Math Problem Statement

Sent by you: A two-compartment vessel contains a fluid in each section. For the first compartment the fluid density is 1200kg/m3 and for the second is 500 kg/m3. The partition is removed and the fluids mix. Determine the difference in densities of the mixture for fluid with equal masses to that of fluid with equal volumes. a.) the fluid mass are equal b.) the fluid volume are equal

Solution

To solve this problem, we need to consider the two cases separately: when the masses of the fluids are equal and when the volumes of the fluids are equal.

Given:

  • Density of fluid in the first compartment, ρ1=1200kg/m3\rho_1 = 1200 \, \text{kg/m}^3
  • Density of fluid in the second compartment, ρ2=500kg/m3\rho_2 = 500 \, \text{kg/m}^3

Case 1: Equal Masses of the Fluids

Let the mass of each fluid be mm.

  • Volume of Fluid 1: V1=mρ1V_1 = \frac{m}{\rho_1}
  • Volume of Fluid 2: V2=mρ2V_2 = \frac{m}{\rho_2}

Total mass of the mixture M=m+m=2mM = m + m = 2m.

Total volume of the mixture V=V1+V2=mρ1+mρ2V = V_1 + V_2 = \frac{m}{\rho_1} + \frac{m}{\rho_2}.

The density of the mixture ρm\rho_m can be found by:

ρm=Total MassTotal Volume=2mmρ1+mρ2\rho_m = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{2m}{\frac{m}{\rho_1} + \frac{m}{\rho_2}}

Simplifying:

ρm=2mm(1ρ1+1ρ2)=21ρ1+1ρ2\rho_m = \frac{2m}{m \left( \frac{1}{\rho_1} + \frac{1}{\rho_2} \right)} = \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}}

ρm=211200+1500=25+126000=2176000=2×600017705.88kg/m3\rho_m = \frac{2}{\frac{1}{1200} + \frac{1}{500}} = \frac{2}{\frac{5 + 12}{6000}} = \frac{2}{\frac{17}{6000}} = \frac{2 \times 6000}{17} \approx 705.88 \, \text{kg/m}^3

Case 2: Equal Volumes of the Fluids

Let the volume of each fluid be VV.

  • Mass of Fluid 1: m1=ρ1×Vm_1 = \rho_1 \times V
  • Mass of Fluid 2: m2=ρ2×Vm_2 = \rho_2 \times V

Total mass of the mixture M=m1+m2=ρ1V+ρ2V=V(ρ1+ρ2)M = m_1 + m_2 = \rho_1 V + \rho_2 V = V(\rho_1 + \rho_2).

Total volume of the mixture Vtotal=V+V=2VV_{\text{total}} = V + V = 2V.

The density of the mixture ρm\rho_m' can be found by:

ρm=Total MassTotal Volume=V(ρ1+ρ2)2V=ρ1+ρ22\rho_m' = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{V(\rho_1 + \rho_2)}{2V} = \frac{\rho_1 + \rho_2}{2}

ρm=1200+5002=17002=850kg/m3\rho_m' = \frac{1200 + 500}{2} = \frac{1700}{2} = 850 \, \text{kg/m}^3

Difference in Densities

  • For equal masses: ρm705.88kg/m3\rho_m \approx 705.88 \, \text{kg/m}^3
  • For equal volumes: ρm=850kg/m3\rho_m' = 850 \, \text{kg/m}^3

The difference in densities is:

Δρ=ρmρm850705.88144.12kg/m3\Delta \rho = \rho_m' - \rho_m \approx 850 - 705.88 \approx 144.12 \, \text{kg/m}^3

Final Answer:

  • (a) When the fluid masses are equal: ρm705.88kg/m3\rho_m \approx 705.88 \, \text{kg/m}^3
  • (b) When the fluid volumes are equal: ρm=850kg/m3\rho_m' = 850 \, \text{kg/m}^3
  • Difference: Δρ144.12kg/m3\Delta \rho \approx 144.12 \, \text{kg/m}^3

Would you like more details or have any questions?
Here are 5 related questions for further exploration:

  1. How does changing the mass ratio affect the final density of the mixture?
  2. What would happen if the fluids had different temperatures when mixed?
  3. How would the density change if one fluid were a gas and the other a liquid?
  4. Can you derive a general formula for the density of a mixture with arbitrary fluid densities and mass/volume ratios?
  5. What assumptions are made in the calculation of mixture densities, and how might they affect real-world scenarios?

Tip: Always consider the conservation of mass when dealing with fluid mixtures.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Fluid Mechanics
Density
Mixture Problems

Formulas

Density of the mixture with equal masses: ρ_m = 2 / (1/ρ_1 + 1/ρ_2)
Density of the mixture with equal volumes: ρ_m' = (ρ_1 + ρ_2) / 2
Difference in densities: Δρ = ρ_m' - ρ_m

Theorems

Conservation of Mass

Suitable Grade Level

Grades 11-12