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One thousand liters of a 95 wt\% glycerol- \(5 \%\) water solution is to be diluted to \(60 \%\) glycerol by adding a \(35 \%\) solution pumped from a large storage tank through a \(5-\mathrm{cm}\) ID pipe at a steady rate. The pipe discharges at a point 23 m higher than the liquid surface in the storage tank. The operation is carried out isothermally and takes 13 min to complete. The friction loss ( \(\hat{F}\) of Equation \(7.7-2\) ) is \(50 \mathrm{J} / \mathrm{kg}\). Calculate the final solution volume and the shaft work in \(\mathrm{kW}\) that the pump must deliver, assuming that the surface of the stored solution and the pipe outlet are both at 1 atm. Data: \(\quad \rho_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{kg} / \mathrm{L}, \rho_{\mathrm{gly}}=1.26 \mathrm{kg} / \mathrm{L} .\) (Use to estimate solution densities.)

Step by step solution

01

Initial Volume Calculation

Firstly, calculate the initial volume of glycerol (substance A) and water (substance B) in the solution. The given solution is 1000 L and composed of 95 % glycerol and 5 % water. Meaning there is \(0.95 \times 1000\) L of glycerol and \(0.05 \times 1000\) L of water.

02

Fluid solution Calculation

Next, consider the solution added from the storage tank. It consists of 35 \% glycerol (substance A) and 65 \% water (substance B) and the final solution is to be 60 \% glycerol. The mass balance for substance A is: \[ \text{initial of A} + \text{input of A} = \text{final of A} \] Where, \[ \text{initial of A} = \text{Volume of A initially}* \text{Density of A}, \text{input of A} = X*\text{Density of A}*0.35 \] Where X is the total volume of the solution input and \[ \text{final of A} = \text{Volume of A initially} + X \] Solving the above equation provides the value of X.

03

Final volume Calculation

The final volume of solution (F) is the sum of the initial solution and the volume added from the tank, so: \[ \text{F} = 1000 + X \]

04

Shaft work Calculation

In order to find the shaft work, we consider three components: energy required to lift the fluid h height (\(\rho_{fluid} g h\)), loss of energy due to friction (F), and energy required to overcome the potential energy change (\(p_{atm} V_{fluid}/\rho_{fluid}\)). The total shaft work W delivered by the pump will be the sum of these three quantities. \[W = \rho_{fluid} g h + F + p_{atm} V_{fluid}/\rho_{fluid}\] In this equation g is acceleration due to gravity, h is the height of lifting, \(V_{fluid}\) is the volume of fluid being moved, and \(p_{atm}\) is atmospheric pressure. Using the values given in the exercise, the final value of shaft work can then be calculated. This value must then be converted from J to kW and then divided by the time the operation took, 13 min = 780 s, to find the power (P) the pump needs to deliver in kW. \[P = W / 780\]

05

Density of the mixture

In order to find the densities of the initial and final solutions, you should use the rule of mixtures which states that the density of a mixture is equal to the sum of the fraction of each component times its respective density. It can be written as \[\rho_{\text{mix}} = \text{wt \% A} * \rho_A + \text{wt \% B} * \rho_B\] Here, A represents glycerol and B represents water. Using the provided densities of glycerol and water, the densities of the initial and final mixtures can be found.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Balance

Mass balance is a fundamental principle used in chemical process calculations to maintain equilibrium in a system. In the context of this exercise, it involves calculating the amount of a particular substance before and after a mixture process.

• Initially, we have 1000 L of a solution that is 95% glycerol and 5% water. This step calculates the initial amounts of each component. For glycerol, it is 0.95 times 1000 L, and for water, 0.05 times 1000 L.
• We then introduce a second solution containing 35% glycerol to achieve a final mixture of 60% glycerol.
• Using mass balances, we set up an equation to find how much additional solution we need to add, ensuring that the amount of glycerol and water in the system reaches the desired levels.

Understanding mass balance helps ensure that chemical processes are efficient and that final product compositions meet specific requirements.

Shaft Work Calculation

Shaft work calculation is crucial in determining the energy required for moving fluids using pumps in chemical processes. In this exercise, we aim to find the power needed to pump a dilute solution through a pipe.

• The height difference between the storage tank and the discharge point is 23 meters, a key factor in the energy required. Gravity acts on the fluid requiring work to move it upward, calculated as \( \rho_{\text{fluid}} g h \).
• Energy loss due to friction, given as 50 J/kg in the exercise, is another component. It affects the effectiveness of the pump.
• The equation accounts for atmospheric pressure, which is consistent at both the storage tank surface and the pipe outlet, thus simplifying calculations.
• Total work is summed and then converted to power by dividing by the total time, 780 seconds, required to complete the operation. This gives us the operational kW necessary from the pump.

Calculating shaft work in this manner illustrates the intricacies of energy management in fluid dynamics, aiding in efficient system design.

Solution Density Estimation

Estimating the density of a solution is a critical part of chemical process calculations. It allows us to determine how solutions behave and interact during mixing.

• The density of a solution is not the simple average of the component densities but a weighted calculation based on their respective mass percentages, often referred to as the rule of mixtures.
• For glycerol and water, given densities are 1.26 kg/L and 1.00 kg/L, respectively.
• The calculation for the final solution density involves summing the product of each component’s weight percentage and its density: \( \rho_{\text{mix}} = (wt\% \text{glycerol}) \times \rho_{\text{glycerol}} + (wt\% \text{water}) \times \rho_{\text{water}} \).
• Understanding these calculations helps predict how the solution will affect subsequent chemical reactions or physical processes.

Estimation of solution density is vital for ensuring consistency and quality in chemical manufacturing processes.

Fluid Dynamics

Fluid dynamics underpins many chemical engineering operations, especially involving movement through pipes and reactors.

• Fluid dynamics is key to understanding how the glycerol solution is transported from the storage tank through the pipe system to the mixing area.
• Pumping fluid up a 23-meter incline requires overcoming both gravitational and frictional forces, which are critical fluid dynamic components.
• The diameter of the pipe, taken as 5 cm here, influences the flow rate and velocity of the solution, impacting the energy required for pumping.
• By evaluating these factors, we can ensure that fluid moves efficiently without causing excess energy expenditure or operational issues.

The principles of fluid dynamics enable engineers to design systems that optimize flow paths, minimize energy uses, and enhance process efficiency.