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A gas stream containing \(n\) -hexane in nitrogen with a relative saturation of \(90 \%\) is fed to a condenser at \(75^{\circ} \mathrm{C}\) and 3.0 atm absolute. The product gas emerges at \(0^{\circ} \mathrm{C}\) and 3.0 atm at a rate of \(746.7 \mathrm{m}^{3} / \mathrm{h}\). (a) Calculate the percentage condensation of hexane (moles condensed/mole fed) and the rate \((\mathrm{kW})\) at which heat must be transferred from the condenser. (b) Suppose the feed stream flow rate and composition and the heat transfer from the condenser are the same as in Part (a), but the condenser and outlet stream pressure is only 2.5 atm instead of 3.0 atm. How would the outlet stream temperatures and flow rates and the percentage condensations of hexane calculated in Parts (a) and (b) change (increase, decrease, no change, no way to tell)? Don't do any calculations, but explain your reasoning.

Short Answer

Expert verified
The percentage condensation of hexane and the rate of heat transfer from the condenser can be calculated using the given information and relevant formulae. Decreasing the pressure will increase the percentage of hexane that condenses out of the gas phase.

Step by step solution

01

Calculation of moles of hexane

Using the relative saturation, we can write a formula. For 90% saturation, the moles of hexane is equal to 0.9 times the saturation limit moles of hexane at 75 C and 3 atm. The saturation limit can be found from vapor pressure data for n-hexane.
02

Determination of moles condensed

After passing through the condenser, the temperature of the gas stream drops to 0 degrees Celsius. At this reduced temperature and a pressure of 3 atm, the saturation limit for n-hexane will be lower than the number of moles we determined in Step 1. Therefore, some hexane will condense out of the gas phase.
03

Percentage condensation calculation

The percentage condensation of hexane is calculated by subtracting the number of moles of hexane at 0 degrees Celsius from the number of moles of hexane at 75 degrees Celsius, and then dividing by the number of moles at 75 degrees Celsius. Multiply by 100 to convert to a percentage.
04

Heat transfer rate calculation

The rate of heat transfer from the condenser can be calculated using the formula Q = mcp(T1-T2), where m is the mass flow rate of the gas, cp is the specific heat of the gas, and T1 and T2 are the initial and final temperatures of the gas in Kelvin, respectively.
05

Conceptual understanding on effect of pressure change

Lowering the pressure will decrease the saturation limit for n-hexane at 0 degrees Celsius, leading to an increase in the percentage of hexane that condenses out of the gas phase.

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Key Concepts

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

Percentage Condensation Calculation
Understanding the percentage condensation calculation is crucial for students working on chemical process problems. This concept involves determining the proportion of a vapor that becomes liquid during condensation.

In the context of the exercise, this calculation is performed by comparing the moles of n-hexane before and after passing through a condenser. Initially, the gas stream is saturated with n-hexane vapor at a given temperature and pressure. As the gas stream cools down in the condenser, the amount of n-hexane that can remain vapor decreases, and excess n-hexane condenses out of the vapor phase.

The formula used for this calculation is as follows: \[ \text{Percentage condensation of hexane} = \frac{\text{moles of hexane before condensation} - \text{moles of hexane after condensation}}{\text{moles of hexane before condensation}} \times 100\% \]
It's important for students to note that precise percentage calculations require accurate vapor pressure data for the specific compound, in this case, n-hexane. Furthermore, incorporating the concept of relative saturation into the calculation allows for a more realistic and applicable result when dealing with industrial processes.
Heat Transfer Rate Calculation
The heat transfer rate reflects the amount of heat energy transferred per unit time and is a fundamental concept in chemical engineering thermodynamics. The calculation of this rate is pivotal when assessing the efficiency of condensers and other heat-exchanging devices within a chemical process.

The formula to calculate the rate at which heat must be transferred from the condenser, denoted as \( Q \) in kilowatts (kW), can be represented as: \[ Q = \frac{m \cdot c_p \cdot (T_1 - T_2)}{1000} \] where \( m \) is the mass flow rate, \( c_p \) is the specific heat capacity of the gas at constant pressure, and \( T_1 \) and \( T_2 \) are the initial and final temperatures of the gas stream in Celsius, respectively. The division by 1000 is required to convert the energy from Joules (J) to kilowatts (kW).

When performing this calculation, it is vital for students to remember to convert all temperature measurements to Kelvin by adding 273.15 to the Celsius values. This step ensures that all thermodynamic equations utilized in the process are consistent with the Kelvin scale, which is the standard scale used in scientific heat calculations.
Vapor Pressure and Temperature Relationship
The relationship between vapor pressure and temperature is fundamental to understanding phase changes such as condensation and boiling. This relationship is especially important in the context of chemical processes that involve temperature changes under constant pressure conditions.

For a given substance, its vapor pressure increases with temperature. This behavior is described by the Clausius-Clapeyron equation, which relates the change in vapor pressure to temperature. As part of the improved exercise advice, it's critical to highlight this dependence: at higher temperatures, the amount of vapor that pressure can support grows, while at lower temperatures, the vapor pressure decreases, and more of the substance condenses into liquid.

In the exercise provided, as the temperature of the n-hexane and nitrogen gas stream decreases in the condenser, the vapor pressure of n-hexane falls, leading to a higher percentage of n-hexane condensing out of the gas phase. Students should also grasp that changes in system pressure can affect the vapor pressure of a substance, which subsequently impacts condensation behavior and the temperature at which the phase change occurs.

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Most popular questions from this chapter

Propane is to be burned with \(25.0 \%\) excess air. Before entering the furnace, the air is preheated from \(32^{\circ} \mathrm{F}\) to \(575^{\circ} \mathrm{F}\) (a) At what rate (B tu/h) must heat be transferred to the air if the feed rate of propane is \(1.35 \times 10^{5}\) SCFH (ft \(^{3} / \mathrm{h}\) at \(\mathrm{STP}\) )? (b) The stack gas leaves the furnace at \(855^{\circ} \mathrm{F}\). How is the air likely to be preheated?

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The heat required to raise the temperature of \(m\) (kg) of a liquid from \(T_{1}\) to \(T_{2}\) at constant pressure is $$ Q=\Delta H=m \int_{T_{1}}^{T_{2}} C_{p}(T) d T $$ In high school and in first-year college physics courses, the formula is usually given as $$ Q=m C_{p} \Delta T=m C_{p}\left(T_{2}-T_{1}\right) $$ (a) What assumption about \(C_{p}\) is required to go from Equation 1 to Equation \(2 ?\) (b) The heat capacity \(\left(C_{p}\right)\) of liquid \(n\) -hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well- insulated vessel containing \(2.00 \mathrm{L}\) of liquid \(n-\mathrm{C}_{6} \mathrm{H}_{14}\) at \(T=300 \mathrm{K} .\) A combustion reaction known to release \(16.73 \mathrm{kJ}\) of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be \(3.10 \mathrm{K}\). In a separate experiment, it is found that \(6.14 \mathrm{kJ}\) of heat is required to raise the temperature of everything in the system except the hexane by \(3.10 \mathrm{K}\). Use these data to estimate \(C_{p}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]\) for liquid \(n\) -hexane at \(T \approx 300 \mathrm{K},\) assuming that the condition required for the validity of Equation 2 is satisfied. Compare your result with a tabulated value.

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