The Setup: Two vertical plates separated by distance $L_c$ with a temperature difference.
The Solution Manual Insight:
The physics here involves conduction and radiation competing with natural convection. The effective thermal conductivity is key.
Problem Statement:
Consider a vertical 0.2 m high, 0.5 m wide plate maintained at a uniform surface temperature of $T_s = 80^\circ C$. The plate is exposed to quiescent air at $T_\infty = 20^\circ C$. Determine the rate of heat transfer from the plate by natural convection.
Solution:
1. Assumptions:
2. Properties:
The film temperature is:
$$ T_f = \fracT_s + T_\infty2 = \frac80 + 202 = 50^\circ C $$
From the thermophysical property tables (Table A-15 for Air at $50^\circ C$):
3. Analysis:
Step A: Calculate the Rayleigh Number ($Ra_L$)
The characteristic length $L$ for a vertical plate is its height ($L = 0.2 , \textm$).
$$ Ra_L = \fracg \beta (T_s - T_\infty) L^3\nu^2 Pr $$
Substituting values:
$$ Ra_L = \frac(9.81)(0.003096)(80 - 20)(0.2)^3(1.798 \times 10^-5)^2 (0.7228) $$
$$ Ra_L = \frac9.81 \times 0.003096 \times 60 \times 0.0083.233 \times 10^-10 (0.7228) $$
$$ Ra_L \approx 3.27 \times 10^7 $$
Step B: Select Correlation
Since $Ra_L < 10^9$, the flow is laminar.
We use the correlation for a vertical isothermal plate (Churchill and Chu):
$$ Nu = \frachLk = \left 0.68 + \frac0.670 Ra_L^1/4[1 + (0.492/Pr)^9/16]^4/9 \right $$
Step C: Calculate Nusselt Number and $h$
For air, $Pr \approx 0.72$, so the denominator term $[1 + (0.492/Pr)^9/16]^4/9 \approx 1.06$.
Simplifying for air (or solving strictly):
$$ Nu = 0.68 + \frac0.670 (3.27 \times 10^7)^1/4[1 + (0.492/0.7228)^9/16]^4/9 $$
$$ Nu = 0.68 + \frac0.670 \times 75.361.06 $$
$$ Nu = 0.68 + 47.63 = 48.31 $$
Now, solve for $h$:
$$ h = \fracNu \cdot kL = \frac48.31 \times 0.027350.2 $$
$$ h \approx 6.61 , \textW/m^2 \cdot \textK $$
Step D: Calculate Heat Transfer Rate
$$ Q = h A_s (T_s - T_\infty) $$
Area $A_s = (\textheight)(\textwidth) = 0.2 \times 0.5 = 0.1 , \textm^2$.
$$ Q = (6.61)(0.1)(80 - 20) $$
$$ Q = 39.66 , \textW $$
Result: The rate of heat transfer is approximately 39.7 W.
The 5th edition of Cengel’s text is renowned for its clear examples, but Chapter 9 introduces a distinct shift in problem-solving strategy. In forced convection, you typically calculate the Reynolds number first. In natural convection, the Grashof number (Gr) takes center stage. It represents the ratio of buoyancy force to viscous force.
Key topics covered in this chapter include:
Before delving into specific lifestyle applications, the solution manual establishes the core physics required to solve these problems:
The solution manual provides step-by-step derivations for these formulas, which are then applied to the "Lifestyle and Entertainment" problems to demonstrate how thermal management works in everyday objects.
A significant portion of the lifestyle problems focuses on the heating and cooling of residential spaces.
Window Insulation: Problems often involve calculating heat loss through single or double-pane windows.
When you locate the correct solution manual heat and mass transfer cengel 5th edition chapter 9, you will find solutions for approximately 50–70 problems, ranging from conceptual discussions to complex numerical analyses. Here is a breakdown of the typical problem categories and how the manual approaches them.
Solution Manual Heat And Mass Transfer Cengel 5th Edition Chapter 9 Today
The Setup: Two vertical plates separated by distance $L_c$ with a temperature difference.
The Solution Manual Insight:
The physics here involves conduction and radiation competing with natural convection. The effective thermal conductivity is key.
Problem Statement:
Consider a vertical 0.2 m high, 0.5 m wide plate maintained at a uniform surface temperature of $T_s = 80^\circ C$. The plate is exposed to quiescent air at $T_\infty = 20^\circ C$. Determine the rate of heat transfer from the plate by natural convection.
Solution:
1. Assumptions:
2. Properties:
The film temperature is:
$$ T_f = \fracT_s + T_\infty2 = \frac80 + 202 = 50^\circ C $$
From the thermophysical property tables (Table A-15 for Air at $50^\circ C$):
3. Analysis:
Step A: Calculate the Rayleigh Number ($Ra_L$)
The characteristic length $L$ for a vertical plate is its height ($L = 0.2 , \textm$).
$$ Ra_L = \fracg \beta (T_s - T_\infty) L^3\nu^2 Pr $$
Substituting values:
$$ Ra_L = \frac(9.81)(0.003096)(80 - 20)(0.2)^3(1.798 \times 10^-5)^2 (0.7228) $$
$$ Ra_L = \frac9.81 \times 0.003096 \times 60 \times 0.0083.233 \times 10^-10 (0.7228) $$
$$ Ra_L \approx 3.27 \times 10^7 $$
Step B: Select Correlation
Since $Ra_L < 10^9$, the flow is laminar.
We use the correlation for a vertical isothermal plate (Churchill and Chu):
$$ Nu = \frachLk = \left 0.68 + \frac0.670 Ra_L^1/4[1 + (0.492/Pr)^9/16]^4/9 \right $$
Step C: Calculate Nusselt Number and $h$
For air, $Pr \approx 0.72$, so the denominator term $[1 + (0.492/Pr)^9/16]^4/9 \approx 1.06$.
Simplifying for air (or solving strictly): The Setup: Two vertical plates separated by distance
$$ Nu = 0.68 + \frac0.670 (3.27 \times 10^7)^1/4[1 + (0.492/0.7228)^9/16]^4/9 $$
$$ Nu = 0.68 + \frac0.670 \times 75.361.06 $$
$$ Nu = 0.68 + 47.63 = 48.31 $$
Now, solve for $h$:
$$ h = \fracNu \cdot kL = \frac48.31 \times 0.027350.2 $$
$$ h \approx 6.61 , \textW/m^2 \cdot \textK $$
Step D: Calculate Heat Transfer Rate
$$ Q = h A_s (T_s - T_\infty) $$
Area $A_s = (\textheight)(\textwidth) = 0.2 \times 0.5 = 0.1 , \textm^2$.
$$ Q = (6.61)(0.1)(80 - 20) $$
$$ Q = 39.66 , \textW $$
Result: The rate of heat transfer is approximately 39.7 W.
The 5th edition of Cengel’s text is renowned for its clear examples, but Chapter 9 introduces a distinct shift in problem-solving strategy. In forced convection, you typically calculate the Reynolds number first. In natural convection, the Grashof number (Gr) takes center stage. It represents the ratio of buoyancy force to viscous force. $Pr \approx 0.72$
Key topics covered in this chapter include:
Before delving into specific lifestyle applications, the solution manual establishes the core physics required to solve these problems:
The solution manual provides step-by-step derivations for these formulas, which are then applied to the "Lifestyle and Entertainment" problems to demonstrate how thermal management works in everyday objects.
A significant portion of the lifestyle problems focuses on the heating and cooling of residential spaces.
Window Insulation: Problems often involve calculating heat loss through single or double-pane windows.
When you locate the correct solution manual heat and mass transfer cengel 5th edition chapter 9, you will find solutions for approximately 50–70 problems, ranging from conceptual discussions to complex numerical analyses. Here is a breakdown of the typical problem categories and how the manual approaches them.
