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Chapter 11: Problem 151

A shell-and-tube heat exchanger is used for cooling \(47 \mathrm{~kg} / \mathrm{s}\) of a process stream flowing through the tubes from \(160^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). This heat exchanger has a total of 100 identical tubes, each with an inside diameter of \(2.5 \mathrm{~cm}\) and negligible wall thickness. The average properties of the process stream are: \(\rho=950 \mathrm{~kg} / \mathrm{m}^{3}, k=0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=3.5 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and \(\mu=2.0 \mathrm{mPa} \cdot \mathrm{s}\). The coolant stream is water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a flow rate of \(66 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(10^{\circ} \mathrm{C}\), which yields an average shell-side heat transfer coefficient of \(4.0 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the tube length if the heat exchanger has \((a)\) a 1 -shell pass and a 1 -tube pass and (b) a 1-shell pass and 4-tube passes.

Short Answer

Expert verified
Answer: The tube length for the heat exchanger with a 1-shell pass and 1-tube pass is 2.63 meters, while the tube length for the heat exchanger with a 1-shell pass and 4-tube passes is 4.38 meters.

Step by step solution

01

Calculate the heat transfer

First, we need to calculate the heat transfer required for the process stream to cool down from 160°C to 100°C. For that, we use the formula: \(Q = \dot{m}_p c_{p, p} \delta T_p\) Where: \(Q\) = heat transfer (kW) \(\dot{m}_p\) = mass flow rate of process stream (kg/s) \(c_{p, p}\) = specific heat capacity of process stream (kJ/kg·K) \(\delta T_p\) = temperature difference of process stream (\(^{\circ}\mathrm{C}\)) Now, we plug in the given values: \(Q=47 \mathrm{~kg/s} \times 3.5 \mathrm{~kJ/(kg\cdot K)} \times (160-100)^{\circ} \mathrm{C}\) \(Q=12390 \mathrm{~kW}\)
02

Calculate the tube-side heat transfer coefficient

Now we need to calculate the tube-side heat transfer coefficient (\(h_t\)) using the Sieder-Tate equation: \(h_t = \frac{k_t}{D_i} \times Nu\) Where: \(k_t\) = thermal conductivity of the process stream (W/m·K) \(D_i\) = inside diameter of the tube (m) \(Nu\) = Nusselt number, calculated using the Sieder-Tate equation: \(Nu = 1.86 \times Re^{1/3} \times Pr^{1/3} \times \frac{L}{D_i}^{1/3}\) First, we need to calculate the Reynolds number (\(Re\)) and Prandtl number (\(Pr\)): \(Re=\frac{\rho \cdot V \cdot D_i}{\mu}\) \(Pr=\frac{c_{p, p} \cdot \mu}{k_t}\) The tube-side velocity (\(V\)) can be calculated using the process flow rate and the number of tubes: \(V = \frac{\dot{m}_p}{100 \times \rho \times \frac{\pi}{4} D_i^2}\) Now, we plug in the given values: \(Re=\frac{950 \mathrm{~kg/m^3} \times (47 \mathrm{~kg/s}) (2.5 \times 10^{-2} \mathrm{~m})}{100 \times 2.0 \times 10^{-3} \mathrm{~kg/(m\cdot s)} \times \frac{\pi}{4} (2.5 \times 10^{-2} \mathrm{~m})^2}\) \(Re=3359.03\) \(Pr=\frac{3.5 \times 10^3 \mathrm{~J/(kg\cdot K)} \times 2.0 \times 10^{-3} \mathrm{~kg/(m\cdot s)}}{0.5 \mathrm{~W/(m\cdot K)}}\) \(Pr=14\) Now we can calculate \(Nu\), using \(L\) in meters as the length of the tube: \(Nu = 1.86 \times Re^{1/3} \times Pr^{1/3} \times \frac{L}{D_i}^{1/3}\) Which can be rewritten as: \(L = D_i \times \left(\frac{Nu}{1.86 \times Re^{1/3} \times Pr^{1/3}}\right)^{3/1}\) Next, let's calculate the heat transfer coefficients for both scenarios: (a) 1-shell pass and 1-tube pass: \(Nu = 1\) \(L_{1} = (2.5 \times 10^{-2} \mathrm{~m}) \times \left(\frac{1}{1.86 \times (3359.03)^{1/3} \times (14)^{1/3}}\right)^{3/1}\) \(L_{1} = 2.63 \mathrm{~m}\) (b) 1-shell pass and 4-tube passes: \(Nu = 4\) \(L_{4} = (2.5 \times 10^{-2} \mathrm{~m}) \times \left(\frac{4}{1.86 \times (3359.03)^{1/3} \times (14)^{1/3}}\right)^{3/1}\) \(L_{4} = 4.38 \mathrm{~m}\) Therefore, the tube length for the heat exchanger with a 1-shell pass and 1-tube pass is 2.63 meters, while the tube length for the heat exchanger with a 1-shell pass and 4-tube passes is 4.38 meters.

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Key Concepts

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

Shell-and-Tube Heat Exchanger
A shell-and-tube heat exchanger is a commonly used heat transfer device that effectively cools or heats fluids by allowing them to flow through a series of tubes enclosed within a shell. The fluid flowing inside the tubes is generally called the tube-side fluid, and the fluid outside the tubes but inside the shell is referred to as the shell-side fluid.

The design of a shell-and-tube heat exchanger involves several components such as baffles, which direct the shell-side fluid flow and improve heat transfer, and tube bundles, which contain multiple tubes to maximize the surface area for heat exchange. In our example, we consider an exchanger cooling a process stream using water as the coolant on the shell side. One critical aspect of design is determining the most efficient tube length, which ensures that the desired temperature change is achieved without excessive construction costs or pressure losses.
Heat Transfer Calculations
Heat transfer calculations are vital in determining the overall performance of a heat exchanger. These calculations take into account the amount of heat energy that needs to be transferred between the two fluids to achieve the required temperature changes. The basic formula used in these calculations is the heat transfer equation,

\(Q = \dot{m}_p c_{p, p} \Delta T_p\),
where \(Q\) represents the heat transfer rate, \(\dot{m}_p\) is the mass flow rate of the process stream, \(c_{p, p}\) is the specific heat capacity of the process stream, and \(\Delta T_p\) is the temperature difference of the process stream. The result gives us a quantitative measure of the cooling capacity required by the exchanger and is the starting point for further design decisions.
Nusselt Number
The Nusselt number (Nu) is a dimensionless number representing the ratio of convective to conductive heat transfer at a boundary in a fluid medium. In essence, it provides a measure of the relative effectiveness of these two mechanisms in transferring heat.

Higher Nusselt numbers indicate greater convective heat transfer relative to conduction, which typically happens with turbulent flow or when certain geometries enhance fluid mixing. When designing heat exchangers, engineers use the Nusselt number to compute the convective heat transfer coefficient. The formula for Nu can vary depending on the flow conditions and physical properties of the fluid, but a common way to calculate it for a tube flow, as shown in our example, is using the Sieder-Tate correlation. This correlation factors in the Reynolds and Prandtl numbers of the fluid, which reflect its flow dynamics and thermal properties, respectively.
Reynolds Number
The Reynolds number (Re) is another dimensionless number that is used to predict the flow regime of the fluid within the tubes of the heat exchanger. It fundamentally describes whether the flow is laminar or turbulent.

The Reynolds number is calculated using the formula:

\(Re = \frac{\rho \cdot V \cdot D_i}{\mu}\),
where \(\rho\) is the fluid density, \(V\) is the velocity of the fluid inside the tube, \(D_i\) is the inside diameter of the tube, and \(\mu\) is the dynamic viscosity of the fluid. Laminar flow typically occurs at low Reynolds numbers and is characterized by smooth, orderly fluid motion, whereas turbulent flow occurs at high Reynolds numbers and involves chaotic, irregular fluid motion, which enhances heat transfer.
Prandtl Number
The Prandtl number (Pr) is a dimensionless number that relates the momentum diffusivity (kinematic viscosity) to the thermal diffusivity of a fluid. It essentially compares the relative thickness of the velocity boundary layer to the thermal boundary layer at a wall.

It is defined as:

\(Pr = \frac{c_{p} \mu}{k}\),
where \(c_p\) is the specific heat of the fluid at constant pressure, \(\mu\) is the dynamic viscosity, and \(k\) is the thermal conductivity of the fluid. The Prandtl number gives insight into the behavior of the fluid's heat transfer; fluids with high Pr numbers have a relatively low thermal diffusivity resulting in a thicker thermal boundary compared to their velocity boundary.
Heat Transfer Coefficient
The heat transfer coefficient is a critical parameter in calculating the heat exchange in a heat exchanger. It relates to the ease with which heat is transferred between the fluid and the surface of the tubes. In our example, the tube-side heat transfer coefficient is estimated using the Sieder-Tate equation that includes the Nusselt number.

As the equation shows, the heat transfer coefficient (h_t) is a function of the tube's thermal conductivity (k_t), the inside diameter (D_i), and the Nusselt number (Nu). With a higher heat transfer coefficient, more heat can be transferred per unit area at a given temperature difference, which is desirable for efficient heat exchanger operation. When both turbulent flow (high Reynolds number) and good thermal properties (high Prandtl number) are present, larger heat transfer coefficients are achieved, optimizing the design and efficiency of the heat exchanger.

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Most popular questions from this chapter

In a parallel-flow heat exchanger, the NTU is calculated to be \(2.5\). The lowest possible effectiveness for this heat exchanger is (a) \(10 \%\) (b) \(27 \%\) (c) \(41 \%\) (d) \(50 \%\) (e) \(92 \%\)

Consider a recuperative cross flow heat exchanger (both fluids unmixed) used in a gas turbine system that carries the exhaust gases at a flow rate of \(7.5 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(500^{\circ} \mathrm{C}\). The air initially at \(30^{\circ} \mathrm{C}\) and flowing at a rate of \(15 \mathrm{~kg} / \mathrm{s}\) is to be heated in the recuperator. The convective heat transfer coefficients on the exhaust gas and air side are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Due to long term use of the gas turbine the recuperative heat exchanger is subject to fouling on both gas and air side that offers a resistance of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) each. Take the properties of exhaust gas to be the same as that of air \(\left(c_{p}=1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). If the exit temperature of the exhaust gas is \(320^{\circ} \mathrm{C}\) determine \((a)\) if the air could be heated to a temperature of \(150^{\circ} \mathrm{C}(b)\) the area of heat exchanger \((c)\) if the answer to part (a) is no, then determine what should be the air mass flow rate in order to attain the desired exit temperature of \(150^{\circ} \mathrm{C}\) and \((d)\) plot variation of the exit air temperature over a temperature range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\) with air mass flow rate assuming all the other conditions remain the same.

The condenser of a room air conditioner is designed to reject heat at a rate of \(15,000 \mathrm{~kJ} / \mathrm{h}\) from refrigerant-134a as the refrigerant is condensed at a temperature of \(40^{\circ} \mathrm{C}\). Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows across the finned condenser coils, entering at \(25^{\circ} \mathrm{C}\) and leaving at \(35^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the refrigerant side is \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer area on the refrigerant side.

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(20^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(4.5 \mathrm{~kg} / \mathrm{s}\). Heat is supplied by hot oil \(\left(c_{p}=2300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(170^{\circ} \mathrm{C}\) at a rate of \(10 \mathrm{~kg} / \mathrm{s}\). For a tube-side overall heat transfer coefficient of \(350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area on the tube side.

A counterflow double-pipe heat exchanger with \(A_{s}=\) \(9.0 \mathrm{~m}^{2}\) is used for cooling a liquid stream \(\left(c_{p}=3.15 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a rate of \(10.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(90^{\circ} \mathrm{C}\). The coolant \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the heat exchanger at a rate of \(8.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(10^{\circ} \mathrm{C}\). The plant data gave the following equation for the overall heat transfer coefficient in W/m \({ }^{2} \cdot \mathrm{K}: U=600 /\left(1 / \dot{m}_{c}^{0.8}+2 / \dot{m}_{h}^{0.8}\right)\), where \(\dot{m}_{c}\) and \(\dot{m}_{h}\) are the cold-and hot-stream flow rates in kg/s, respectively. (a) Calculate the rate of heat transfer and the outlet stream temperatures for this unit. (b) The existing unit is to be replaced. A vendor is offering a very attractive discount on two identical heat exchangers that are presently stocked in its warehouse, each with \(A_{s}=5 \mathrm{~m}^{2}\). Because the tube diameters in the existing and new units are the same, the above heat transfer coefficient equation is expected to be valid for the new units as well. The vendor is proposing that the two new units could be operated in parallel, such that each unit would process exactly one-half the flow rate of each of the hot and cold streams in a counterflow manner; hence, they together would meet (or exceed) the present plant heat duty. Give your recommendation, with supporting calculations, on this replacement proposal.
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