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Liquid water enters a cooling tower operating at steady state at \(40^{\circ} \mathrm{C}\) with a mass flow rate of \(10^{5} \mathrm{~kg} / \mathrm{h}\). Cooled water at \(25^{\circ} \mathrm{C}\) exits the cooling tower at the same mass flow rate. Makeup water is supplied at \(23^{\circ} \mathrm{C}\). Atmospheric air enters the tower at \(30^{\circ} \mathrm{C}, 1\) bar, \(35 \%\) relative humidity. A saturated moist air stream exits at \(34^{\circ} \mathrm{C}, 1\) bar. Determine (a) the mass flow rates of the dry air and makeup water, each in \(\mathrm{kg} / \mathrm{h}\). (b) the rate of exergy destruction within the cooling tower, in \(\mathrm{kW}\), for \(T_{0}=23^{\circ} \mathrm{C}\). Ignore kinetic and potential energy effects.

Short Answer

Expert verified
Calculate specific enthalpies and find mass flow rates using steady-state mass and energy balances. Finally, compute exergy destruction using the appropriate formula.

Step by step solution

01

- Determine Enthalpy of Entering and Exiting Water

Calculate the specific enthalpy of water at the given temperatures. The water enters at 40°C and exits at 25°C. Use water enthalpy values: \(h_{in} = 167.5 \, \mathrm{kJ/kg}\) at 40°C and \(h_{out} = 104.9 \, \mathrm{kJ/kg}\) at 25°C.
02

- Calculate Heat Removed from Water

Find the heat removed from the water using the formula: \[ Q_{w} = \dot{m}_{w} (h_{in} - h_{out}) \] Substitute \(\dot{m}_{w} = 10^{5} \, \mathrm{kg/h}\) , \(h_{in} = 167.5 \, \mathrm{kJ/kg}\) and \(h_{out} = 104.9 \, \mathrm{kJ/kg}\).
03

- Calculate Makeup Water Flow Rate

Assume enthalpy of makeup water at 23°C is \(h_{m} = 95.5 \, \mathrm{kJ/kg}\). Use the steady-state assumption where the mass flow rate of water out must equal the mass flow rate of water in. Apply mass and energy balance: \[ \dot{m}_{m} = \frac{\dot{m}_{w} (h_{in} - h_{out})}{h_{m}} \]
04

- Determine Properties of Moist Air

Identify the properties of humid air entering and exiting the tower from psychrometric charts or tables given temperature, relative humidity, and pressure conditions.
05

- Calculate Mass Flow Rate of Dry Air

Using the properties identified, leverage the relationship between humid air, specific humidity, and dry air mass. Apply steady-state mass balance for dry air.
06

- Calculate Rate of Exergy Destruction

Apply the formula for exergy destruction and compute it using outlet and inlet conditions. The general exergy destruction formula: \[ \dot{X}_{d} = \dot{m} ( (h_{e} - T_{0}s_{e}) - (h_{i} - T_{0}s_{i}) ) \]

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Key Concepts

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

enthalpy calculation
Enthalpy is a measure of the total heat content in a thermodynamic system. For your cooling tower analysis, enthalpy calculation is vital as it allows you to understand the energy changes during the cooling process.
When water enters the cooling tower at 40°C, you need to determine its specific enthalpy. This is commonly denoted as \(h_{in}\), and at 40°C, it is 167.5 kJ/kg. Similarly, when water exits the tower at 25°C, the specific enthalpy is denoted as \(h_{out}\) and is 104.9 kJ/kg.
We use these enthalpy values to calculate the heat removed from the water using the formula:
\[ Q_{w} = \dot{m}_{w} (h_{in} - h_{out}) \]
Where \(\dot{m}_{w}\) represents the mass flow rate of the water. In this case, \(\dot{m}_{w} = 10^{5} \mathrm{kg/h}\). Plugging in the values, we can determine how much energy is removed from the water during the cooling process.
Understanding enthalpy calculations ensures that you can carry out complex heat transfer evaluations in thermodynamic systems like cooling towers efficiently.
mass flow rate
The mass flow rate is a measure of the mass of a substance passing through a given surface per unit time. In cooling tower analysis, it's crucial for balancing both mass and energy.
Initially, the water enters the tower at \(40^{\circ} \mathrm{C}\) with a mass flow rate of \(10^{5} \mathrm{kg/h}\). This water gets cooled and exits at \(25^{\circ} \mathrm{C}\) with the same mass flow rate.
But what about the makeup water? Makeup water compensates for the loss due to evaporation and drift. To find the mass flow rate of the makeup water, assume its enthalpy at \(23^{\circ} \mathrm{C}\) (which is 95.5 kJ/kg). You then apply the mass and energy balance equation:
\[ \dot{m}_{m} = \frac{\dot{m}_{w} (h_{in} - h_{out})}{h_{m}} \]
By inserting the known values, you will find the makeup water's mass flow rate. Additionally, understanding dry air's mass flow rate is essential. Given the properties of moist air provided, you use psychrometric calculations, linking humidity ratios, and applying steady-state mass balance to achieve precise results.
Accurate calculation of mass flow rate ensures the efficiency and effectiveness of thermal systems, helping to maintain operational stability and performance.
exergy destruction
Exergy destruction is a measure of the irreversibility in a thermodynamic process. It reflects the lost potential to do work, often due to entropy generation.
To determine the exergy destruction rate within the cooling tower, you'll use the general exergy destruction formula:
\[ \dot{X}_{d} = \dot{m} ((h_{e} - T_{0}s_{e}) - (h_{i} - T_{0}s_{i})) \]
Here, \(h_{e}\) and \(h_{i}\) refer to the enthalpy at the exit and inlet, respectively, while \(s_{e}\) and \(s_{i}\) represent entropy at these states. \(T_{0}\) is the reference temperature, assumed to be \(23^{\circ} \mathrm{C}\).
By evaluating these terms, you can calculate how much exergy—or energy potential—has been destroyed due to irreversibilities present in the system. These calculations help in identifying inefficiencies in the cooling tower.
Recognizing and minimizing exergy destruction leads to more sustainable and cost-effective operations by reducing wasted energy and improving overall system performance.

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Most popular questions from this chapter

\mathrm{~A}\( system consists initially of \)n_{\mathrm{A}}\( moles of gas \)\mathrm{A}\( at pressure \)p\( and temperature \)T\( and \)n_{\mathrm{B}}\( moles of gas \)\mathrm{B}\( separate from gas \)A\( but at the same pressure and temperature. The gases are allowed to mix with no heat or work interactions with the surroundings. The final equilibrium pressure and temperature are \)p\( and \)T\(, respectively, and the mixing occurs with no change in total volume. (a) Assuming ideal gas behavior, obtain an expression for the entropy produced in terms of \)\bar{R}, n_{\mathrm{A}}\(, and \)n_{\mathrm{B}}$ - (b) Using the result of part (a), demonstrate that the entropy produced has a positive value. (c) Would entropy be produced when samples of the same gas at the same temperature and pressure mix? Explain.

Outside air at \(50^{\circ} \mathrm{F}, 1\) atm, and \(40 \%\) relative humidity enters an air-conditioning device operating at steady state. Liquid water is injected at \(45^{\circ} \mathrm{F}\) and a moist air stream exits with a volumetric flow rate of \(1000 \mathrm{ft}^{3} / \mathrm{min}\) at \(90^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and a relative humidity of \(40 \%\). Neglecting kinetic and potential energy effects, determine (a) the rate water is injected, in \(\mathrm{lb} / \mathrm{min}\). (b) the rate of heat transfer to the moist air, in Btu/h.

A device is being designed to separate a natural gas having a molar analysis of \(94 \% \mathrm{CH}_{4}\) and \(6 \% \mathrm{C}_{2} \mathrm{H}_{6}\) into components. The device will receive natural gas at \(20^{\circ} \mathrm{C}\), \(1 \mathrm{~atm}\) with a volumetric flow rate of \(100 \mathrm{~m}^{3} / \mathrm{s}\). Separate streams of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) will exit, each at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The device will operate isothermally at \(20^{\circ} \mathrm{C}\). Ignoring kinetic and potential energy effects and assuming ideal gas behavior, determine the minimum theoretical work input required at steady state, in kW.

Helium at \(400 \mathrm{~K}, 1\) bar enters an insulated mixing chamber operating at steady state, where it mixes with argon entering at \(300 \mathrm{~K}, 1\) bar. The mixture exits at a pressure of 1 bar. If the argon mass flow rate is \(x\) times that of helium, plot versus \(x\) (a) the exit temperature, in \(\mathrm{K}\). (b) the rate of exergy destruction within the chamber, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of helium entering. Kinetic and potential energy effects can be ignored. Let \(T_{0}=300 \mathrm{~K}\).

Using the ideal gas model with constant specific heats, determine the mixture temperature, in \(\mathrm{K}\), for each of two cases: (a) Initially, \(0.6 \mathrm{kmol}\) of \(\mathrm{O}_{2}\) at \(500 \mathrm{~K}\) is separated by a partition from \(0.4 \mathrm{kmol}\) of \(\mathrm{H}_{2}\) at \(300 \mathrm{~K}\) in a rigid insulated vessel. The partition is removed and the gases mix to obtain a final equilibrium state. (b) Oxygen \(\left(\mathrm{O}_{2}\right)\) at \(500 \mathrm{~K}\) and a molar flow rate of \(0.6 \mathrm{kmol} / \mathrm{s}\) enters an insulated control volume operating at steady state and mixes with \(\mathrm{H}_{2}\) entering as a separate stream at \(300 \mathrm{~K}\) and a molar flow rate of \(0.4 \mathrm{kmol} / \mathrm{s}\). A single mixed stream exits. Kinetic and potential energy effects can be ignored.

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