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Steam is to be condensed on the shell side of a 2-shell-passes and 8-tube- passes condenser, with 20 tubes in each pass. Cooling water enters the tubes a rate of \(2 \mathrm{~kg} / \mathrm{s}\). If the heat transfer area is \(14 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is \(1800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the effectiveness of this condenser is (a) \(0.70\) (b) \(0.80\) (c) \(0.90\) (d) \(0.95\) (e) \(1.0\)

Short Answer

Expert verified
Answer: The effectiveness of the given condenser is approximately 0.90.

Step by step solution

01

Write down the given parameters

The given parameters are: - Number of shell passes: 2 - Number of tube passes: 8 - Number of tubes per pass: 20 - Mass flow rate of cooling water: \(m = 2 \mathrm{~kg} / \mathrm{s}\) - Heat transfer area: \(A = 14 \mathrm{~m}^{2}\) - Overall heat transfer coefficient: \(U = 1800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
02

Calculate the theoretical temperature difference

We will use the mass flow rate of cooling water and its specific heat to calculate the theoretical temperature difference. Assuming the specific heat of water is \(c_p = 4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), the theoretical temperature difference for a condenser can be calculated with the formula: \(\Delta T_{th} = \dfrac{Q_{th}}{mc_p}\) where \(Q_{th}\) is the heat transfer rate and \(\Delta T_{th}\) is the theoretical temperature difference. We will calculate \(Q_{th}\) in the next step.
03

Calculate the heat transfer rate

Use the formula \(Q=UA\Delta T_{lm}\) to calculate the heat transfer rate, where \(Q\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient (\(U = 1800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), \(A\) is the heat transfer area (\(A = 14 \mathrm{~m}^{2}\)), and \(\Delta T_{lm}\) is the logarithmic mean temperature difference. We can assume \(\Delta T_{lm} \approx \Delta T_{th}\) for this calculation. \(Q = 1800 \cdot 14 \cdot \Delta T_{th}\)
04

Calculate the actual temperature difference

Now that we have the heat transfer rate \(Q\), we can use the formula \(\Delta T_{act} = \dfrac{Q}{mc_p}\) to calculate the actual temperature difference, where \(m = 2 \mathrm{~kg} / \mathrm{s}\) and \(c_p = 4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). \(\Delta T_{act} = \dfrac{Q}{(2)(4187)}\)
05

Calculate the effectiveness of the condenser

Finally, we can calculate the effectiveness of the condenser by dividing the actual temperature difference by the theoretical temperature difference: \(\text{Effectiveness} = \dfrac{\Delta T_{act}}{\Delta T_{th}}\). Plug in the values from Steps 3 and 4: \(\text{Effectiveness} = \dfrac{\dfrac{1800 \cdot 14 \cdot \Delta T_{th}}{(2)(4187)}}{\Delta T_{th}} = \dfrac{1800 \cdot 14}{(2)(4187)} \approx 0.90\) The effectiveness of the condenser is approximately 0.90, so the correct answer is (c) \(0.90\).

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Key Concepts

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

Condenser Effectiveness
Understanding condenser effectiveness is key in evaluating how well a condenser performs in transferring heat. It measures the actual heat transfer in comparison to the theoretical maximum possible. This is expressed as a ratio:
  • High effectiveness means the condenser is transferring heat close to its maximum capability.
  • A perfect effectiveness of 1.0 is rare in practical systems, where losses occur.
To find the effectiveness, we use the formula:\[\text{Effectiveness} = \frac{\Delta T_{\text{actual}}}{\Delta T_{\text{theoretical}}}\]This helps highlight any inefficiencies and design improvements needed.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient (\(U\)) is crucial in determining how well heat is transferred across different mediums. It considers factors like material, surface area, and temperature differences.
  • Measured in \(\text{W}/\text{m}^2 \cdot \text{K}\), it combines convection and conduction resistances.
  • Higher values imply efficient heat transfer across the system.
  • It is essential in designing systems to ensure they meet the required heat transfer needs.
Logarithmic Mean Temperature Difference
The logarithmic mean temperature difference (\(\Delta T_{\text{lm}}\)) allows for the average temperature difference to be calculated in heat exchangers.
  • It accounts for variations in temperature difference across the device.
  • This method is favored because it more accurately reflects the varying conditions compared to simple averages.
The formula to find \(\Delta T_{\text{lm}}\) is:\[\Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}\]Where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences at each end of the exchanger.
Mass Flow Rate
Mass flow rate (\(m\)) is the amount of mass moving through a surface per time unit, typically measured in \(\text{kg}/\text{s}\).
  • In systems involving heat transfer, mass flow rate influences how much heat is absorbed or released.
  • A higher mass flow rate can improve heat transfer, but also increases energy consumption.
  • Balance must be struck to optimize both efficiency and performance.
Thermodynamic Calculations
Thermodynamic calculations are essential for understanding energy transformations in systems.
  • Involves the use of equations to predict how changes in conditions affect performance, efficiency, and design.
  • Key parameters include temperature, pressure, and specific heat.
  • These calculations ensure the system meets specifications and operates within safe limits.
Mastering these calculations helps in optimizing thermal systems for maximum efficiency.

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Most popular questions from this chapter

Steam is to be condensed on the shell side of a 1 -shellpass and 8-tube-passes condenser, with 50 tubes in each pass, at \(30^{\circ} \mathrm{C}\left(h_{f g}=2431 \mathrm{~kJ} / \mathrm{kg}\right)\). Cooling water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(15^{\circ} \mathrm{C}\) at a rate of \(1800 \mathrm{~kg} / \mathrm{h}\). The tubes are thin-walled, and have a diameter of \(1.5 \mathrm{~cm}\) and length of \(2 \mathrm{~m}\) per pass. If the overall heat transfer coefficient is \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the rate of heat transfer and \((b)\) the rate of condensation of steam.

An air-cooled condenser is used to condense isobutane in a binary geothermal power plant. The isobutane is condensed at \(85^{\circ} \mathrm{C}\) by air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(2.4 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1.25 \mathrm{~m}^{2}\), respectively. The outlet temperature of air is (a) \(45.4^{\circ} \mathrm{C}\) (b) \(40.9^{\circ} \mathrm{C}\) (c) \(37.5^{\circ} \mathrm{C}\) (d) \(34.2^{\circ} \mathrm{C}\) (e) \(31.7^{\circ} \mathrm{C}\)

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in the shell side of a 1 -shell and 2-tube heat exchanger with a surface area of \(0.5 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(0.5 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(15^{\circ} \mathrm{C}\), determine the outlet temperature of the cold water and the heat transfer rate for the heat exchanger.

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hotand cold-water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same.

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