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.

Step by step solution

01

Calculate the overall heat transfer coefficient (U) for the heat exchanger before fouling.

Before fouling, the overall heat transfer coefficient U can be calculated using the convection heat transfer coefficients of the glycerin (shell) side and the water (tube) side: \(\frac{1}{U} = \frac{1}{h_{shell}} + \frac{1}{h_{tube}}\) Given: \(h_{shell} = 4 \ \mathrm{Btu / h \cdot ft^2 \cdot ^\circ F}\) \(h_{tube} = 50 \ \mathrm{Btu / h \cdot ft^2 \cdot ^\circ F}\) Let's plug in the values and solve for U: \(\frac{1}{U} = \frac{1}{4} + \frac{1}{50}\) \(=> U \approx 3.846 \ \mathrm{Btu / h \cdot ft^2 \cdot ^\circ F}\) The overall heat transfer coefficient before fouling is approximately 3.846 Btu/h·ft²·°F.

02

Calculate the heat transfer area (A) for the heat exchanger.

We are given the total length of tubes in the heat exchanger, 500 ft, and the diameter of each tube, 0.5 in (converted to ft: 0.04167 ft). To determine the total heat transfer area A, we need to multiply the total length of the tubes by their perimeter, which can be calculated by multiplying the diameter by pi: \(A = L \cdot \text{(Perimeter)}\) We can calculate the heat transfer area: \(A = 500 \cdot (0.04167 \cdot \pi) \approx 65.45 \ \mathrm{ft^2}\) The heat transfer area of the heat exchanger is approximately 65.45 ft².

03

Calculate the rate of heat transfer (Q) in the heat exchanger before fouling.

Now that we have the overall heat transfer coefficient U and the heat transfer area A, we can calculate the rate of heat transfer Q using the temperature differences between the fluids: \(Q = U \cdot A \cdot (T_{hot,inlet} - T_{cold,outlet})\) Given: \(T_{hot,inlet} = 175^{\circ} F\) \(T_{cold,outlet} = 140^{\circ} F\) Plugging the values into the equation: \(Q = 3.846 \cdot 65.45 \cdot (175 - 140) \approx 13896 \ \mathrm{Btu / h}\) The rate of heat transfer in the heat exchanger before fouling is approximately 13,896 Btu/h.

04

Calculate the overall heat transfer coefficient (U') for the heat exchanger after fouling.

To calculate the overall heat transfer coefficient after fouling, we need to consider the fouling factor (Rf): \(\frac{1}{U'} = \frac{1}{h_{shell}'} + \frac{1}{h_{tube}'} + R_f\) Given: \(fouling \ factor = 0.002 \ \mathrm{h \cdot ft^2 \cdot ^\circ F / Btu}\) Let's plug in the values: \(\frac{1}{U'} = \frac{1}{4} + \frac{1}{50} + 0.002\) \(=> U' = 3.437 \ \mathrm{Btu / h \cdot ft^2 \cdot ^\circ F}\) The overall heat transfer coefficient after fouling with a fouling factor is approximately 3.437 Btu/h·ft²·°F.

05

Calculate the rate of heat transfer (Q') in the heat exchanger after fouling.

Using the overall heat transfer coefficient with the fouling factor included (U') and the heat transfer area A, we can calculate the rate of heat transfer after fouling: \(Q' = U' \cdot A \cdot (T_{hot,inlet} - T_{cold,outlet})\) Plugging the values into the equation: \(Q' = 3.437 \cdot 65.45 \cdot (175 - 140) \approx 12336 \ \mathrm{Btu / h}\) The rate of heat transfer in the heat exchanger after fouling is approximately 12,336 Btu/h. In summary: (a) Before fouling, the rate of heat transfer in the heat exchanger is approximately 13,896 Btu/h. (b) After fouling with a fouling factor of 0.002 h·ft²·°F/Btu, the rate of heat transfer in the heat exchanger is approximately 12,336 Btu/h.

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Key Concepts

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

Overall Heat Transfer Coefficient

The overall heat transfer coefficient (\textbf{U}) is a measure that helps determine the heat exchange effectiveness between fluid streams separated by a solid barrier. It is a combined measure of the heat transfer by conduction and convection and includes the thermal resistance of the solid wall and the resistance to convection on each side of the wall.

In the example problem, the calculation of the overall heat transfer coefficient before fouling takes into account the heat transfer coefficients of both the glycerin and water sides of the exchanger. Using the formula \( \frac{1}{U} = \frac{1}{h_{shell}} + \frac{1}{h_{tube}} \) and the given values (where \( h_{shell} \) and \( h_{tube} \) represent the convection heat transfer coefficients on the shell and tube sides respectively), students can solve for \( U \) to understand the heat exchanger's performance under ideal conditions.

Fouling Factor Impact

Fouling in heat exchangers refers to the accumulation of unwanted materials on the heat transfer surfaces, which impedes the flow of heat. The fouling factor (\textbf{Rf}) quantifies this resistance. It is a reflector of how much the heat transfer is affected by this accumulation.

To comprehend the impact, the overall heat transfer coefficient (\textbf{U'}) after fouling is recalculated, incorporating the fouling factor. The updated formula, \( \frac{1}{U'} = \frac{1}{h_{shell}'} + \frac{1}{h_{tube}'} + R_f \) where \( R_f \) is the fouling factor, shows decreased efficiency. From the step by step solution, after considering the fouling factor, \( U' \) is lower than \( U \) signifying that fouling has caused the thermal performance of the heat exchanger to deteriorate.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient (\textbf{h}) is critical in calculating the overall heat transfer coefficient (\textbf{U}). It represents the amount of heat transferred per unit area, time, and temperature difference due to fluid motion.

In the given problem, \( h \) has different values for the shell and tube sides, reflecting the distinct characteristics of the fluids and their flow within the heat exchanger. The larger \( h \) value on the water (tube) side indicates that water more efficiently transfers heat compared to glycerin (on the shell side), likely due to the higher flow rates and better thermal properties. This discrepancy illustrates why each fluid's convection heat transfer coefficient must be carefully evaluated to accurately predict heat exchanger performance.