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Chapter 11: Problem 67

A shell-and-tube heat exchanger is used for heating \(10 \mathrm{~kg} / \mathrm{s}\) of oil \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(46^{\circ} \mathrm{C}\). The heat exchanger has 1 -shell pass and 6-tube passes. Water enters the shell side at \(80^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). The overall heat transfer coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the rate of heat transfer and the heat transfer area.

Short Answer

Expert verified
Based on the given information and calculations, the rate of heat transfer (Q) in the shell-and-tube heat exchanger is approximately 420,000 W, and the required heat transfer area (A) is approximately 14.19 m².

Step by step solution

01

Calculate the heat transfer rate (Q)

To compute the heat transfer rate, apply the formula \(Q=m_{o} \cdot c_{p} \cdot \Delta T_{o}\), where \(\Delta T_{o}\) is the temperature difference of the oil between the inlet and outlet. \(\Delta T_{o} = T_{2} - T_{1} = 46^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 21^{\circ} \mathrm{C}\) \(Q = 10\ \mathrm{kg/s} \cdot 2.0\ \mathrm{kJ/(kg\cdot K)} \cdot 21\ \mathrm{K}\) Since \(1\ \mathrm{kJ/s} = 1000\ \mathrm{W}\), we can convert the units: \(Q = 10 \cdot 2.0 \cdot 21 \cdot 1000\ \mathrm{W} = 420,000\ \mathrm{W}\)
02

Calculate temperature differences (ΔT)

Calculate the temperature difference between inlet and outlet temperatures on both sides: \(\Delta T_{1} = T_{3} - T_{1} = 80^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 55^{\circ} \mathrm{C}\) \(\Delta T_{2} = T_{4} - T_{2} = 60^{\circ} \mathrm{C} - 46^{\circ} \mathrm{C} = 14^{\circ} \mathrm{C}\)
03

Calculate Log Mean Temperature Difference (LMTD)

To compute the LMTD, use the formula \(\mathrm{LMTD} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\): \(\mathrm{LMTD} = \frac{55^{\circ} \mathrm{C} - 14^{\circ} \mathrm{C}}{\ln(\frac{55^{\circ} \mathrm{C}}{14^{\circ} \mathrm{C}})} = \frac{41}{\ln(\frac{55}{14})} \approx 29.59^{\circ} \mathrm{C}\)
04

Calculate the heat transfer area (A)

To calculate the heat transfer area, use the formula \(A=\frac{Q}{U \cdot \mathrm{LMTD}}\): \(A = \frac{420,000\ \mathrm{W}}{1000\ \mathrm{W/m^{2}\cdot K} \cdot 29.59^{\circ} \mathrm{C}} \approx 14.19\ \mathrm{m^{2}}\) The rate of heat transfer is approximately \(420,000\ \mathrm{W}\), and the heat transfer area required is approximately \(14.19\ \mathrm{m^{2}}\).

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Most popular questions from this chapter

A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

Glycerin \(\left(c_{p}=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(20^{\circ} \mathrm{C}\) and \(0.3 \mathrm{~kg} / \mathrm{s}\) is to be heated by ethylene glycol \(\left(c_{p}=2500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(60^{\circ} \mathrm{C}\) and the same mass flow rate in a thin-walled double-pipe parallel-flow heat exchanger. If the overall heat transfer coefficient is \(380 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area is \(5.3 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer and \((b)\) the outlet temperatures of the glycerin and the glycol.

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.

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) to 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, the effectiveness of the radiator is \(0.4\). Determine \((a)\) the outlet temperature of the air and (b) the rate of heat transfer between the two fluids.

The cardiovascular counter-current heat exchanger mechanism is to warm venous blood from \(28^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) at a mass flow rate of \(2 \mathrm{~g} / \mathrm{s}\). The artery inflow temperature is \(37^{\circ} \mathrm{C}\) at a mass flow rate of \(5 \mathrm{~g} / \mathrm{s}\). The average diameter of the vein is \(5 \mathrm{~cm}\) and the overall heat transfer coefficient is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the overall blood vessel length needed to warm the venous blood to \(35^{\circ} \mathrm{C}\) if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).
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