Solution Manual Heat And Mass Transfer Cengel 5th Edition Chapter 9

The Solution Manual for Chapter 9 of Çengel’s Heat and Mass Transfer (5th Ed.)

focuses on the complex topic of Natural Convection, where fluid motion is driven by buoyancy forces rather than external fans or pumps. Natural Convection: A Physical Overview

Natural convection, or free convection, occurs when a fluid (like air or water) contacts a surface at a different temperature. The resulting temperature gradient causes density variations; for example, air near a hot radiator becomes less dense and rises, while cooler, denser air sinks to take its place. This motion, known as a natural convection current, significantly enhances heat transfer compared to pure conduction. Key Concepts and Mathematical Foundations

Chapter 9 details the governing equations—continuity, momentum, and energy—which must often be solved simultaneously because fluid velocity depends directly on the temperature field. The manual relies on three critical dimensionless numbers to characterize these flows:

Mod-01 Lec-35 Introduction to Natural Convection Heat Transfer

Introduction

For engineering students worldwide, Heat and Mass Transfer: Fundamentals and Applications by Yunus A. Cengel and Afshin J. Ghajar is the gold standard textbook. Among its many challenging sections, Chapter 9: Natural Convection often stands as a significant hurdle. Unlike forced convection, where fans or pumps dictate fluid motion, natural convection relies on buoyancy forces driven by temperature gradients—a concept that is physically intuitive but mathematically complex.

If you are searching for the "solution manual heat and mass transfer cengel 5th edition chapter 9," you are likely struggling with the transition from theory to problem-solving. This article serves three purposes:

1. To explain the core concepts of Chapter 9 that the solutions manual builds upon.
2. To guide you on how to effectively use the solution manual for deep learning, not just homework copying.
3. To highlight the typical pitfalls in natural convection problems and how the official solutions address them.

Part 6: Sample Problem Walkthrough (From the Solution Manual)

Let us replicate the logic of the official solution manual for a classic Chapter 9 problem: A vertical plate 1.5 m high is maintained at 85°C in quiescent air at 25°C. Determine the heat transfer coefficient. The Solution Manual for Chapter 9 of Çengel’s

Solution Manual Approach (5th Edition):

1. Assumptions: Steady state, ideal gas, uniform surface temperature, negligible radiation.
2. Film Temperature: (T_f = (85+25)/2 = 55°C = 328 K).
3. Properties (Table A-15, 328 K):
   (k = 0.0283 , W/m·K), (\nu = 1.86 \times 10^-5 , m²/s), (Pr = 0.708), (\beta = 1/328 = 0.00305 , K^-1).
4. Grashof Number:
   (g = 9.81 , m/s²), (L = 1.5 , m), (\Delta T = 60 , K).
   (Gr = \frac9.81 \times 0.00305 \times 60 \times (1.5)^3(1.86 \times 10^-5)^2 = 2.18 \times 10^10).
5. Rayleigh Number: (Ra = Gr \times Pr = (2.18 \times 10^10) \times 0.708 = 1.54 \times 10^10).
6. Nusselt Correlation (Eq. 9-26, valid for all Ra):
   (Nu = \left(0.825 + \frac0.387 (1.54 \times 10^10)^1/6[1 + (0.492/0.708)^9/16]^8/27\right)^2).
7. Solve:
   (Ra^1/6 = (1.54e10)^0.1667 \approx 30.6).
   Denominator bracket: ((0.492/0.708) = 0.695), to the (9/16) power ≈ 0.79, plus 1 = 1.79.
   (1.79^8/27 \approx 1.19).
   (0.387 \times 30.6 / 1.19 = 9.96).
   Add 0.825 → (10.785). Square → (Nu \approx 116.3).
8. Heat Transfer Coefficient: (h = Nu \times k / L = 116.3 \times 0.0283 / 1.5 = 2.19 , W/m²·K).

Why this is valuable: The solution manual would provide all intermediate rounding and comment: "Note that if we assumed laminar only (Nu = 0.59 Ra^1/4), we would get Nu=67, a 42% error." This comparative insight is what separates a manual from a simple answer key.

Category 4: Natural Convection in Enclosed Spaces

Typical Problem: Double-pane window with air gap.

What the Solution Manual Shows:

• Effective thermal conductivity method ((k_eff)).
• Calculating the aspect ratio ((H/\delta)) and using the Hollands correlation for vertical enclosures.
• Handling concentric cylinders and spheres (a rare but challenging problem).

Problem 9-2: Horizontal Cylinder Analysis (Sample Problem)

Problem Statement: A 2-m-long, 0.5-m-diameter horizontal steam pipe passes through a large room. The surface temperature of the pipe is $150^\circ C$, and the room air temperature is $20^\circ C$. Determine the rate of heat loss from the pipe by natural convection.

Solution:

1. Properties: Film Temperature: $$ T_f = \frac150 + 202 = 85^\circ C = 358 , \textK $$ Properties of Air at $85^\circ C$ (interpolated from Table A-15): To explain the core concepts of Chapter 9

• $k = 0.0305 , \textW/m \cdot \textK$
• $\nu = 2.14 \times 10^-5 , \textm^2/\texts$
• $Pr = 0.705$
• $\beta = \frac1358 , \textK = 0.00279 , \textK^-1$

2. Analysis:

Step A: Rayleigh Number Characteristic length $L_c = D = 0.5 , \textm$.

$$ Ra_D = \fracg \beta (T_s - T_\infty) D^3\nu^2 Pr $$ $$ Ra_D = \frac(9.81)(0.00279)(150 - 20)(0.5)^3(2.14 \times 10^-5)^2 (0.705) $$ $$ Ra_D \approx 1.55 \times 10^9 $$

Step B: Correlation Since $Ra_D > 10^9$, the flow is turbulent. We use the correlation for a horizontal cylinder (Churchill and Chu):

$$ Nu = \left 0.6 + \frac0.387 Ra_D^1/6[1 + (0.559/Pr)^9/16]^8/27 \right^2 $$

Step C: Calculation Solving the denominator for air ($Pr = 0.705$): $$ [1 + (0.559/0.705)^9/16]^8/27 \approx 1.09 $$

Calculate the main term: $$ Nu = \left 0.6 + \frac0.387 (1.55 \times 10^9)^1/61.09 \right^2 $$ $$ Nu = \left 0.6 + \frac0.387 \times 17.781.09 \right^2 $$ $$ Nu = 0.6 + 6.31 ^2 = (6.91)^2 = 47.75 $$ which yield different heat transfer coefficients.

Solve for $h$: $$ h = \fracNu \cdot kD = \frac47.75 \times 0.03050.5 $$ $$ h \approx 2.91 , \textW/m^2 \cdot \textK $$

Step D: Heat Transfer Area $A_s = \pi D L = \pi(0.5)(2) = 3.14 , \textm^2$. $$ Q = h A_s (T_s - T_\infty) $$ $$ Q = (2.91)(3.14)(150 - 20) $$ $$ Q \approx 1189 , \textW $$

Result: The heat loss is approximately 1.19 kW.

Step 1: Attempt the Problem Blind

Spend 30 minutes on a problem with only the textbook and a NIST properties table. Write down what you know: (T_s), (T_\infty), geometry, (L_c). Identify the unknown: (h), (Q), or (T_s).

B. Lighting and Decor

With the shift from incandescent to LED lighting, thermal management in bulbs is a major textbook theme.

• Light Bulbs: Several problems analyze the heat dissipation from a spherical or cylindrical light bulb.
  • Solution Methodology: The manual solves for the surface temperature of a bulb assuming it dissipates heat purely by natural convection (and radiation). It contrasts the high heat output of incandescent bulbs with the lower heat output of LEDs, often citing lifestyle benefits (cool to the touch, energy savings).
• Electronic Entertainment Systems: Problems involving the cooling of TV components or gaming consoles (modeled as rectangular boxes) are common.
  • Key Insight: The solutions differentiate between a "hot surface facing up" (top of the console) and "hot surface facing down" (bottom), which yield different heat transfer coefficients.

Type 2: Horizontal Cylinder (Pipe without insulation)

The Setup: A hot steam pipe horizontally suspended in a cold room.

The Solution Manual Trick: The correlation changes.

• $Nu_D = \left(0.6 + \frac0.387 Ra_D^1/6[1 + (0.559/Pr)^9/16]^8/27\right)^2$ (for all $Ra_D$)

Students searching for "Chapter 9 solutions" often misplace exponents here. The solution manual clarifies that the exponent inside the denominator is $8/27$, not $1/3$.