Question

A single-pass heat exchanger is to be designed to heat \(100,000 \mathrm{lbm}\) of water in an hour from \(60^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) by condensation of water vapor at \(230^{\circ} \mathrm{F}\) on the shell side. Each tube has an inner diameter of \(1.2\) in and a wall thickness of \(0.12\) in. The inner surface convection heat transfer coefficient is $480 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$ and the outer surface convection heat transfer coefficient is $2000 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. If the inlet velocity of water $\left(c_{p c}=1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right.\( and \)\left.\rho=62.3 \mathrm{lbm} / \mathrm{ft}^{3}\right)$ on the tube side is \(4 \mathrm{ft} / \mathrm{s}\), determine the required number of tubes and length of tubes. Assume the thermal resistance of the tubes is negligible.

Short answer

The designed single-pass heat exchanger will require 139 tubes with a length of 91.50 ft per tube to heat the given water flow from 60°F to 100°F using condensing water vapor.

Step-by-step solution

Calculate the heat transfer needed

First, we need to calculate the heat transfer that needs to occur within the heat exchanger to achieve the desired temperature change in the water. For this, we will use the formula: $$Q = mc_p\Delta T$$ Where: - \(Q\) - Heat transfer [Btu/h] - \(m\) - Mass flow rate of water [lbm/h] - \(c_p\) - Specific heat capacity of water [Btu/lbm °F] - \(\Delta T\) - Temperature difference [°F] Given values: - Mass flow rate: \(100,000\) lbm/h - Initial temperature: \(60\) °F - Final temperature: \(100\) °F - Specific heat capacity: \(1\) Btu/lbm °F (constant) Now, we can calculate the heat transfer needed: $$Q = (100,000\, \mathrm{lbm/h})(1\, \mathrm{Btu/lbm\,°F})(100\, ^\circ\mathrm{F} - 60\,^\circ\mathrm{F}) = 4,000,000\, \mathrm{Btu/h}$$

Calculate the heat transfer rate per tube

Now we need to determine the heat transfer rate per tube for both the tube and shell sides. For this, we will use the formula: $$Q_\mathrm{tube}=h_\mathrm{i} A_\mathrm{i} (\Delta T_\mathrm{i})$$ \(Q_\mathrm{tube}\) is the heat transfer rate per tube [Btu/h] \(h_\mathrm{i}\) is the inner convection heat transfer coefficient [Btu/h·ft²·°F] \(A_\mathrm{i}\) is the inner heat transfer area per tube [ft²] \(\Delta T_\mathrm{i}\) is the temperature difference between the two fluids on the tube side [°F] Given values: - Inner surface convection heat transfer coefficient: \(480\, \mathrm{Btu/h\cdot ft^2\cdot °F}\) - Outer surface convection heat transfer coefficient: \(2000\, \mathrm{Btu/h\cdot ft^2\cdot °F}\) - Inlet velocity of water on the tube side: \(4\, \mathrm{ft/s}\) - \(\Delta T_\mathrm{i}\) = \(230\,^\circ\mathrm{F} - 100\,^\circ\mathrm{F} = 130\,^\circ\mathrm{F}\) The calculation for the heat transfer area per tube (A) should consider the tube inner diameter and the length of tubes. However, since the required number of tubes would be determined based on their velocities, and length is unknown yet, we don't need to calculate the actual inner tube surface area at this point. Now, we need to find the volumetric flow rate (\(\dot{V}\)) per tube: $$\dot{V}=\frac{\rho \cdot \dot{m}}{\text{Number of tubes}}$$ Also: $$A_i = \pi D_i L$$ $$\dot{V}=\frac{\pi D_i^2 v}{4}$$ $$\text{Number of tubes} = \frac{4 \cdot \dot{V} \cdot L}{\pi D_i^2v}$$ Now suppose at any time, \(n\) tubes have temperature \(T\), so heat transfer is taking place: $$Q_\mathrm{total} = Q_\mathrm{tube} \cdot \text{Number of tubes}$$ $$ \Rightarrow Q_\mathrm{tube} = \frac{Q_\mathrm{total}\cdot \pi D_i^2v}{4 \cdot \dot{V} \cdot L}$$ Finally, we can substitute the given values and solve for \(\text{Number of tubes}\) and the length \(L\) of the tubes: $$480 \cdot \frac{Q_\mathrm{total}\cdot \pi (1.2\times 10^{-1})^2(4)}{4 \cdot \dot{V} \cdot L} = 130$$ Now, \(L\) is a variable at any stage of the tube, and ultimately: $$\frac{4,000,000\cdot \pi (1.2\times 10^{-1})^2(4)}{4 \cdot \dot{V} \cdot L} = 130$$ Now we have the main equation to find the variable \(\dot{V}\) and ultimately the number of tubes and the length \(L\).

Calculate the volumetric flow rate

Given the inlet velocity of water (\(v = 4\, \mathrm{ft/s}\)) and the known tube side density of water (\(\rho=62.3\, \mathrm{lbm/ft^3}\)), we can calculate the volumetric flow rate (\(\dot{V}\)) for water on the tube side: $$\dot{V} = \frac{\dot{m}}{\rho} = \frac{100,000\, \mathrm{lbm/h}}{62.3\, \mathrm{lbm/ft^3}} = 1,\!605.47\, \mathrm{ft^3/h}=0.44624\, \mathrm{ft^3/min}$$

Determine the required number of tubes and length of tubes

Now that we have the volumetric flow rate, we will revisit the main equation and determine the required number of tubes and the length of the tubes: $$\frac{4,000,000\, \pi (1.2\times 10^{-1})^2(4)}{4 \cdot 0.44624 \cdot L} = 130$$ We can now solve the equation for \(L\): $$L = \frac{4,000,000\, \pi (1.2\times 10^{-1})^2(4)}{4 \cdot 0.44624 \cdot 130} = 91.503\, \mathrm{ft}$$ As the length \(L\) is given only for four digits of precision, we'll round it up to \(91.50\, \mathrm{ft}\) for the final answer. Now, we can find the number of tubes needed using the L and volumetric flow rate: $$\text{Number of tubes} = \frac{4 \cdot \dot{V} \cdot L}{\pi D_i^2v} = \frac{4 \cdot 0.44624\cdot 91.5}{\pi (1.2\times 10^{-1})^2(4)}$$ The result is: $$\text{Number of tubes} = 138.87$$ Since we cannot have partial tubes, we round up the number of tubes to 139. #Final Answers# The required number of tubes for the heat exchanger design is 139, and the length of each tube is 91.50 ft.