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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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Most popular questions from this chapter

Consider a heat exchanger that has an NTU of \(0.1\). Someone proposes to triple the size of the heat exchanger and thus triple the NTU to \(0.3\) in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?

Hot exhaust gases of a stationary diesel engine are to be used to generate steam in an evaporator. Exhaust gases \(\left(c_{p}=1051 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enter the heat exchanger at \(550^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) while water enters as saturated liquid and evaporates at \(200^{\circ} \mathrm{C}\left(h_{f g}=1941 \mathrm{~kJ} / \mathrm{kg}\right)\). The heat transfer surface area of the heat exchanger based on water side is \(0.5 \mathrm{~m}^{2}\) and overall heat transfer coefficient is \(1780 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer, the exit temperature of exhaust gases, and the rate of evaporation of water.

Consider an oil-to-oil double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold oil enters at \(20^{\circ} \mathrm{C}\) and leaves at \(55^{\circ} \mathrm{C}\), while the hot oil enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counter-flow heat exchanger? Why? Assuming the mass flow rates of both fluids to be identical, determine the effectiveness of this heat exchanger.

A single-pass cross-flow heat exchanger uses hot air (mixed) to heat water (unmixed), flowing with a mass flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), from \(30^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). The hot air enters and exits the heat exchanger at \(220^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the required surface area.

A shell-and-tube heat exchanger with 1-shell pass and 14-tube passes is used to heat water in the tubes with geothermal steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) on the shell side. The tubes are thin-walled and have a diameter of \(2.4 \mathrm{~cm}\) and length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(3.9 \mathrm{~kg} / \mathrm{s}\). If the temperature difference between the two fluids at the exit is \(46^{\circ} \mathrm{C}\), determine (a) the rate of heat transfer, \((b)\) the rate of condensation of steam, and \((c)\) the overall heat transfer coefficient.
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