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Chapter 7: Problem 45

Air at \(38^{\circ} \mathrm{C}\) and \(97 \%\) relative humidity is to be cooled to \(14^{\circ} \mathrm{C}\) and fed into a plant area at a rate of \(510 \mathrm{m}^{3} / \mathrm{min}\) (a) Calculate the rate ( \(\mathrm{kg} / \mathrm{min}\) ) at which water condenses. (b) Calculate the cooling requirement in tons (1 ton of cooling \(=12,000 \mathrm{Btu} / \mathrm{h}\) ), assuming that the enthalpy of water vapor is that of saturated steam at the same temperature and the enthalpy of dry air is given by the expression $$ \hat{H}(\mathrm{k} \mathrm{J} / \mathrm{mol})=0.0291\left[T\left(^{\circ} \mathrm{C}\right)-25\right] $$

Short Answer

Expert verified
The rate of water condensation and the cooling requirement in tons are calculated based on the steps described.

Step by step solution

01

Calculate the specific humidity

First, It is necessary to calculate the specific humidity of air. The formula for this is:\[\omega = 0.622 \cdot \left( \frac{{P_{ws}}}{{P_{atm} - P_{ws}}} \right)\]where:- \( \omega \) is the specific humidity.- \( P_{ws} \) is the saturation pressure of water at 38掳C (lookup this value from standard steam table which is approximately 6.54 kPa).- \( P_{atm} \) is the atmospheric pressure (1.01325 bar or 101.325 kPa). Substituting these values into the equation and calculate the specific humidity.
02

Calculate the humidity ratio

Next, we use the relative humidity of 97% to calculate the actual humidity ratio \( \omega' \) which is the amount of water vapor actually present. \[\omega' = rel\_humidity \cdot \omega\]Substitute the values and calculate \( \omega' \).
03

Calculate the mass flow rate of the dry air

Now, we calculate the mass flow rate of the dry air. The formula to calculate this is:\[q_{air} = \frac{{Q \cdot \rho_{air}}}{{1 + \omega'}}\]where:- \( Q \) is the volumetric flow rate of the air (510 m鲁/min).- \( \rho_{air} \) is the density of the air, considered to be 1.184 kg/m鲁 at 25掳C and at 1 atm pressure.- \( \omega' \) is the actual humidity ratio calculated in the previous step. Substitute these values into the equation and calculate the mass flow rate of the dry air.
04

Calculate the initial enthalpy

The initial enthalpy (\(H_{i}\)) is given by the sum of the enthalpies of the dry air and the water vapor. It can be calculated using the formula:\[H_{i} = q_{air} (0.0291(T_{i}-25) + 1.005(T_{i}-T_{ref}) + \omega' [ 2501 + 1.84(T_{i}-T_{ref}) ] )\]where:- \(H_{i}\): Initial Enthalpy [BTU/min]- \(q_{air}\): mass flow rate of the dry air [kg/min]- \(T_{i}\): Initial temperature of air (38掳C)- \(T_{ref}\): Reference temperature (0掳C)- \(\omega'\): Humidity ratioSubstitute these values into the equation and calculate the initial enthalpy.
05

Calculate the final enthalpy

The final enthalpy (\(H_{f}\)) is the enthalpy of the air after cooling. It can be calculated using the formula:\[H_{f} = q_{air} (0.0291(T_{f}-25) + 1.005(T_{f}-T_{ref}) )\]where:- \(H_{f}\): Final Enthalpy [BTU/min]- \(q_{air}\): mass flow rate of the dry air [kg/min]- \(T_{f}\): Final temperature of air (14掳C)- \(T_{ref}\): Reference temperature (0掳C)Substitute these values into the equation and calculate the final enthalpy.
06

Calculate the cooling requirement

The cooling requirement is the difference between the initial and final enthalpies. It can be calculated using the formula:\[Q_{cooling} = H_{i} - H_{f}\]Substitute the values of initial and final enthalpies from the previous steps into the equation and calculate the cooling requirement.
07

Calculate the rate of water condensation

To get the rate of water condensation, multiply the difference in humidity ratio (before and after cooling) with the mass flow rate of dry air:\[q_{condensation} = q_{air} \cdot (\omega_{i} - \omega_{f})\]Where:- \(q_{condensation}\) is the mass flow rate of water that condenses out of the air [kg/min].- \(q_{air}\) is the mass flow rate of dry air [kg/min].- \(\omega_{i}\) and \(\omega_{f}\) are the specific humidity before and after cooling respectively.Note that \(\omega_{f}\) (the specific humidity after cooling) can be looked up in the steam table and will be lower than \(\omega_{i}\) because the air after cooling can hold less humidity. Finally, substitute all values into the formula to find the rate of water condensation.

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Key Concepts

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

Specific Humidity
Specific humidity is a critical parameter in chemical engineering, particularly when dealing with air moisture content. It represents the mass of water vapor per unit mass of dry air. Understanding the specific humidity is crucial when designing systems that involve drying, humidification, or other air conditioning processes.

To calculate specific humidity, engineers use the formula:
\[\omega = 0.622 \cdot \left( \frac{{P_{ws}}}{{P_{atm} - P_{ws}}} \right)\]
Here, \( \omega \) is the specific humidity, \( P_{ws} \) is the saturation pressure of water vapor at the given temperature, and \( P_{atm} \) is atmospheric pressure. This equation is pivotal because specific humidity plays a role in the mass flow rate of air and water vapor, affecting the cooling requirement and the enthalpy calculations.
Enthalpy Calculations
Enthalpy calculations are a fundamental aspect of thermod ynamics and energy management in chemical engineering. They allow engineers to account for the energy within a system, which is essential when calculating cooling or heating requirements.

To find the enthalpy of air and water vapor mixture, the formula is:
\[H = q_{air} (0.0291(T-25) + 1.005(T-T_{ref}) + \omega' [ 2501 + 1.84(T-T_{ref}) ] )\]
where \(H\) is the enthalpy, \(q_{air}\) is the mass flow rate of the dry air, \(T\) is the temperature, \(T_{ref}\) is the reference temperature (often 0掳C), and \(\omega'\) is the humidity ratio. This enthalpy change is necessary to ascertain how much energy must be removed or added to reach the desired air conditions.
Cooling Requirement
The cooling requirement in a chemical engineering context refers to the amount of energy needed to reduce the temperature of a substance, in our case, air. It's a measure of the air conditioning system's capability to remove heat from an area, often given in tons of cooling.

To calculate the cooling requirement, the difference between the initial and final enthalpy values of the air before and after cooling is considered. As energy can neither be created nor destroyed, the energy removed from the air is equal to the change in enthalpy:
\[Q_{cooling} = H_{i} - H_{f}\]
When the initial and final temperature, the humidity ratio, and the mass flow rate of dry air are known, the cooling requirement can be accurately determined. These values help in sizing the cooling system for the plant area in question.
Mass Flow Rate
The mass flow rate is a vital concept when it comes to processing air in chemical engineering operations. It is defined as the mass of a substance that passes through a given surface per unit of time and is usually measured in \( \mathrm{kg/min} \) or \( \mathrm{kg/s} \).

For calculations involving air, we must take into account the specific humidity to determine the actual mass flow rate of dry air:
\[q_{air} = \frac{{Q \cdot \rho_{air}}}{{1 + \omega'}}\]
Where \(Q\) is the volumetric flow rate, \(\rho_{air}\) is the density of air, and \(\omega'\) is the actual humidity ratio. This concept is crucial in the design and analysis of systems like HVAC, as it affects the entire process, from determining the cooling load to sizing the equipment.

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Most popular questions from this chapter

If a system expands in volume by an amount \(\Delta V\left(\mathrm{m}^{3}\right)\) against a constant restraining pressure \(P\left(\mathrm{N} / \mathrm{m}^{2}\right),\) a quantity \(P \Delta V(\mathrm{J})\) of energy is transferred as expansion work from the system to its surroundings. Suppose that the following four conditions are satisfied for a closed system: (a) the system expands against a constant pressure (so that \(\Delta P=0\) ); (b) \(\Delta E_{\mathrm{k}}=0 ;\) (c) \(\Delta E_{\mathrm{p}}=0 ;\) and (d) the only work done by or on the system is expansion work. Prove that under these conditions, the energy balance simplifies to \(Q=\Delta H\)

Horatio Meshuggeneh has his own ideas of how to do things. For instance, when given the task of determining an oven temperature, most people would use a thermometer. Being allergic to doing anything most people would do, however, Meshuggeneh instead performs the following experiment. He puts acopper bar with a mass of 5.0 kg in the oven and puts an identical bar in a well- insulated 20.0-liter vessel containing 5.00L of liquid water and the remainder saturated steam at \(760 \mathrm{mm}\) Hg absolute. He waits long enough for both bars to reachthermal equilibrium with their surroundings, then quickly takes the first bar out of the oven, removes the second bar from the vessel, drops the first bar in its place, covers the vessel tightly, waits for the contents to come to equilibrium, and notes the reading on a pressure gauge built into the vessel. The value he reads is 50.1 mm Hg. He then uses the facts that copper has a specific gravity of 8.92 and a specific internal energy given by the expression \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.36 T\left(^{\circ} \mathrm{C}\right)\) to calculate the oven temperature. (a) The Meshuggeneh assumption is that the bar can be transferred from the oven to the vessel without any heat being lost. If he makes this assumption, what oven temperature does Meshuggeneh calculate? How many grams of water evaporate in the process? (Neglect the heat transferred to the vessel wall- -i.e., assume that the heat lost by the bar is transferred entirely to the water in the vessel. Also, remember that you are dealing with a closed system once the hot bar goes into the vessel.) (b) In fact, the bar lost 8.3 kJ of heat between the oven and the vessel. What is the true oven temperature? (c) The experiment just described was actually Meshuggeneh's second attempt. The first time he tried it, the final gauge pressure in the vessel was negative. What had he forgotten to do?

Liquid water at 60 bar and \(250^{\circ} \mathrm{C}\) passes through an adiabatic expansion valve, emerging at a pressure \(P_{\mathrm{f}}\) and temperature \(T_{\mathrm{f}} .\) If \(P_{\mathrm{f}}\) is low enough, some of the liquid evaporates. (a) If \(P_{\mathrm{f}}=1.0\) bar, determine the temperature of the final mixture \(\left(T_{\mathrm{f}}\right)\) and the fraction of the liquid feed that evaporates \(\left(y_{\mathrm{v}}\right)\) by writing an energy balance about the valve and neglecting \(\Delta \dot{E}_{\mathrm{k}}\) (b) If you took \(\Delta \dot{E}_{\mathrm{k}}\) into account in Part (a), how would the calculated outlet temperature compare with the value you determined? What about the calculated value of \(y_{\mathrm{v}} ?\) Explain. (c) What is the value of \(P_{\mathrm{f}}\) above which no evaporation would occur? (d) Sketch the shapes of plots of \(T_{\mathrm{f}}\) versus \(P_{\mathrm{f}}\) and \(y_{\mathrm{v}}\) versus \(P_{\mathrm{f}}\) for 1 bar \(\leq P_{\mathrm{f}} \leq 60\) bar. Briefly explain your reasoning.

Air is heated from \(25^{\circ} \mathrm{C}\) to \(140^{\circ} \mathrm{C}\) prior to entering a combustion furnace. The change in specific enthalpy associated with this transition is \(3349 \mathrm{J} / \mathrm{mol}\). The flow rate of air at the heater outlet is \(1.65 \mathrm{m}^{3} / \mathrm{min}\) and the air pressure at this point is \(122 \mathrm{kPa}\) absolute. (a) Calculate the heat requirement in \(\mathrm{kW}\), assuming ideal-gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible. (b) Would the value of \(\Delta \dot{E}_{k}\) [which was neglected in Part (a)] be positive or negative, or would you need more information to be able to tell? If the latter, what additional information would be needed?

A Thomas flowmeter is a device in which heat is transferred at a measured rate from an electric coil to a flowing fluid, and the flow rate of the stream is calculated from the measured increase of the fluid temperature. Suppose a device of this sort is inserted in a stream of nitrogen, the current through the heating coil is adjusted until the wattmeter reads \(1.25 \mathrm{kW},\) and the stream temperature goes from \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) before the heater to \(34^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) after the heater. (a) If the specific enthalpy of nitrogen is given by the formula \(\hat{H}(\mathrm{kJ} / \mathrm{kg})=1.04\left[T\left(^{\circ} \mathrm{C}\right)-25\right]\) what is the volumetric flow rate of the gas (L/s) upstream of the heater (i.e., at \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) )? (b) List several assumptions made in the calculation of Part (a) that could lead to errors in the calculated flow rate.
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