Department Of Chemical Engineering

3rd Semester 2024

Assignment No 2
Submitted To: Engr. Syed Kamal Zafar
Course Title: Heat Transfer-I
Course Code: CHE-234
Submitted By:
Name: Soban Babar
Roll No: 23013123-010
 Pr
oblem 1:
Consider a 3 m high, 5 m wide, and 0.3 m thick wall whose thermal conductivity is k =0.9
W/m.°C. On a certain day, the temperatures of the inner and the outer surfaces of the wall are
measured to be 16°C and 2°C, respectively. Determine the rate of heat loss through the wall on
that day.

Solution:

To determine the rate of heat loss through the wall, we can use Fourier’s law of heat conduction:

kA ( T 1−T 2 )
Q=
L

Where:

 Q is the rate of heat loss (in watts, W),

W
 K is the thermal conductivity of the material (in ),
m℃
 A is the surface area of the wall (in square meters, m²),
 T1is the temperature on the inner surface (in °C),
 T2 is the temperature on the outer surface (in °C),
 L is the thickness of the wall (in meters, m).

Given:

 k=0.9 W /m°Ck
 T1=16℃
 T2=2℃
 L=0.3 m
 The wall is 3 m high and 5 m wide.

Step 1: Calculate the surface area A

The surface area of the wall can be calculated as:

A=height×width=3 m×5 m=15 m2

Step 2: Apply Fourier's Law

Now, substitute the known values into the heat conduction formula:

0.9 ×15 × ( 16−2 )

Q=
0.3

Q = 630W

Final Answer:

The rate of heat loss through the wall is 630 W

Problem 2:
At steady-state conditions, a slab with thickness of 30 mm conducts heat at the rate of 10 W. The
width and height of the slab are 10 in and 12 in, respectively. If the temperatures at the cold and
hot ends are maintained at 30°C and 50°C, calculate the thermal conductivity of the material in
W cal Btu
the units of , , and
mk s .cm ℃ hft .° F

Solution:

We are tasked with finding the thermal conductivity k of the material of a slab based on the heat
transfer rate and the given dimensions.

The equation for steady-state heat conduction is given by Fourier’s law:

kA ( T 1−T 2 )
Q=
L

Where:

 Q is the rate of heat transfer (in watts, W),

 k is the thermal conductivity of the material (in W/m·K),
 A is the cross-sectional area through which heat is conducted (in m²),
 T1 is the temperature at the hot end (in °C),
 T2 is the temperature at the cold end (in °C),
 L is the thickness of the slab (in meters, m).

Given:

 Q=10 W
 T1=50℃
  T2=30℃
 L=30 mm = 0.003m
 The width of the slab is 10 in
 The height of the slab is 12 in

We will need to calculate the area A in square meters, and then solve for k.

Step 1: Convert dimensions to meters

 Width=10 in=10×0.0254 m = 0.254 m

 Height=12 in =12×0.0254 m = 0.3048 m

Step 2: Calculate the area A

The cross-sectional area A is:

A = width×height = 0.254 m×0.3048 m = 0.0775 m2

Step 3: Solve for thermal conductivity k

Rearranging Fourier’s law to solve for k:

QL
k=
A ( T 1−T 2 )

Substitute the known values:

10 ×0.03
k=
0.0775 (50−30 )

k ≈ 0.1935W/mK

Step 4: Convert k to other units

cal
(a) In
s .cm ℃

1 calorie = 4.184 joules, and 1 meter = 100 cm.

W cal
 To convert from to
mk s .cm ℃

cal
1 W=0.239 1 m=100 cm
s
Thus,

w cal 1 1 cm cal
k=0.1935 ×0.239 × k = 0.000462
mK s W 100 m s .cm ℃

Btu
(b) In
hft .° F

Btu
1 W = 3.412 , 1 m = 3.2808 ft , and 1 K = 1 °F for temperature differences.
h

w Btu 1 ft
k=0.1935 ×3.412 ×W×
mK h 3.2808 m
Btu
k ≈ 0.204
hft .° F

Final Results:

 k≈0.1935 W/mK
cal
 k≈0.000462
s .cm ℃
Btu
 k≈0.204
hft .° F

Problem 3:
A layer of pulverized cork 6 in thick is used as a layer of thermal insulation in a flat wall. The
temperature of the cold side of the cork is 40 °F and that of the warm side is 180 °F. The thermal
conductivity of the cork at 32 °F is 0.021 Btu/ft-h-°F and that at 200 °F is 0.032 Btu/ft-h.°F. The
area of the wall is 25 ft². What is the rate of heat flow through the wall in Btu per hour?

Solution:

To determine the rate of heat loss through the wall, we can use Fourier’s law of heat conduction:

kA ( T 1−T 2 )
Q=
L

Where:

 Q is the rate of heat loss (in watts, W),

W
 K is the thermal conductivity of the material (in ),
m℃
  A is the surface area of the wall (in square meters, m²),
 T1is the temperature on the inner surface (in °C),
 T2 is the temperature on the outer surface (in °C),
 L is the thickness of the wall (in meters, m).

Given:

 Thickness of the cork, L= 6 in = 0.5 ft

 Temperature on the warm side, T1=180 °F
 Temperature on the cold side, T2=40 °F
 Area of the wall, A=25 ft 2
Btu
 Thermal conductivity at 32 °F, k 1=0.021
ft −h−° F
Btu
 Thermal conductivity at 200 °F, k2=0.032
ft −h−° F

Step 1: Average the thermal conductivity

The thermal conductivity of the cork changes with temperature, so we'll estimate the average
thermal conductivity between the temperatures of 40 °F (cold side) and 180 °F (warm side).

k 1+k 2 0.021+0.032 Btu

kavg= = =¿0.0265
2 2 ft −h−° F

Step 2: Apply Fourier’s law

Now, we can use the Fourier's law formula to calculate the rate of heat flow:

k avg A (T 1−T 2 )
Q=
L

0.0265 ×25 × ( 180−40 )

Q=
0.5

Btu
Q=259.35
h

Problem 4:
An industrial furnace wall is composed of three plane walls, namely 0.10 m thick fireclay wall,
0.15 m wide insulation wall, and 0.20 m wide common brick wall. Thermal conductivities of
fireclay brick, insulating brick, and common brick are 1.5 W/mK, 0.20 W/mK, and 1.0 W/m.K,
respectively. The inside composite wall temperature is 1300 K and the outside composite wall
temperature is 300 K. Calculate the heat loss per unit area and the temperatures at the junctions
of the individual walls. Also calculate the thermal resistance of each wall.
Solution:

Step 1: Formula for heat transfer rate

The heat transfer through a plane wall can be calculated using Fourier’s law, which in terms of
thermal resistance is given by:

ΔT
Q=
Rtotal

Where:

 Q is the rate of heat transfer per unit area (W/m²),

 ΔT is the temperature difference between the inside and outside (K),
 Rtotal is the total thermal resistance of the composite wall (m²·K/W).

The total thermal resistance is the sum of the individual thermal resistances of each layer:

Rtotal=R fireclay + R insulation + Rcommon brick

The thermal resistance of a single layer is given by:

L
R=
k

Where:

 L is the thickness of the wall (m),

 k is the thermal conductivity of the material ( mWK )
Step 2: Calculate the thermal resistance of each wall

1. Fireclay wall:

 Thickness: Lfireclay =0.10 m

W
 Thermal conductivity: K fireclay =1.5
m, K

0.10 2 K
R fireclay= = 0.0667 m ⋅
1.5 W

2. Insulation wall
  Thickness Lfireclay =0.20 m
W
 Thermal conductivity K insulation =0.20
mK
2
0.15 m K
Rinsulation = =0.75
0.20 W

3. Common brick wall:

 Thickness: Lbrick =0.20 m

W
 Thermal conductivity:k Brick =1.0
mK
2
0.20 m K
R Brick= =0.20
1.0 W

Step 3: Calculate the total thermal resistance

Now, we add up the thermal resistances of all three walls:

Rtotal=R fireclay + R insulation + Rcommon brick

Rtotal=0.0667+0.75+0.20

2
1.0167 m K
Rtotal=
W

Step 4: Calculate the heat loss per unit area

The temperature difference across the entire composite wall is:

ΔT =T inside −T outside=1300 K−300 K =1000 K

Now, we can calculate the heat loss per unit area using:

ΔT
Q=
Rtotal

1000 W
Q= =983.5 2
1.016 m

Step 5: Calculate the temperatures at the junctions of the individual walls

To find the temperatures at the junctions of the walls, we will calculate the temperature drop
across each layer of the wall using the formula:

∆ T l ayer =Q × Rlayer

1. Temperature drop across the fireclay wall:

2
W m K
∆ T l ayer =983.5 2
×0.0667 = 65.7K
m W

Thus, the temperature at the junction between the fireclay and insulation walls is:

T Junction ,1 =1300 K −65.7 K = 1234.3K

2. Temperature drop across the insulation wall:

2
W m K
T insulation=983.5 2 × 0.7 5 = 737.6K
m W

Thus, the temperature at the junction between the insulation and common brick walls is:

T Junction ,2 =1300 K −737.6 K = 496.7K

3. Temperature drop across the common brick wall:

2
W m K
T Brick =983.5 2
× 0. 20 = 196.7K
m W

Thus, the temperature at the outside surface of the common brick wall is:

T outside=496.7 K−196.7 K = 300K

Final Answer:

W
 Heat loss per unit area: 983.5 2
m
 Temperature at the junction between fireclay and insulation walls: 1234.3 K
 Temperature at the junction between insulation and common brick walls: 496.7 K
2
m K
 Thermal resistance of fireclay wall: 0.0667
W
2
m K
 Thermal resistance of insulation wall: 0.75
W
2
m K
 Thermal resistance of common brick wall: 0.20
W