Quick Recap:

The team engaged in a detailed discussion about heat transfer concepts, including convection and the calculation of heat transfer coefficients. Janarth presented and solved several heat transfer problems involving steam condensers, thermo-pan windows, and jacketed vessels, while explaining the relevant mathematical approaches and assumptions. The session concluded with Karthikeyan demonstrating how to solve various heat exchanger problems and explaining the relationship between heat transfer coefficients and characteristic lengths.

Next Steps:

  • Students to review and clarify doubts on convection, uniform heat plates, and evaporators.
  • All attendees to practice solving heat transfer problems, especially those related to conduction, radiation shields, and heat exchangers.
  • All attendees to focus on mastering heat transfer concepts, as it has equal weightage to thermodynamics and reaction engineering in the curriculum.
  • All attendees to prepare for potential heat transfer questions in upcoming GATE exams.

SUMMARY:

Heat Transfer and Convection Discussion

The team discusses various topics related to heat transfer and convection. Smrithi expresses doubts about uniform heat plates and convection, particularly regarding calculating \(k\) values for different surfaces. Janarth prepares to present questions from previous GATE exams to provide insights to the group. Karthikeyan facilitates the discussion, asking about doubts and encouraging participation from team members.

Steam Condenser Design Calculations

Janarth explains how to solve a problem involving a shell and tube steam condenser. He outlines the steps to calculate the tube length and the rate of steam condensation. The process involves finding the Reynolds number, Prandtl number, and Nusselt number to determine the internal heat transfer coefficient. This is then used to calculate the overall heat transfer coefficient, which is applied in the heat transfer equation along with the log mean temperature difference (LMTD) to find the tube length. For the second part, Janarth explains how to use the latent heat of condensation to determine the rate of steam condensation. He then briefly introduces a new problem about a double pipe heat exchanger, asking for ideas on how to calculate the LMTD given temperature profile equations for hot and cold fluids.

Heat Transfer Problem Solutions

Janarth explains how to solve two heat transfer problems. The first involves calculating the percentage reduction in heat loss for a thermo-pan window compared to a single glass sheet, using given thermal conductivity values and dimensions. The second problem requires determining the time it takes for a solid sphere to reach a specific temperature when exposed to a gas stream, using the lumped system analysis approach. Janarth provides step-by-step instructions for both problems, including relevant equations and data interpretation.

Heat Transfer in Jacketed Vessel

The discussion focuses on solving a heat transfer problem involving a jacketed vessel. Janarth explains that since the vessel is well-mixed, it can be treated as a continuous stirred tank reactor (CSTR). The hot fluid inside the vessel maintains a uniform temperature of 100°C, while the cold water in the jacket increases from 20°C to 64.8°C. Janarth emphasizes that understanding the well-mixed assumption is crucial for correctly calculating the temperature difference and heat transfer area. Karthikeyan then presents a question about heat conduction in a slab, explaining that the Fourier number concept should be used instead of lumped heat transfer analysis. He demonstrates that the time required for heat transfer is proportional to the square of the characteristic length.

Heat Exchanger Temperature Analysis

Karthikeyan explains how to solve a heat exchanger problem involving an hot oil and a cold fluid. He outlines the steps to find the end temperature of the hot oil, which is calculated to be 68°C. Karthikeyan then describes how to determine the log mean temperature difference (LMTD), the heat transfer rate (Q), and finally the heat transfer area, which is found to be 17.95 square meters. He emphasizes the importance of identifying whether the heat exchanger is in counter-current or co-current arrangement, as it significantly affects the LMTD calculation.

Heat Transfer Coefficient Calculations

Karthikeyan explains how to solve a problem involving heat transfer coefficients in different directions. He demonstrates that the heat transfer coefficient is proportional to the characteristic length raised to the power of 0.5, and introduces a constant C1 to account for other factors. The group calculates the ratio of average heat transfer coefficients in the \(x\) and \(y\) directions, arriving at an answer of 2. Arputhaselvi confirms the importance of including the constant in the calculations. The session concludes with a brief discussion about the importance of heat transfer in their curriculum.

Last modified: Sunday, 29 June 2025, 6:53 PM