Question One In The Olden Days We Used To Have Milk In Bottl
Question Onein The Olden Days We Used To Have Milk In Bottles Delivere
Compare the milk bottle where the cream is still uniformly mixed with the milk, to the same milk bottle later on when the cream has separated and risen to the top. Does the hydrostatic pressure at the bottom of the milk bottle remain the same during the phase separation of the cream? Assume the volume of the cream is 4.6 x 10-6 m3, the volume of milk (minus the cream) is 4.554 x 10-4 m3, the density of cream is 994 kg/m3, and the density of the milk is 1033 kg/m3. The diameter of the bottle at the top is 26 mm (where the cream separates) and the average diameter for the rest of the bottle is 75 mm. Find the difference in hydrostatic pressure at the bottom of the bottle after the phase separation of the cream.
Paper For Above instruction
The process of cream separation in milk bottles is a classical demonstration of fluid stratification and hydrostatic pressure variations within a contained fluid system. The core of this analysis involves understanding how phase separation affects the hydrostatic pressure at the bottom of the bottle, which is primarily dependent on the fluid's density and height, not on the distribution of the components within the fluid. This paper explores the underlying physics principles, performing calculations to determine whether the hydrostatic pressure at the bottom remains constant during the separation of cream from milk.
Initially, milk with uniformly mixed cream and milk would exert a certain hydrostatic pressure at the bottom of the bottle determined by the overall density and height of the fluid column. As the cream rises to the top due to differences in density, the composition and distribution of fluids within the bottle change. The cream density (994 kg/m3) and the milk density (1033 kg/m3) influence how the phases separate over time. Since the total volume of the milk and cream remains constant, the separation results in two distinct layers, with the cream at the top and the milk at the bottom.
The hydrostatic pressure at any depth in a fluid is given by the equation:
P = ρgh
where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity (approximated as 9.81 m/s2), and h is the height of the fluid column above the point of measurement.
In the case of uniform mixture, the effective density of the entire fluid column influences this pressure. As phase separation occurs, the less dense cream layer sits atop the denser milk layer; thus, the pressure at the bottom depends on the cumulative height and densities of the respective layers.
To analyze the change, we calculate the hydrostatic pressure before and after separation, considering the different densities and fluid heights.
Calculations
**Step 1: Calculate total heights corresponding to the different volumes and cross-sectional areas. The bottle's cross-sectional area varies at different sections and is derived from diameters provided.
At the wider section (radius = 75 mm):
A1 = π (r1)2 = π (0.075 m/2)2 ≈ 4.42 x 10-3 m2
At the narrower top section (diameter = 26 mm):
A2 = π * (0.026 m/2)2 ≈ 5.3 x 10-4 m2
Since the volume of the milk (minus cream) and the cream are known, the heights in different sections are determined by dividing volume by area:
Height of the mixture before separation:
- Volume of mixture Vtotal ≈ 4.6 x 10-6 + 4.554 x 10-4 ≈ 4.599 x 10-4 m3
- Using area A1: htotal ≈ Vtotal / A1 ≈ (4.599 x 10-4) / (4.42 x 10-3) ≈ 0.104 m
Height of the cream layer after separation:
- Volume of cream Vcream = 4.6 x 10-6 m3
- Height hcream ≈ Vcream / A1 ≈ (4.6 x 10-6) / (4.42 x 10-3) ≈ 0.00104 m
Height of the milk layer after separation:
- Volume of milk Vmilk ≈ 4.554 x 10-4 m3
- Height hmilk ≈ Vmilk / A1 ≈ (4.554 x 10-4) / (4.42 x 10-3) ≈ 0.103 m
Hydrostatic Pressure Difference Calculation
Post-separation, the key is the distribution of the layers. The total height of the fluid column remains approximately the same (~0.104 m). The bottom pressure during the uniform mixture and during phase separation primarily depends on the average density of the combined layers.
In the mixed state:
Average density, ρavg = (ρcream Vcream + ρmilk Vmilk) / (Vcream + Vmilk)
Calculating:
ρavg = (994 4.6 x 10-6 + 1033 4.554 x 10-4) / 4.599 x 10-4 ≈ (0.00457 + 4.702 x 10-1) / 4.599 x 10-4 ≈ 0.4707 / 0.0004599 ≈ 1023.6 kg/m3
This is close to the initial density of milk, indicating the initial uniform density was around this value.
After separation:
The pressure at the bottom is dominated by the weight of the entire column, which now consists of a top cream layer (density ≈ 994 kg/m3) overlying a larger volume of milk (density ≈ 1033 kg/m3).
The hydrostatic pressure difference between the bottom and a point just below the top layer is calculated as:
ΔP = g (ρmilk hmilk + ρcream hcream - total height ρavg)
Given the complex layering, the dominant factor remains the weight of the entire fluid column, which, by the principle of fluid statics, depends on the total mass and height, not on the internal distribution. Thus, the change in pressure at the bottom can be approximated by the difference in the total weight of the fluid before and after separation.
Conclusion
Because the total volume and overall average density of the fluid remain nearly constant, the hydrostatic pressure at the bottom of the bottle remains essentially unchanged during phase separation. The processes of cream rising and separation alter the internal composition but do not significantly affect the net hydrostatic pressure exerted at the bottom. Therefore, the hydrostatic pressure at the bottom of the bottle remains approximately the same during the phase separation of the cream.
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