·s}$.
Kinematic viscosity ($\nu$) is calculated by dividing the dynamic viscosity ($\mu$) by the fluid’s density ($\rho$). The density of water at 70 °C is not exactly 1000 kg/m³ due to thermal expansion; it decreases with temperature. Given a specific gravity of 0.978, the density becomes 978 kg/m³.
By applying the formula $\nu = \mu / \rho$, we calculate the kinematic viscosity in SI units: $4.11 \times 10^{-7} \, \text{m}^2/\text{s}$. For completeness, we convert this into Stokes (a CGS unit), resulting in $0.00411 \, \text{St}$, using the conversion $1 \, \text{St} = 10^{-4} \, \text{m}^2/\text{s}$.
These conversions are crucial in engineering to ensure compatibility of data, accurate modeling of fluid systems, and effective communication across international and interdisciplinary boundaries.