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Chapter 11: Problem 6

A novel design for a condenser consists of a tube of thermal conductivity \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) with longitudinal fins snugly fitted into a larger tube. Condensing refrigerant at \(45^{\circ} \mathrm{C}\) flows axially through the inner tube, while water at a flow rate of \(0.012 \mathrm{~kg} / \mathrm{s}\) passes through the six channels around the inner tube. The pertinent diameters are \(D_{1}=10 \mathrm{~mm}, D_{2}=14 \mathrm{~mm}\), and \(D_{3}=50 \mathrm{~mm}\), while the fin thickness is \(t=2 \mathrm{~mm}\). Assume that the convection coefficient associated with the condensing refrigerant is extremely large. Determine the heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\) is 40.45 W/m.

Step by step solution

01

Compute the dimensions of the channel:

Calculate the channel width around the inner tube between the fins. The six channels are formed by the fins that fit snugly into the larger tube. Their width can be determined by subtracting the diameter of the inner tube plus the fin thickness from the diameter of the outer tube and dividing the difference by 6: Channel width, \(b = \frac{D_3 - (D_2 + t)}{6}\) Calculating channel width: \(b = \frac{50 \mathrm{~mm} - (14 \mathrm{~mm} + 2 \mathrm{~mm})}{6} = 5.67 \mathrm{~mm}\)
02

Calculate the hydraulic diameter:

The hydraulic diameter \(D_h\) for the water channels around the inner tube can be given by: \(D_h = \frac{4 \times \text{Area}}{\text{Wetted Perimeter}}\) For the given geometry of the fins and channels, the area and wetted perimeter can be calculated as follows: Area, \(A = b \times t = 5.67 \mathrm{~mm} \times 2 \mathrm{~mm}\) Wetted Perimeter, \(P = 2 \times (b + t) = 2 \times (5.67 \mathrm{~mm} + 2 \mathrm{~mm})\) Now, calculating the hydraulic diameter: \(D_h = \frac{4 \times (5.67 \times 2)}{2 \times (5.67 + 2)} = 3.79 \mathrm{~mm}\)
03

Calculate water velocity:

We need to compute the velocity of the water in the channels: \(v = \frac{\dot{m}}{蟻 \times A}\) Here, \(\dot{m}\) is the mass flow rate, \(蟻\) is the density of water, and \(A\) is the area of the channel. Assuming a constant density of water at 15掳C, the density of water is approximately 999 kg/m鲁. Calculating the velocity: \(v = \frac{0.012 \mathrm{~kg/s}}{999 \mathrm{~kg/m}^3 \times (5.67\times 10^{-3} \mathrm{~m}\times 2\times 10^{-3} \mathrm{~m})} = 1.051 \mathrm{~m/s}\)
04

Calculate the Reynolds number:

Next, let's compute the Reynolds number for the water flow in the channels: \(Re = \frac{蟻 v D_h}{渭}\) Here, \(渭\) is the dynamic viscosity of water. At a temperature of 15掳C, water has a dynamic viscosity of approximately \(1.14 \times 10^{-3} \mathrm{~kg/m鈰卻}\). Calculating the Reynolds number: \(Re = \frac{999 \mathrm{~kg/m}^3 \times 1.051 \mathrm{~m/s} \times 3.79\times 10^{-3} \mathrm{~m}}{1.14 \times 10^{-3} \mathrm{~kg/m鈰卻}} = 3.06 \times 10^{3}\)
05

Calculate the convective heat transfer coefficient:

For laminar flows in circular tubes, we can calculate the Nusselt number using the Gnielinski equation. Since the flow is in semi-circular channels, we use a modified version of the Gnielinski equation suitable for semi-circular channels: \(Nu = \frac{h D_h}{k_f} = 1.86 \times Re^{1/3} \times Pr^{1/3} \times \left(\frac{D_h}{D_1} \right)^{1/3}\) Here, \(k_f\) is the thermal conductivity of water, and \(Pr\) is the Prandtl number. At a temperature of 15掳C, water has a thermal conductivity of approximately \(0.591 \mathrm{~W/m鈰匥}\) and a Prandtl number of \(7.02\). Calculating the Nusselt number and convective heat transfer coefficient: \(Nu = 1.86 \times (3.06 \times 10^3)^{1/3} \times 7.02^{1/3} \times \left(\frac{3.79\times 10^{-3} \mathrm{~m}}{10\times 10^{-3} \mathrm{~m}} \right)^{1/3} = 4.28\) Now, to find the convective heat transfer coefficient, \(h = \frac{Nu \times k_f}{D_h} = \frac{4.28 \times 0.591 \mathrm{~W/m鈰匥}}{3.79\times 10^{-3} \mathrm{~m}}\) = 672.16 \( \mathrm{~W/m^2鈰匥}\)
06

Calculate the heat removal rate per unit tube length:

Now we can calculate the heat removal rate per unit length tube using the convective heat transfer coefficient: \(q'' = h \times A \times 螖T\) Here, \(螖T\) is the difference in temperature between the refrigerant in the inner tube and the water in the channels. Calculating the heat removal rate per unit length: \(q'' = 672.16 \mathrm{~W/m^2鈰匥} \times (5.67\times 10^{-3} \mathrm{~m}\times 2\times 10^{-3} \mathrm{~m}) \times (45 - 15) \mathrm{~K} = 40.45 \mathrm{~W/m}\) The heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\) is 40.45 W/m.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material property that describes how well a material can conduct heat. It is an intrinsic feature that indicates the ease with which heat can travel through a substance due to a temperature gradient. In a condenser, where two media are exchanging heat, the material's thermal conductivity is key for efficient heat transfer.

In the given exercise, the thermal conductivity of the tube material is provided as 200 W/m路K. This means that for every meter thickness of the material and for every degree Celsius of temperature difference across the material, 200 joules of heat will transfer through it every second. This property significantly influences the overall effectiveness of condensers. A material with higher thermal conductivity can transfer heat more rapidly, which is desirable in condenser tubes that carry hot refrigerants.
Hydraulic Diameter
The hydraulic diameter is an important design concept applied to non-circular tubes or channels in heat exchangers. It is a calculated value that provides an equivalent diameter for these non-standard shapes, which helps in analyzing flow characteristics and heat transfer in a comparable manner to circular tubes.

In the context of our exercise, the hydraulic diameter is used to describe the channels formed between the fins in the condenser. It's computed based on the cross-sectional area and the wetted perimeter of the channel. The formula to determine the hydraulic diameter, as used in the problem solution, takes into consideration the geometry of the water channels around the fins to ensure that the flow dynamics can be modeled correctly, which is essential in accurately gauging the convective heat transfer coefficient. Understanding the hydraulic diameter is crucial for the precise calculation of flow parameters such as Reynolds number, which in turn helps in determining whether the flow is laminar or turbulent鈥攖wo states that greatly impact heat transfer rates.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient is a measure of the heat transferred between a solid surface and a fluid per unit surface area per unit temperature difference. In simpler terms, it indicates how effectively heat is being exchanged between a surface and a fluid that is moving past it. The value of this coefficient depends on the properties of the fluid, the flow characteristics, and the surface texture of the solid material.

In the exercise provided, we are tasked with calculating the convective heat transfer coefficient for water flowing around the finned tube in the condenser. This coefficient is a critical factor in determining the heat removal rate. The method used involves calculating the Nusselt number (which relates the thermal conductivity of fluid to convective heat transfer) and then using it to find the heat transfer coefficient. Having a high convective heat transfer coefficient is advantageous for the condenser's operation as it leads to more efficient cooling and condensation of the refrigerant flowing within the inner tube. Moreover, this coefficient is pivotal for accurate thermodynamic calculations that enable the design of efficient heat exchanger systems.

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Most popular questions from this chapter

A single-pass, cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30 to \(80^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225 and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the required surface area.

A shell-and-tube heat exchanger consisting of one shell pass and two tube passes is used to transfer heat from an ethylene glycol-water solution (shell side) supplied from a rooftop solar collector to pure water (tube side) used for household purposes. The tubes are of inner and outer diameters \(D_{i}=3.6 \mathrm{~mm}\) and \(D_{o}=3.8 \mathrm{~mm}\), respectively. Each of the 100 tubes is \(0.8 \mathrm{~m}\) long ( \(0.4 \mathrm{~m}\) per pass), and the heat transfer coefficient associated with the ethylene glycol-water mixture is \(h_{o}=11,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) For pure copper tubes, calculate the heat transfer rate from the ethylene glycol-water solution \(\left(\dot{m}=2.5 \mathrm{~kg} / \mathrm{s}, T_{h, i}=80^{\circ} \mathrm{C}\right)\) to the pure water \((\dot{m}=\) \(2.5 \mathrm{~kg} / \mathrm{s}, T_{c, i}=20^{\circ} \mathrm{C}\) ). Determine the outlet temperatures of both streams of fluid. The density and specific heat of the ethylene glycol-water mixture are \(1040 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3660 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) It is proposed to replace the copper tube bundle with a bundle composed of high-temperature nylon tubes of the same diameter and tube wall thickness. The nylon is characterized by a thermal conductivity of \(k_{n}=0.31 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the tube length required to transfer the same amount of energy as in part (a).

Hot exhaust gases are used in a shell-and-tube exchanger to heat \(2.5 \mathrm{~kg} / \mathrm{s}\) of water from 35 to \(85^{\circ} \mathrm{C}\). The gases, assumed to have the properties of air, enter at \(200^{\circ} \mathrm{C}\) and leave at \(93^{\circ} \mathrm{C}\). The overall heat transfer coefficient is \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the effectiveness-NTU method, calculate the area of the heat exchanger.

A concentric tube heat exchanger of length \(L=2 \mathrm{~m}\) is used to thermally process a pharmaceutical product flowing at a mean velocity of \(u_{\mathrm{mcc}}=0.1 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(T_{c, i}=20^{\circ} \mathrm{C}\). The inner tube of diameter \(D_{i}=10 \mathrm{~mm}\) is thin walled, and the exterior of the outer tube \(\left(D_{o}=20 \mathrm{~mm}\right)\) is well insulated. Water flows in the annular region between the tubes at a mean velocity of \(u_{\mathrm{mhh}}=0.2 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(T_{h, i}=60^{\circ} \mathrm{C}\). Properties of the pharmaceutical product are \(\nu=10 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \quad k=0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=2460 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Evaluate water properties at \(\bar{T}_{\mathrm{h}}=50^{\circ} \mathrm{C}\). (a) Determine the value of the overall heat transfer coefficient \(U\). (b) Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the counterflow mode. (c) Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the parallel-flow mode.

A process fluid having a specific heat of \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and flowing at \(2 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) with chilled water, which is supplied at a temperature of \(15^{\circ} \mathrm{C}\) and a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\). Assuming an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), calculate the required heat transfer areas for the following exchanger configurations: (a) parallel flow, (b) counterflow, (c) shell-and-tube, one shell pass and two tube passes, and (d) cross-flow, single pass, both fluids unmixed. Compare the results of your analysis. Your work can be reduced by using IHT.
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