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 enter 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.

sad

To calculate the mixing velocity and density of the mixture at the outlet, we need to first understand how the velocities and densities of the fluids combine.

1. Velocity Combination:
Since the total volume flow rate at the inlet is equal to the total volume flow rate at the outlet, we can express the mixing velocity using the conservation of mass equation:

(Volume flow rate of water at inlet 1) + (Volume flow rate of methanol at inlet 2) = (Volume flow rate of mixture at outlet)

We can use the formula for the volume flow rate (Q) of a fluid through a pipe, which is given by:

Q = (Area of the pipe) * (Velocity of the fluid)

Given that the diameter of inlet 1 is 4 cm and the velocity of water at inlet 1 is 9 m/s, we can calculate the volume flow rate of water at inlet 1 as:

Q1 = (pi * (diameter1/2)^2) * velocity1

Now, let's calculate the volume flow rate of water at inlet 1:

diameter1 = 4 cm = 0.04 m
velocity1 = 9 m/s

Q1 = (pi * (0.04/2)^2) * 9
= (pi * 0.02^2) * 9
= 0.0018 m^3/s

Similarly, we can calculate the volume flow rate of methanol at inlet 2:

diameter2 = 3 cm = 0.03 m
velocity2 = 7 m/s

Q2 = (pi * (0.03/2)^2) * 7
= (pi * 0.015^2) * 7
= 0.00111 m^3/s

Since the total volume flow rate at the inlet is equal to the total volume flow rate at the outlet, we have:

Q1 + Q2 = Qout

0.0018 + 0.00111 = Qout
0.00291 = Qout

Now, we can find the mixing velocity of the outlet:

Qout = (pi * (diameter_outlet/2)^2) * velocity_outlet

Since the diameter of the outlet is the same as inlet 2, we have:

diameter_outlet = diameter2 = 0.03 m

Let's calculate the velocity_outlet:

0.00291 = (pi * (0.03/2)^2) * velocity_outlet
velocity_outlet = 0.00291 / ((pi * (0.03/2)^2))
≈ 7.23 m/s

Therefore, the mixing velocity of the fluid at the outlet is approximately 7.23 m/s.

2. Density Combination:
The density of the mixture at the outlet can be determined using the conservation of mass equation:

(Mass flow rate of water at inlet 1) + (Mass flow rate of methanol at inlet 2) = (Mass flow rate of mixture at outlet)

The mass flow rate of a fluid can be calculated by multiplying the volume flow rate with the density.

Given that the specific gravity (SG) of methanol is 0.85 (which is the ratio of its density to the density of water), we can deduce that the density of methanol (ρmethanol) is:

ρmethanol = SG * ρwater

Given that ρwater is the density of water, which is approximately 1000 kg/m^3, we can calculate the density of methanol:

ρmethanol = 0.85 * 1000
≈ 850 kg/m^3

Now, we can calculate the mass flow rate of water at inlet 1:

Mass flow rate of water at inlet 1 = Q1 * ρwater

Mass flow rate of water at inlet 1 = 0.0018 * 1000
= 1.8 kg/s

Similarly, we can calculate the mass flow rate of methanol at inlet 2:

Mass flow rate of methanol at inlet 2 = Q2 * ρmethanol

Mass flow rate of methanol at inlet 2 = 0.00111 * 850
= 0.9435 kg/s

Since the total mass flow rate at the inlet is equal to the total mass flow rate at the outlet, we have:

Mass flow rate of mixture at outlet = Mass flow rate of water at inlet 1 + Mass flow rate of methanol at inlet 2

Mass flow rate of mixture at outlet = Required Mass flow rate at outlet

Now, we can calculate the density of the mixture at the outlet:

Required Mass flow rate at outlet = Mass flow rate of mixture at outlet

Required Mass flow rate at outlet = Qout * ρoutlet

Qout * ρoutlet = 1.8 + 0.9435

ρoutlet = (1.8 + 0.9435)/ Qout

ρoutlet = (1.8 + 0.9435)/ 0.00291
≈ 1158.43 kg/m^3

Therefore, the density of the mixture at the outlet is approximately 1158.43 kg/m^3.