raw sugar cane is taken into a process to create sugar, which is essentially sucrose. the raw cane is approximately 16% sucrose, 63% water, and the rest fiber by mass. juice from the cane is extracted by passing the cane through a series of crushers. about 5% extra mass of water is added to the sugar cane prior to this step to help in the extraction process. the crushed cane and liquid juice is sent to a filter press that creates a cake that contains 4% of the weight of the cane juice, which has a composition similar to the overall non-fiber content of the raw cane. the filtrate is sent to an evaporator where enough water is evaporated to obtain a pale yellow juice that is 41% water. a

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Answer 1

The process of creating sugar from raw sugar cane involves several steps. First, the raw cane is passed through a series of crushers to extract the juice from the cane.

What is crusher?

A crusher is a machine designed to reduce large rocks into smaller rocks, gravel, or rock dust.

This step also adds about 5% extra mass of water to the sugar cane to help with the extraction process. The crushed cane and liquid juice are then sent to a filter press, which creates a cake containing 4% of the weight of the cane juice. This cake has a composition similar to the overall non-fiber content of the raw cane. The filtrate is then sent to an evaporator, where enough water is evaporated to obtain a pale yellow juice that is 41% water. Finally, the liquid is crystallized to produce sugar crystals, which are typically about 99.8% sucrose.

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Related Questions

Raw sugar cane is taken into a process to create sugar, which is essentially sucrose. the raw cane is approximately 16% sucrose, 63% water, and the rest fiber by mass. juice from the cane is extracted by passing the cane through a series of crushers. about 5% extra mass of water is added to the sugar cane prior to this step to help in the extraction process. the crushed cane and liquid juice is sent to a filter press that creates a cake that contains 4% of the weight of the cane juice, which has a composition similar to the overall non-fiber content of the raw cane. the filtrate is sent to an evaporator where enough water is evaporated to obtain a pale yellow juice that is 41% water. A series of vacuum processes removes enough water without damaging the sugars until you obtain a solution that is 91% sucrose. At this point, the mixture is fed to a crystallizer to produce a final product of sucrose that is 97.8% crystal. You control the process by measuring the flowrate of the solution into the crystallizer using a manometer that has mercury as the working fluid. The open-ended manometer shows a mercury height difference of 6.3 inches, , while the height of the sucrose solution in the manometer between the mercury and the pipe is 15.3 in.
The flowrate of the mixture into the crystallizer is related to the pressure with the following equation:
q(m^3/s) = 0.0307 m^3/hr atm^1/2 √ Pmanometer
What is the mass of sugar cane being fed to the process with this flowrate, in kg/s?

Answers

The mass of sugar cane being fed to the process with this flowrate is 0.0014 kg/s.

What is flowrate?

Flowrate is the rate at which a fluid or gas passes through a given space or container over a period of time. It is usually expressed in terms of volume per unit of time, such as liters per second or gallons per minute. Flowrate is an important factor in many engineering and scientific applications, such as fluid dynamics, hydroelectric power generation, and chemical processing. It is also used to measure the flow of liquids and gases through pipes, valves, and other components of a system.

The mass of sugar cane being fed to the process can be calculated using the following equation:
Mass (kg/s) = Flowrate (m^3/s) * Density of the mixture (kg/m^3)
The flowrate can be calculated using the equation given:
q(m^3/s) = 0.0307 m^3/hr atm^1/2 √ Pmanometer
Therefore, the flowrate is:
q(m^3/s) = 0.0307 m^3/hr * (atm^1/2 * (6.3 in/12 in/ft)^1/2)
q(m^3/s) = 0.0015 m^3/s
The density of the mixture can be calculated using the following equation:
Density of the mixture (kg/m^3) = Mass of sugar (kg) / Volume of the mixture (m^3)
The mass of sugar can be calculated using the following equation:
Mass of sugar (kg) = Mass of the mixture (kg) * (Mass fraction of sugar (kg/kg) / Mass fraction of the mixture (kg/kg))
The mass of the mixture can be calculated using the following equation:
Mass of the mixture (kg) = Volume of the mixture (m^3) * Density of the mixture (kg/m^3)
The volume of the mixture can be calculated using the following equation:
Volume of the mixture (m^3) = Flowrate (m^3/s)
The mass fraction of sugar can be calculated using the following equation:
Mass fraction of sugar (kg/kg) = Mass of Sugar (kg) / Mass of the mixture (kg)
The mass fraction of the mixture can be calculated using the following equation:
Mass fraction of the mixture (kg/kg) = Mass of the mixture (kg) / Mass of the mixture (kg)
Substituting the values into the equation, we get:
Mass (kg/s) = 0.0015 m^3/s * (0.91 kg/m^3)
Mass (kg/s) = 0.0014 kg/s
Therefore, the mass of sugar cane being fed to the process with this flowrate is 0.0014 kg/s.

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Find the weight W needed to hold the wall shown in Fig. P2.76 upright. The wall is 10 m wide.

Answers

The weight needed to hold the wall is equal to 14.9285 N.

What is force equilibrium?

The force equilibrium can be described as the addition of the forces about x, y, and z-axis will be equal to zero and the sum of the moment also equal to zero.

Given, the width of the wall, b = 10 m

The height of the water, d = 4 m

The height of the wall, l = 7m

The area of the wall, A = bd = 10(4) = 40 m²

The expression of the centroid of the wall,  [tex]\displaystyle \bar x = \frac{d}{2}[/tex]

[tex]\displaystyle \bar x = \frac{4}{2} = 2 m[/tex]

The hydrostatic force, F = ρgAx

F = (1000) (9.81) (40) (2)

F = 784,000 N

The expression for  the moment of inertia: [tex]\displaystyle I_C =\frac{bd^3}{12}[/tex]

[tex]I_c = 10\times 4/12 = 53.34 m^4[/tex]

The relation of the center of the pressure gate can be written as:

[tex]\displaystyle \bar h = \bar x +\frac{I_C}{A\bar x}[/tex]

[tex]\bar h = 2 + \frac{53.34}{40\times 2} = 2.667 \; m[/tex]

The moment about point: l × W - F(d - h) = 0

Substitute the values in the above equation:

7 × W - 78400 (4 - 2.667) = 0

7 W = 78400 (4 - 2.667)

W = 14.92 kN

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