An ideal mixing (total volume inlet = total volume outlet) of two incompressible fluids
which is between water and methanol (SG =0.85) is coming from two inlets. At inlet 1,
water enter at 9 m/s and at inlet 2, methanol enters at 7 m/s. The diameter of inlet 1 and
2 is 4 cm and 3 cm. If the outlet diameter is same as inlet 2, calculate the mixing velocity
and density of the mixture at the outlet.
To calculate the mixing velocity and density of the mixture at the outlet, we need to first apply the principle of conservation of mass. According to this principle, the total mass entering the system must be equal to the total mass leaving the system.
Since the fluids are incompressible and the total volume inlet is equal to the total volume outlet, we can assume that the mass flow rate entering the system is equal to the mass flow rate leaving the system.
The mass flow rate (ṁ) can be calculated using the equation:
ṁ = ρ * A * v
Where:
ṁ = mass flow rate (kg/s)
ρ = density (kg/m^3)
A = cross-sectional area of flow (m^2)
v = velocity (m/s)
Let's calculate the mass flow rates at inlets 1 and 2 first:
For inlet 1 (water):
ρ1 = density of water = 1000 kg/m^3 (approximately)
A1 = π * (diameter1/2)^2 = π * (0.04/2)^2 = 0.0012566 m^2 (approximately)
v1 = velocity of water at inlet 1 = 9 m/s
ṁ1 = ρ1 * A1 * v1
= 1000 kg/m^3 * 0.0012566 m^2 * 9 m/s
≈ 11.3094 kg/s
For inlet 2 (methanol):
ρ2 = density of methanol = 0.85 * 1000 kg/m^3 = 850 kg/m^3
A2 = π * (diameter2/2)^2 = π * (0.03/2)^2 = 0.0007069 m^2 (approximately)
v2 = velocity of methanol at inlet 2 = 7 m/s
ṁ2 = ρ2 * A2 * v2
= 850 kg/m^3 * 0.0007069 m^2 * 7 m/s
≈ 4.8 kg/s
Now, with the assumption that the total mass flow rate entering the system is equal to the total mass flow rate leaving the system, we can calculate the mixing velocity (v_mixture) and density (ρ_mixture) at the outlet.
Total mass flow rate = ṁ1 + ṁ2 = 11.3094 kg/s + 4.8 kg/s = 16.1094 kg/s
Using the conservation of mass principle:
ṁ_mixture = ρ_mixture * A_outlet * v_mixture
Since the outlet diameter is the same as inlet 2:
A_outlet = π * (diameter2/2)^2 = π * (0.03/2)^2 = 0.0007069 m^2 (approximately)
Now we can solve for the mixing velocity:
16.1094 kg/s = ρ_mixture * 0.0007069 m^2 * v_mixture
v_mixture = 16.1094 kg/s / (ρ_mixture * 0.0007069 m^2)
We need to find the value of ρ_mixture in order to calculate v_mixture.
To find the density of the mixture, we can use the principle of mixture density. The mixture density is given by the equation:
ρ_mixture = (ṁ1 * ρ1 + ṁ2 * ρ2) / (ṁ1 + ṁ2)
ρ_mixture = (11.3094 kg/s * 1000 kg/m^3 + 4.8 kg/s * 850 kg/m^3) / (11.3094 kg/s + 4.8 kg/s)
Now, by substituting the values we have:
ρ_mixture = (11309.4 kg*m/s + 4080 kg*m/s) / 16.1094 kg/s
ρ_mixture ≈ 11389.4 kg/m^3
Now, we can substitute this value back into the formula for the mixing velocity:
v_mixture = 16.1094 kg/s / (11389.4 kg/m^3 * 0.0007069 m^2)
v_mixture ≈ 2.147 m/s
Therefore, the mixing velocity of the mixture at the outlet is approximately 2.147 m/s, and the density of the mixture at the outlet is approximately 11389.4 kg/m^3.