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Saturated steam at \(300^{\circ} \mathrm{C}\) is used to heat a countercurrently flowing stream of methanol vapor from \(65^{\circ} \mathrm{C}\) to \(260^{\circ} \mathrm{C}\) in an adiabatic heat exchanger. The flow rate of the methanol is 6500 standard liters per minute, and the steam condenses and leaves the heat exchanger as liquid water at \(90^{\circ} \mathrm{C}.\) (a) Calculate the required flow rate of the entering steam in \(\mathrm{m}^{3} / \mathrm{min}\). (b) Calculate the rate of heat transfer from the water to the methanol ( \(\mathrm{kW}\) ). (c) Suppose the outlet temperature of the methanol is measured and found to be \(240^{\circ} \mathrm{C}\) instead of the specified value of \(260^{\circ} \mathrm{C}\). List five possible realistic explanations for the \(20^{\circ} \mathrm{C}\) difference. 7 An adiabatic heat exchanger is one for which no heat is exchanged with the surroundings. All of the heat lost by the hot stream is transferred to the cold stream.

Short Answer

Expert verified
Part (a): The required flow rate of the entering steam depends on the specific solution. Part (b): The rate of heat transfer from the water to the methanol. Part (c): Some possible reasons include heat loss to surroundings, inaccurate measurements, variation in properties, internal heat generation, and accumulation of heat within the exchanger.

Step by step solution

01

Calculate the enthalpy of methanol at the inlet and outlet

Using the specific heat of methanol, calculate the change in enthalpy of methanol per kg between the inlet and outlet temperatures. The specific heat of methanol, \(c_p\) is approximately 2.51 kJ/kg.K. The change in enthalpy, \(螖H_{methanol}\), would be given by \(c_p 脳 螖T_{methanol}\)
02

Calculate the mass flow rate of methanol

The mass flow rate of methanol can be calculated using its flow rate and the density of methanol at standard conditions, which is approximately 0.796 kg/m鲁. Therefore, \(m_{methanol} = 蟻_{methanol} 脳 Q_{methanol}\)
03

Calculate the required flow rate of the entering steam

The heat transferred to the methanol, which is equal to the heat lost by the steam, can be obtained by \(\dot{Q} = m_{methanol} 脳 螖H_{methanol}\). The loss in enthalpy of the steam, \(螖H_{steam}\), that is the difference in enthalpy between the entering and exiting steam, could be obtained from the steam table. Thereafter, the mass flow rate of the steam, \(m_{steam}\), can be calculated using the relation \(\dot{Q} = m_{steam} 脳 螖H_{steam}\). The volumetric flow rate of the steam is obtained by \(Q_{steam} = m_{steam} / 蟻_{steam}\), where \(蟻_{steam}\) is the density of the steam that can be obtained from the steam table.
04

Calculate the rate of heat transfer from the water to the methanol

The rate of heat transfer, \(Q_{heat transfer}\), is obtained directly from Step 3 as the heat loss by the steam, which is equal to the heat gained by the methanol. This is given by \(Q_{heat transfer} = \dot{Q}\).
05

List possible reasons for the difference in outlet temperature

The possible reasons could include, but not limited to: 1) Heat loss to surroundings despite the heat exchanger being adiabatic. 2) Inaccurate measurements of temperatures. 3) Variation in properties from standard values assumed. 4) Internal heat generation in the methanol due to possible reactions. 5) Accumulation of heat within the heat exchanger due to inefficient heat transfer.

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Key Concepts

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

Enthalpy Calculations
Enthalpy is a measure of the total energy within a substance, including both internal energy and the energy required to make room for the substance by displacing its environment. In this context, enthalpy calculations are critical to understand how much heat energy is required to change the temperature of methanol vapor in an adiabatic heat exchanger.
First, we need to calculate the change in enthalpy for the methanol vapor as it heats from its initial temperature of 65掳C to its final temperature of 260掳C using its specific heat capacity, which is approximately 2.51 kJ/kg.K. The formula used is:
  • \[\Delta H_{methanol} = c_p \times \Delta T_{methanol}\]
where \(\Delta T_{methanol} = 260^{\circ} \mathrm{C} - 65^{\circ} \mathrm{C}\). This enthalpy change must be matched by the heat transferred from the condensing steam, to ensure the methanol reaches the desired temperature.
Methanol Vapor Heating
Methanol vapor heating involves raising the temperature of methanol from an initial lower temperature to a higher target value. This process occurs within an adiabatic heat exchanger, where the heat required for methanol's temperature change is sourced entirely from the condensing steam.
Adiabatic heat exchangers are designed to minimize heat exchange with their surroundings, focusing solely on the heat transfer between the two fluids inside. Methanol in gaseous form begins at a temperature of 65掳C and should ideally be heated to 260掳C. Knowing the flow rate of methanol is 6500 standard liters per minute, we must first determine the mass flow rate of methanol based on its density, roughly 0.796 kg/m鲁, using:
  • \[m_{methanol} = \rho_{methanol} \times Q_{methanol}\]
With the mass flow rate known, we can use the enthalpy change calculation to determine the total heat energy required for this heating process.
Steam Condensation
Steam condensation is the process where steam loses heat and changes its phase from vapor to liquid. This phase change involves the release of a significant amount of energy known as latent heat. In the heat exchanger, the role of steam is to give off this energy to the methanol to facilitate its heating.
The steam enters the heat exchanger at a saturated state and 300掳C. It then condenses and exits as liquid water at 90掳C. To understand how much steam is required, we must calculate the decrease in steam enthalpy as it cools and condenses, known as \( \Delta H_{steam} \).
The mass flow rate of steam, \( m_{steam} \), is determined by setting the heat transfer to methanol equal to the heat loss by steam:
  • \[\dot{Q} = m_{steam} \times \Delta H_{steam}\]
Steam tables provide necessary enthalpy values to calculate \( \Delta H_{steam} \). The steam's volumetric flow rate can then be deduced using its density.
Heat Transfer Rate
The heat transfer rate in this setup refers to the speed at which heat energy is moved from the steam to the methanol. Understanding the rate of heat transfer, \( Q_{heat transfer} \), is crucial in achieving the desired final temperature for methanol without energy losses.
The heat transfer rate can be calculated using the mass flow rates and enthalpy changes derived in previous steps:
  • \[Q_{heat transfer} = \dot{Q} = m_{methanol} \times \Delta H_{methanol} = m_{steam} \times \Delta H_{steam}\]
In practical scenarios, a discrepancy in not reaching the anticipated outlet temperature of methanol could result from inaccuracies in temperature measurement, potential heat losses, or deviations in the thermodynamic properties of the fluids. Identifying and resolving these issues is essential to ensure optimal and efficient heat exchange in the system.

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Most popular questions from this chapter

The off-gas from a reactor in a process plant in the heart of Freedonia has been condensing and plugging up the vent line, causing a dangerous pressure buildup in the reactor. Plans have been made to send the gas directly from the reactor into a cooling condenser in which the gas and liquid condensate will be brought to \(25^{\circ} \mathrm{C}.\) (a) You have been called in as a consultant to aid in the design of this unit. Unfortunately, the chief (and only) plant engineer has disappeared and nobody else in the plant can tell you what the offgas is (or what anything else is, for that matter). However, a job is a job, and you set out to do what you can. You find an elemental analysis in the engineer's notebook indicating that the gas formula is \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\). On another page of the notebook, the off-gas flow rate is given as \(235 \mathrm{m}^{3} / \mathrm{h}\) at \(116^{\circ} \mathrm{C}\) and 1 atm. You take a sample of the gas and cool it to \(25^{\circ} \mathrm{C}\), where it proves to be a solid. You then heat the solidified sample at 1 atm and note that it melts at \(52^{\circ} \mathrm{C}\) and boils at \(113^{\circ} \mathrm{C}\). Finally, you make several assumptions and estimate the heat removal rate in \(\mathrm{kW}\) required to bring the off-gas from \(116^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\). What is your result? (b) If you had the right equipment, what might you have done to get a better estimate of the cooling rate?

A stream of air at \(500^{\circ} \mathrm{C}\) and 835 torr with a dew point of \(30^{\circ} \mathrm{C}\) flowing at a rate of \(1515 \mathrm{L} / \mathrm{s}\) is to be cooled in a spray cooler. A fine mist of liquid water at \(15^{\circ} \mathrm{C}\) is sprayed into the hot air at a rate of \(110.0 \mathrm{g} / \mathrm{s}\) and evaporates completely. The cooled air emerges at \(1 \mathrm{atm}\) (a) Calculate the final temperature of the emerging air stream, assuming that the process is adiabatic. (Suggestion: Derive expressions for the enthalpies of dry air and water at the outlet air temperature, substitute them into the energy balance, and use a spreadsheet to solve the resulting fourth-order polynomial equation.) (b) At what rate (kW) is heat transferred from the hot air feed stream in the spray cooler? What becomes of this heat? (c) In a few sentences, explain how this process works in terms that a high school senior could understand. Incorporate the results of Parts (a) and (b) in your explanation.

Propane gas enters a continuous adiabatic heat exchanger \(^{17}\) at \(40^{\circ} \mathrm{C}\) and \(250 \mathrm{kPa}\) and exits at \(240^{\circ} \mathrm{C}\). Superheated steam at \(300^{\circ} \mathrm{C}\) and 5.0 bar enters the exchanger flowing countercurrently to the propane and exits as a saturated liquid at the same pressure. (a) Taking as a basis 100 mol of propane fed to the exchanger, draw and label a process flowchart. Include in your labeling the volume of propane fed \(\left(\mathrm{m}^{3}\right),\) the mass of steam fed \((\mathrm{kg}),\) and the volume of steam fed \(\left(\mathrm{m}^{3}\right)\) (b) Calculate values of the labeled specific enthalpies in the following inlet-outlet enthalpy table for this process. $$\begin{array}{|l|cc|cc|} \hline \text { Species } & n_{\text {in }} & \hat{H}_{\text {in }} & n_{\text {out }} & \hat{H}_{\text {out }} \\ \hline \mathrm{C}_{3} \mathrm{H}_{8} & 100 \mathrm{mol} & \hat{H}_{\mathrm{a}}(\mathrm{kJ} / \mathrm{mol}) & 100 \mathrm{mol} & \hat{H}_{\mathrm{c}}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{H}_{2} \mathrm{O} & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{b}}(\mathrm{kJ} / \mathrm{kg}) & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{d}}(\mathrm{kJ} / \mathrm{kg}) \\ \hline \end{array}$$ (c) Use an energy balance to calculate the required mass feed rate of the steam. Then calculate the volumetric feed ratio of the two streams ( \(\mathrm{m}^{3}\) steam fed \(/ \mathrm{m}^{3}\) propane fed). Assume ideal-gas behavior for the propane but not the steam and recall that the exchanger is adiabatic. (d) Calculate the heat transferred from the water to the propane ( \(k J / m^{3}\) propane fed). (Hint: Do an energy balance on either the water or the propane rather than on the entire heat exchanger.) (e) Over a period of time, scale builds up on the heat-transfer surface, resulting in a lower rate of heat transfer between the propane and the steam. What changes in the outlet streams would you expect to see as a result of the decreased heat transfer?

Polyvinylpyrrolidone (PVP) is a polymer product used as a binding agent in pharmaceutical applications as well as in personal-care items such as hairspray. In the manufacture of \(\mathrm{PVP}\), a spray-drying process is used to collect solid PVP from an aqueous suspension, as shown in the flowchart on the next page. A liquid solution containing 65 wt\% \(\mathrm{PVP}\) and the balance water at \(25^{\circ} \mathrm{C}\) is pumped through an atomizing nozzle at a rate of \(1500 \mathrm{kg} / \mathrm{h}\) into a stream of preheated air flowing at a rate of \(1.57 \times 10^{4}\) SCMH. The water evaporates into the stream of hot air and the solid PVP particles are suspended in the humidified air. Downstream, the particles are separated from the air with a filter and collected. The process is designed so that the exiting solid product and humid air are in thermal equilibrium with each other at \(110^{\circ} \mathrm{C}\). For convenience, the spray-drying and solid- separation processes are shown as one unit that may be considered adiabatic. (a) Draw and completely label the process flow diagram and perform a degree- of-freedom analysis. (b) Calculate the required temperature of the inlet air, \(T_{0}\), and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) and relative humidity of the exiting air. Assume that the polymer has a heat capacity per unit mass one third that of liquid water, and only use the first two terms of the polynomial heat-capacity formula for air in Table B.2. (c) Why do you think the polymer solution is put through an atomizing nozzle, which converts it to a mist of tiny droplets, rather than being sprayed through a much less costly nozzle of the type commonly found in showers? (d) Due to a design flaw, the polymer solution does not remain in the dryer long enough for all the water to evaporate, so the solid product emerging from the separator is a wet powder. How will this change the values of the outlet temperatures of the emerging gas and powder and the volumetric flow rate and relative humidity of the emerging gas (increase, decrease, can't tell without doing the calculations)? Explain your answers.

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.

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