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Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at \(220^{\circ} \mathrm{C}\) flows in parallel over the upper and lower surfaces of a \(1.5\)-m-long flat plate at a velocity of \(3 \mathrm{~m} / \mathrm{s}\). If the flat plate surface temperature is maintained at \(80^{\circ} \mathrm{C}\), determine (a) the local convection heat transfer coefficient at \(1 \mathrm{~m}\) from the leading edge, \((b)\) the average convection heat transfer coefficient over the entire plate, and \((c)\) the total heat flux transfer to the plate.

Short Answer

Expert verified
Question: Calculate the local convection heat transfer coefficient at 1m from the leading edge, the average convection heat transfer coefficient over the entire plate, and the total heat flux transfer to the plate for hot carbon dioxide gas at 1 atm flowing over flat plates. Given: T_gas = 220°C, V = 3m/s, Plate length = 3m, Plate surface temperature = 195°C. Answer: To solve this problem, follow these steps: Step 1: Calculate the Reynolds number using the given data and properties of carbon dioxide gas. Step 2: Assess the flow regime (laminar or turbulent) based on the calculated Reynolds number. Step 3: Choose the appropriate correlation for calculating the local and average Nusselt numbers based on the flow regime. Step 4: Calculate the local and average convection heat transfer coefficients (h_x and h_avg) using the Nusselt numbers. Step 5: Calculate the total heat flux transfer (q_total) using the average convection heat transfer coefficient (h_avg) and the temperature difference between the gas and the plate surface.

Step by step solution

01

Calculate the Reynolds number

First, we need to calculate the Reynolds number to determine the flow regime (laminar or turbulent). The Reynolds number is given by the formula: Re = \(\frac{\rho V L}{\mu}\), where Re is the Reynolds number, ρ is the gas density, V is the gas velocity, L is the plate length, and μ is the gas dynamic viscosity. We can use the given data (T_gas=220°C, V=3m/s, and Plate length=3m) along with the properties of carbon dioxide gas to compute the Reynolds number.
02

Assess the flow regime

Based on the calculated Reynolds number, we can determine the flow regime (laminar or turbulent) over the flat plate. If Re < 5000, the flow is laminar, and if Re > 5000, the flow is turbulent.
03

Choose the appropriate correlation for Nusselt numbers

Depending on the flow regime obtained in step 2, we need to choose the appropriate correlation for calculating the local and average Nusselt numbers. For laminar flow, we can use the following correlation for a flat plate: Nu_x = 0.332 Re_x^{1/2} Pr^{1/3} For turbulent flow, we can use the following correlation for a flat plate: Nu_x = 0.0296 Re_x^{4/5} Pr^{1/3} Here, Nu_x is the local Nusselt number and Re_x is the local Reynolds number at the specified x location.
04

Calculate the local and average convection heat transfer coefficients

Using the correlations obtained in step 3, we can calculate the local Nusselt number at x=1m (Nu_x). To compute the average Nusselt number over the entire plate (Nu_avg), we can use the following formulas: Nu_avg = \(\frac{1}{L} \int_0^L Nu_x dx\) We can then calculate the local and average convection heat transfer coefficients (h_x and h_avg) using the formulas: h_x = \(\frac{k}{L} Nu_x\) h_avg = \(\frac{k}{L} Nu_{avg}\) Here, k is the gas thermal conductivity.
05

Calculate the total heat flux transfer

Finally, using the average convection heat transfer coefficient (h_avg) obtained in step 4, we can calculate the total heat flux transfer (q_total) to the plate. The formula for total heat flux transfer is: q_total = h_avg * A * ΔT Where A is the surface area of the plate and ΔT is the temperature difference between the gas and the plate surface.

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Key Concepts

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

Nusselt Number
The Nusselt Number (Nu) is a vital dimensionless parameter in convection heat transfer, representing the ratio of convective to conductive heat transfer across a boundary. It essentially quantifies the effectiveness of the heat transfer process:
  • A higher Nusselt number indicates more effective convective heat transfer.
  • A Nusselt number of one would imply that heat transfer is purely conductive.
In practical applications, the Nusselt number helps determine how well a fluid, such as gas flowing over a flat plate, is able to carry heat away from a surface. The value of the Nusselt number can be calculated using correlations based on the flow regime, such as:
  • Laminar Flow: Nu_x = 0.332 Re_x^{1/2} Pr^{1/3}
  • Turbulent Flow: Nu_x = 0.0296 Re_x^{4/5} Pr^{1/3}
Here, Pr is the Prandtl number, a property of the fluid, and Re_x is the local Reynolds number, further analyzed next.
Reynolds Number
The Reynolds Number (Re) is another essential dimensionless parameter used to predict flow patterns in different fluid flow situations. It is crucial in determining the flow regime over a flat plate:
  • A Reynolds number less than 5000 indicates laminar flow.
  • A Reynolds number greater than 5000 indicates turbulent flow.
The formula for Reynolds Number is:\[ Re = \frac{\rho V L}{\mu} \]Where:- \(\rho\) is the fluid density- \(V\) is the flow velocity- \(L\) is the characteristic length (e.g., length of the flat plate)- \(\mu\) is the fluid's dynamic viscosityBy calculating the Reynolds number, engineers can determine if the flow will be smooth or chaotic, which influences the subsequent steps in heat transfer analysis, including the choice of Nusselt number correlation.
Flat Plate Heat Transfer
In many practical situations, heat transfer from flat plates is a common engineering challenge. The heat transfer scenario involves cooling or heating gas as it flows over a flat surface. This kind of analysis is crucial for systems like heat exchangers or in cooling processes for hot exhaust gases.The overall goal is to calculate different heat transfer coefficients:- **Local coefficient** at a specific point on the plate- **Average coefficient** over the entire plateUsing the previously calculated Nusselt number, we convert these values to the convection heat transfer coefficients using the formula:\[ h = \frac{k}{L} Nu \]Where \(k\) is the thermal conductivity of the gas. These coefficients help in determining how efficient the plate is at losing or gaining heat, valuable in optimizing design and performance.
Gas Cooling
Gas cooling through flat plates is a specific application where the heat transfer concepts become operational. The gas, typically at high temperatures, transfers heat to a cooler flat surface as it flows parallel above and below it. Key considerations include:
  • Temperature difference between the plate and the gas.
  • Flow velocity of the gas.
  • Material properties such as thermal conductivity of the gas and plate.
The effectiveness of cooling is directly linked to these factors and can be gauged using the calculated heat transfer coefficients. In applications such as industrial exhaust systems, understanding these principles helps in designing efficient cooling mechanisms to manage temperatures within safe operating limits.
Flow Regime Analysis
Analyzing the flow regime is critical to understanding and predicting the behavior of fluid motion over a surface. The flow can be: - **Laminar**: characterized by smooth and orderly layers of fluid, typically at lower Reynolds numbers. - **Turbulent**: showing chaotic and mixed fluid motion, usually at higher Reynolds numbers. Understanding the flow regime helps in making accurate predictions about heat transfer capabilities:
  • **Laminar flow** generally results in higher surface temperatures due to less mixing, a gentle gradient exists from the surface outward.
  • **Turbulent flow** enhances mixing and disrupting the thermal boundary layer, usually leading to better heat transfer.
These insights allow the improvement of efficiency in systems and precisely tailoring solutions based on specific airflow characteristics.

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Most popular questions from this chapter

Kitchen water at \(10^{\circ} \mathrm{C}\) flows over a 10 -cm-diameter pipe with a velocity of \(1.1 \mathrm{~m} / \mathrm{s}\). Geothermal water enters the pipe at \(90^{\circ} \mathrm{C}\) at a rate of \(1.25 \mathrm{~kg} / \mathrm{s}\). For calculation purposes, the surface temperature of the pipe may be assumed to be \(70^{\circ} \mathrm{C}\). If the geothermal water is to leave the pipe at \(50^{\circ} \mathrm{C}\), the required length of the pipe is (a) \(1.1 \mathrm{~m}\) (b) \(1.8 \mathrm{~m}\) (c) \(2.9 \mathrm{~m}\) (d) \(4.3 \mathrm{~m}\) (e) \(7.6 \mathrm{~m}\) (For both water streams, use \(k=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=4.32\), \(\left.\nu=0.658 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\)

Air at \(20^{\circ} \mathrm{C}\) flows over a 4-m-long and 3-m-wide surface of a plate whose temperature is \(80^{\circ} \mathrm{C}\) with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The rate of heat transfer from the surface is (a) \(7383 \mathrm{~W}\) (b) \(8985 \mathrm{~W}\) (c) \(11,231 \mathrm{~W}\) (d) 14,672 W (e) \(20,402 \mathrm{~W}\) (For air, use \(k=0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7228, \nu=1.798 \times\) \(\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

Air \((\operatorname{Pr}=0.7, k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) at \(200^{\circ} \mathrm{C}\) flows across 2-cm-diameter tubes whose surface temperature is \(50^{\circ} \mathrm{C}\) with a Reynolds number of 8000 . The Churchill and Bernstein convective heat transfer correlation for the average Nusselt number in this situation is $$ \mathrm{Nu}=0.3+\frac{0.62 \mathrm{Re}^{0.5} \mathrm{Pr}^{0.33}}{\left[1+(0.4 / \mathrm{Pr})^{0.67}\right]^{0.25}} $$ (a) \(8.5 \mathrm{~kW} / \mathrm{m}^{2}\) (b) \(9.7 \mathrm{~kW} / \mathrm{m}^{2}\) (c) \(10.5 \mathrm{~kW} / \mathrm{m}^{2}\) (d) \(12.2 \mathrm{~kW} / \mathrm{m}^{2}\) (e) \(13.9 \mathrm{~kW} / \mathrm{m}^{2}\)

Warm air is blown over the inner surface of an automobile windshield to defrost ice accumulated on the outer surface of the windshield. Consider an automobile windshield \(\left(k_{w}=\right.\) \(0.8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R})\) with an overall height of 20 in and thickness of \(0.2\) in. The outside air ( \(1 \mathrm{~atm})\) ambient temperature is \(8^{\circ} \mathrm{F}\) and the average airflow velocity over the outer windshield surface is \(50 \mathrm{mph}\), while the ambient temperature inside the automobile is \(77^{\circ} \mathrm{F}\). Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as a flat plate surface.

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, \(1.7\)-m-long pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. Air properties at the average temperature are \(k=0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735\), \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The rate of heat transfer to the pipe is (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

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