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Chapter 9: Problem 113

A horizontal 100-mm-diameter pipe passing hot oil is to be used in the design of an industrial water heater. Based on a typical water draw rate, the velocity over the pipe is \(0.5 \mathrm{~m} / \mathrm{s}\). The hot oil maintains the outer surface temperature at \(85^{\circ} \mathrm{C}\) and the water temperature is \(37^{\circ} \mathrm{C}\). Investigate the effect of flow direction on the heat rate (W/m) for (a) horizontal, (b) downward, and (c) upward flow.

Short Answer

Expert verified
The effect of flow direction on the heat rate (W/m) for the industrial water heater design is as follows: (a) Horizontal flow: \(q_H = 17{,}689.69 \mathrm{W/m}\) (b) Downward flow: \(q_D = 10{,}050.92 \mathrm{W/m}\) (c) Upward flow: \(q_D = 10{,}050.92 \mathrm{W/m}\) The horizontal flow has the highest heat transfer rate compared to downward and upward flow.

Step by step solution

01

Calculate Surface Area of Pipe

Using the formula for the surface area of a pipe, \(A = 2\pi r L\), with the diameter given as 100 mm, we can find the surface area: First, convert diameter to meters: \(D = 0.1 \mathrm{m}\), Then, find the radius: \(r = \frac{D}{2} = \frac{0.1 \mathrm{m}}{2} = 0.05 \mathrm{m}\), The surface area will be calculated per meter length: \(L = 1\mathrm{m}\), \(A = 2\pi (0.05\mathrm{m})(1\mathrm{m}) = 0.314 \mathrm{m}^{2}\).
02

Calculate Reynold's Number

Using the formula \(Re = \frac{VD}{\nu}\), we can find the Reynold's number: \(Re = \frac{(0.5 \mathrm{m/s})(0.1\mathrm{m})}{1 \times 10^{-6} \mathrm{m}^2/\mathrm{s}} = 50{,}000\). Since \(Re > 4000\), the flow is turbulent.
03

Calculate Convection Heat Transfer Coefficient for Horizontal Flow

We will assume natural convection for horizontal flow, using the following relation for turbulent flows: \(Nu_H = 0.27 Re_H^{0.63} Pr_H^{0.43}\), where \(Nu_H\) is the Nusselt number for horizontal flow, \(Re_H\) is the Reynold's number for horizontal flow, and \(Pr_H\) is the Prandtl number for horizontal flow. Assuming \(Re_H = 50{,}000\) and \(Pr_H = 6\), we can calculate \(Nu_H\): \(Nu_H = 0.27 (50{,}000)^{0.63}(6)^{0.43} = 391.71\). Now, we will find the convection heat transfer coefficient for horizontal flow, \(h_H\): \(h_H = \frac{Nu_Hk}{D} = \frac{(391.71)(0.6\mathrm{W/mK})}{0.1\mathrm{m}} = 2350.33 \mathrm{W/m}^2\mathrm{K}\).
04

Calculate the Heat Rate for Horizontal Flow

Now, we can calculate the heat rate for horizontal flow using the heat transfer equation: \(q_H = h_H \cdot A \cdot (T_s - T_\infty) = (2350.33 \mathrm{W/m}^2\mathrm{K})(0.314 \mathrm{m}^{2})(85^{\circ} \mathrm{C}-37^{\circ} \mathrm{C}) = 17{,}689.69 \mathrm{W/m}\).
05

Calculate Convection Heat Transfer Coefficient for Downward and Upward Flow

For both downward and upward flow cases, we will use the Dittus-Boelter equation (forced convection): \(Nu_D = 0.023 Re_D^{0.8} Pr_D^{0.4}\), where \(Nu_D\) is the Nusselt number for downward and upward flow, and \(Re_D\) and \(Pr_D\) are the Reynold's number and Prandtl number for downward and upward flow, respectively. Assuming \(Re_D = 50{,}000\) and \(Pr_D = 6\), for downward and upward flow cases: \(Nu_D = 0.023 (50{,}000)^{0.8}(6)^{0.4} = 1{,}112.99\). Now, we will find the convection heat transfer coefficient for downward and upward flow, \(h_D\): \(h_D = \frac{Nu_Dk}{D} = \frac{(1{,}112.99)(0.6\mathrm{W/mK})}{0.1\mathrm{m}} = 6{,}677.97 \mathrm{W/m}^2\mathrm{K}\).
06

Calculate the Heat Rate for Downward and Upward Flow

Now, we can calculate the heat rate for both downward and upward flow using the heat transfer equation: \(q_D = h_D \cdot A \cdot (T_s - T_\infty) = (6{,}677.97 \mathrm{W/m}^2\mathrm{K})(0.314 \mathrm{m}^{2})(85^{\circ} \mathrm{C} - 37^{\circ} \mathrm{C}) = 10{,}050.92 \mathrm{W/m}\).
07

Final Results

The heat rate (W/m) for each case is as follows: (a) Horizontal flow: \(q_H = 17{,}689.69 \mathrm{W/m}\) (b) Downward flow: \(q_D = 10{,}050.92 \mathrm{W/m}\) (c) Upward flow: \(q_D = 10{,}050.92 \mathrm{W/m}\) (since the heat transfer coefficient is the same for both downward and upward flow). Thus, the flow direction does affect the heat rate, with horizontal flow having the highest heat transfer rate compared to downward and upward flow.

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Key Concepts

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

Convection Heat Transfer Coefficient
Understanding the convection heat transfer coefficient is crucial when it comes to exchanging heat between surfaces and fluids. This coefficient, often denoted as 'h', measures the effectiveness of the convection process, which is the transfer of heat through a fluid in motion, driven by the difference in temperature between the fluid and a surface.

For example, in the case of a pipe carrying hot oil surrounded by cooler water, 'h' gives insight into how well the pipe鈥檚 surface can transfer its heat to the water. A higher 'h' means a better transfer, leading to more efficient heating or cooling operations.

Impact of Flow Type on Heat Transfer

In our exercise, we explore how the coefficient 'h' varies with the flow direction. For horizontal flow, the coefficient is calculated assuming natural convection and is thus lower than for forced convection, which we assume for downward and upward flows. This difference in 'h' results in the horizontal flow having a higher heat transfer rate, as indicated by the higher rate of watts per meter, compared to downward and upward flows where forced convection is at play.

Calculating 'h' accurately is fundamental in designing efficient thermal systems, and many variables like fluid velocity, properties, and flow direction come into play, highlighting the complex interaction involved in heat transfer.
Reynolds Number
The Reynolds number, commonly represented by 'Re', is a dimensionless quantity that plays a pivotal role in the study of fluid flow and heat transfer. It provides a measure to predict the flow regime, whether it will be laminar or turbulent, based on the fluid's velocity, characteristic dimension (such as diameter in case of pipes), and kinematic viscosity.

In our exercise problem, we determined 'Re' to assess the flow inside the pipe. If 'Re' is below 2000, the flow is usually laminar, meaning it is smooth and orderly. On the other hand, an 'Re' above 4000 suggests turbulent flow, characterized by random, chaotic fluid motion that enhances mixing and heat transfer.

Interpreting Reynolds Number in Heat Transfer

For the given oil in the pipe, 'Re' calculation gave us a value of 50,000, indicating turbulent flow. This has important consequences, as turbulent flow promotes greater heat transfer efficiency due to the increased mixing action within the fluid. Hence, 'Re' directly affects the calculation of the heat transfer coefficient and the overall heat transfer rate from the pipe to its surroundings.
Nusselt Number
When it comes to heat transfer in fluid systems, the Nusselt number (Nu) emerges as a central figure. This dimensionless parameter encapsulates the ratio of convective to conductive heat transfer across a boundary. In simpler terms, it's how we compare the effectiveness of heat transfer via fluid motion to heat transfer via simple thermal conduction.

For the pipe scenario we are considering, the Nusselt number helps in evaluating how efficient the convection process is at moving heat from the hot oil out through the pipe wall and into the water. Specific correlations that include the Nusselt number are used, such as the one for natural convection for horizontal flow and the Dittus-Boelter equation for forced convection in downward and upward flows.

Application in Heat Transfer Calculations

'Nu' can be determined from empirical or theoretical relationships that involve 'Re' and the Prandtl number, another vital dimensionless number in heat transfer. With 'Nu' ascertained, the exercise demonstrates how it's used to find the convection heat transfer coefficient. This chain of calculations ultimately feeds into formulas determining the actual heat rate鈥攖he quantity of heat transferred per unit length of the pipe鈥攚hich concludes our study on how flow direction influences heat transfer rates in pipe systems.

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Most popular questions from this chapter

Consider a large vertical plate with a uniform surface temperature of \(130^{\circ} \mathrm{C}\) suspended in quiescent air at \(25^{\circ} \mathrm{C}\) and atmospheric pressure. (a) Estimate the boundary layer thickness at a location \(0.25 \mathrm{~m}\) measured from the lower edge. (b) What is the maximum velocity in the boundary layer at this location and at what position in the boundary layer does the maximum occur? (c) Using the similarity solution result, Equation \(9.19\), determine the heat transfer coefficient \(0.25 \mathrm{~m}\) from the lower edge. (d) At what location on the plate measured from the lower edge will the boundary layer become turbulent?

A computer code is being developed to analyze a 12.5-mm-diameter, cylindrical sensor used to determine ambient air temperature. The sensor experiences free convection while positioned horizontally in quiescent air at \(T_{\infty}=27^{\circ} \mathrm{C}\). For the temperature range from 30 to \(80^{\circ} \mathrm{C}\), derive an expression for the convection coefficient as a function of only \(\Delta T=\) \(T_{s}-T_{\infty}\), where \(T_{s}\) is the sensor temperature. Evaluate properties at an appropriate film temperature and show what effect this approximation has on the convection coefficient estimate.

A rectangular cavity consists of two parallel, \(0.5-\mathrm{m}-\) square plates separated by a distance of \(50 \mathrm{~mm}\), with the lateral boundaries insulated. The heated plate is maintained at \(325 \mathrm{~K}\) and the cooled plate at \(275 \mathrm{~K}\). Estimate the heat flux between the surfaces for three orientations of the cavity using the notation of Figure 9.6: vertical with \(\tau=90^{\circ}\), horizontal with \(\tau=0^{\circ}\), and horizontal with \(\tau=180^{\circ}\).

The space between the panes of a double-glazed window can be filled with either air or carbon dioxide at atmospheric pressure. The window is \(1.5 \mathrm{~m}\) high and the spacing between the panes can be varied. Develop an analysis to predict the convection heat transfer rate across the window as a function of pane spacing and determine, under otherwise identical conditions, whether air or carbon dioxide will yield the smaller rate. Illustrate the results of your analysis for two surface-temperature conditions: winter \(\left(-10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\right)\) and summer \(\left(35^{\circ} \mathrm{C}, 25^{\circ} \mathrm{C}\right)\).

The surfaces of two long, horizontal, concentric thinwalled tubes having radii of 100 and \(125 \mathrm{~mm}\) are maintained at 300 and \(400 \mathrm{~K}\), respectively. If the annular space is pressurized with nitrogen at \(5 \mathrm{~atm}\), estimate the convection heat transfer rate per unit length of the tubes.
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