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Chapter 11: Problem 182

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

In a textile manufacturing plant, the waste dyeing water \(\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to preheat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(15^{\circ} \mathrm{C}\) at the same flow rate in a double-pipe counter-flow heat exchanger. The heat transfer surface area of the heat exchanger is \(1.65 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is \(625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rate of heat transfer in the heat exchanger is \(35 \mathrm{~kW}\), determine the outlet temperature and the mass flow rate of each fluid stream.

A 1-shell-pass and 8-tube-passes heat exchanger is used to heat glycerin \(\left(c_{p}=0.60 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(65^{\circ} \mathrm{F}\) to \(140^{\circ} \mathrm{F}\) by hot water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) that enters the thinwalled \(0.5\)-in-diameter tubes at \(175^{\circ} \mathrm{F}\) and leaves at \(120^{\circ} \mathrm{F}\). The total length of the tubes in the heat exchanger is \(500 \mathrm{ft}\). The convection heat transfer coefficient is \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the glycerin (shell) side and \(50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water (tube) side. Determine the rate of heat transfer in the heat exchanger \((a)\) before any fouling occurs and \((b)\) after fouling with a fouling factor of \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on the outer surfaces of the tubes.

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} /\right.\) \(\mathrm{kg} \cdot \mathrm{K})\) to \(\mathrm{cool}\) the engine coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(30^{\circ} \mathrm{C}\) air through this radiator at a rate of \(10 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, what is the number of transfer units (NTU) of this radiator? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5

Oil in an engine is being cooled by air in a cross-flow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). Determine \((a)\) the heat transfer effectiveness and \((b)\) the outlet temperature of the oil.

Water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) is to be heated by solarheated hot air \(\left(c_{p}=0.24 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) in a double- pipe counterflow heat exchanger. Air enters the heat exchanger at \(190^{\circ} \mathrm{F}\) at a rate of \(0.7 \mathrm{lbm} / \mathrm{s}\) and leaves at \(135^{\circ} \mathrm{F}\). Water enters at \(70^{\circ} \mathrm{F}\) at a rate of \(0.35 \mathrm{lbm} / \mathrm{s}\). The overall heat transfer coefficient based on the inner side of the tube is given to be \(20 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine the length of the tube required for a tube internal diameter of \(0.5 \mathrm{in}\).
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