Answer :
In this case, the viscosity of the sugar solution is approximately 1.33322 × 10⁻⁴Pa·s.
To find the viscosity of the fluid, we can use the Hagen-Poiseuille equation which describes the flow of incompressible, Newtonian fluids in a long cylindrical pipe. The equation for the volumetric flow rate Q through a tube of radius r, length L under a pressure difference ΔP is given by:
Q = (π * r⁴ * ΔP) / (8 * μ * L)
We are looking for the dynamic viscosity (μ), rearranging the equation for μ, we get:
μ = (π * r⁴ * ΔP) / (8 * Q * L)
Let's convert all units to SI units before substituting into the equation.
- The diameter of the tube d = 2 mm = 0.002 m
- The radius of the tube r = d/2 = 0.001 m
- The length of the tube L = 10 cm = 0.1 m
The volumetric flow rate Q = 60 cm³/min.
We need to convert this to m³/s:
- There are 1e6 cm³ in 1 m³.
- There are 60 seconds in one minute.
Therefore:
Q = (60 cm³/min) × (1 m³/1e⁶ cm³) × (1 min/60 s) = 60/1e⁶/60 m³/s = 1e⁻⁵ m³/s
The pressure drop per length of the tube is 1.0 mmHg/cm. First, we convert the pressure from mmHg to Pascals, knowing that 1 mmHg = 133.322 Pa.
Also, we must account for the length over which the pressure drop occurs (since the units are mmHg/cm), we need to multiply by the length of the capillary tube in cm:
ΔP = 1.0 mmHg/cm × 133.322 Pa/mmHg × 0.1 m (since 1 cm = 0.01 m)
ΔP = 133.322 Pa/cm × 0.1 m/cm
ΔP = 13.3322 Pa/m × 0.1 m/m
ΔP = 13.3322 Pa
Now we may insert all these values into the equation for viscosity μ:
μ = (π * (0.001 m)⁴ * 13.3322 Pa) / (8 * 1e⁻⁵ m³/s * 0.1 m)
μ = (π * 1e⁻¹² m⁴ * 13.3322 Pa) / (8e⁻⁶ m³/s * 0.1 m)
μ = (π * 1e⁻¹² m⁴ * 13.3322 Pa) / (8e⁻⁷m⁴/s)
μ = (π * 1e⁻¹² m⁴ * 13.3322 Pa) / (8e⁻⁷ m⁴/s) * (1 / π)
μ = (13.3322 Pa * 1e⁻¹² m⁴) / (8e-7 m⁴/s)
μ = (13.3322 Pa * 1e⁻¹²) / (8e⁻⁷)
μ = (13.3322 * 1e⁻¹²) / 8e⁻⁷
Pa·s μ = 13.3322 * 1e⁻⁵
Pa·s μ = 1.33322 * 1e⁻⁴
Pa·s μ = 1.33322e⁻⁴ Pa·s
So the viscosity of the sugar solution is approximately 1.33322 × 10⁻⁴Pa·s.
