The viscosity of water at 20 0 C (68 0 F) is 1.008 cp (centipoises).
Compute the absolute viscosity in lb-sec/ft2.
If the specific gravity at 20 0 C is 0.998, compute the kinematic viscosity in ft2/sec.
Use:
1 poise = 1 dyne-sec/cm2
1 lb = 444,800 dynes
1 ft = 30.48 cm
Please provide a brief step explanation since a solution has been already provided, but it isn’t clear how they arrived at the final solution. Thank you!
The Correct Answer and Explanation is:
Sure! Let’s break down the problem and clarify each step in detail to understand how to convert viscosity units and calculate kinematic viscosity.
Given data:
- Viscosity of water at 20°C = 1.008 cp (centipoises)
- Specific gravity (SG) at 20°C = 0.998
- 1 poise = 1 dyne·sec/cm²
- 1 lb = 444,800 dynes
- 1 ft = 30.48 cm
Step 1: Convert centipoises (cp) to absolute viscosity in lb·sec/ft²
What is centipoise?
- 1 poise = 1 dyne·sec/cm²
- 1 centipoise = 0.01 poise = 0.01 dyne·sec/cm²
So,
$$
1.008 \text{ cp} = 1.008 \times 0.01 = 0.01008 \text{ poise} = 0.01008 \text{ dyne·sec/cm}^2
$$
Convert dyne·sec/cm² to lb·sec/ft²
- 1 lb = 444,800 dynes
- 1 ft = 30.48 cm → so $1 \text{ ft}^2 = (30.48)^2 \text{ cm}^2 = 929.0304 \text{ cm}^2$
The goal: convert from dyne·sec/cm² to lb·sec/ft².
- 1 dyne·sec/cm² = ?
Since 1 lb = 444,800 dynes, and 1 ft² = 929.0304 cm²:
$$
1 \text{ dyne·sec/cm}^2 = \frac{1}{444,800} \text{ lb·sec/cm}^2
$$
But we want per ft²:
$$
\text{To get lb·sec/ft}^2 = \text{dyne·sec/cm}^2 \times \frac{1 \text{ lb}}{444,800 \text{ dynes}} \times \frac{929.0304 \text{ cm}^2}{1 \text{ ft}^2}
$$
Since dyne·sec/cm² is per cm², to convert to per ft² we multiply by cm²/ft²:
$$
\text{Absolute viscosity} = 0.01008 \times \frac{929.0304}{444,800} \quad \text{lb·sec/ft}^2
$$
Calculate the factor:
$$
\frac{929.0304}{444,800} = 0.002087
$$
So,
$$
\text{Absolute viscosity} = 0.01008 \times 0.002087 = 2.10 \times 10^{-5} \text{ lb·sec/ft}^2
$$
Step 2: Compute kinematic viscosity $\nu$ in ft²/sec
Definition:
$$
\nu = \frac{\mu}{\rho}
$$
Where
- $\nu$ = kinematic viscosity (ft²/sec)
- $\mu$ = absolute viscosity (lb·sec/ft²)
- $\rho$ = density (lb/ft³)
Calculate density $\rho$:
- Specific gravity (SG) = $\frac{\rho_{\text{water}}}{\rho_{\text{reference}}}$
- At 20°C, water density = 62.4 lb/ft³ (approximate)
- So,
$$
\rho = SG \times 62.4 = 0.998 \times 62.4 = 62.28 \text{ lb/ft}^3
$$
Calculate kinematic viscosity:
$$
\nu = \frac{2.10 \times 10^{-5} \text{ lb·sec/ft}^2}{62.28 \text{ lb/ft}^3} = 3.37 \times 10^{-7} \text{ ft}^2/\text{sec}
$$
Summary of final answers:
- Absolute viscosity $\mu = 2.10 \times 10^{-5}$ lb·sec/ft²
- Kinematic viscosity $\nu = 3.37 \times 10^{-7}$ ft²/sec
Explanation:
This problem is about unit conversion and understanding the physical relationship between absolute and kinematic viscosity.
- Viscosity in Centipoises: Viscosity is often given in poise or centipoise in the CGS system (centimeter-gram-second). One poise is 1 dyne·sec/cm². Since the given viscosity is in centipoise, you multiply by 0.01 to convert to poise.
- Absolute Viscosity Conversion: To convert the viscosity to imperial units (lb·sec/ft²), convert dynes to pounds and cm² to ft². The key conversion factor is that 1 lb equals 444,800 dynes and 1 ft equals 30.48 cm, so 1 ft² = (30.48)² cm². This lets us scale the units accordingly.
- Specific Gravity and Density: The specific gravity relates the fluid density to water’s density. Since the density of water at 20°C is about 62.4 lb/ft³, multiply by the specific gravity to find the actual density.
- Kinematic Viscosity: Kinematic viscosity is absolute viscosity divided by density, giving a measure of the fluid’s resistance to flow under gravity. The units ft²/sec correspond to how quickly the fluid spreads or flows.
This detailed breakdown clarifies how the unit conversions and calculations come together to provide the absolute and kinematic viscosity in imperial units.
If you want, I can walk through the math again with more detail or help with related problems!
