Each fin takes up an area on the surface of the pipe

Each fin takes up an area on the surface of the pipe = pi * (3.0/12 ft) * 0.1/12 ft = 0.00654 ft
There are 24 * 40 fins on the pipe = 960.
The total surface area of the original pipe that is now covered by fins is 6.28 ft 2
Each fin has two flat surfaces in the shape of a disc and an edge that is 0.1 inches thick with an outside diameter
of 4 inches.
The disc shapes create new surface area of 960 fins * 2 surfaces/fin * pi/4 * {(4/12 ft) 2 – (3/12 ft) 2 } (flat edges)
= 73.3 ft2
PLUS the edge at 4 inches = 960 * pi * 4/12 ft * 0.1/12 ft = 8.38 ft 2
Total surface area of the finned pipe = 31.4 – 6.28 + 73.3 + 8.38 = 106.8 ft 2
Solving for q = 31.7 * 106.8 ft 2 * (250 – 88 F) = 548,000 Btu/h
And the surface effectiveness, ε
f
= 548,000 / 161,000 = 3.4

CHAPTER 4
2. Given a 2-D square block of solid material with dimensions 20 cm on each side. It is heated on the bottom surface (y =
0), and is insulated on the left side (x = 0) so that no heat is transferred across this surface. On the right side (x =
20 cm), it is held isothermal at 50 o C. On the top surface (y = 20 cm), the material is exposed to air at 25 o C with a
convective heat transfer coefficient of 65 W/m 2 -K. The material’s thermal conductivity is 28 W/m-K. Divide the
material into a grid with Δx = Δy = 10 cm. You may assume constant k, and no heat generation inside the
material.
The following temperatures are known:
T(0,20) = 35
o
C; T(20,20) = 50
o
C; T(10,10) = 46
o
C;
T(20,10) = 50 o C; T(0,0) = 64 o C; T(20,0) = 50 o C.
The temperatures T(10,20), T(0,10) and T(10,0) are unknown.
Use the equations derived
A. What is T(0,10)?
B. What is T(10,20)?
C. What is the steady state rate of heat transfer from the top surface (y = 20 cm) to the air?
(You may assume Δ z = 20 cm; Treat each node exposed to the surface as an independent calculation
using Newton’s Law of Cooling, with either Δx = 10 cm or 20 cm, depending on the node position)
DRAW A DIAGRAM!
35 T2 50
T4 46 50
64 T8 50
A. This node (T4) is at an adiabatic edge with other internal nodes known. (Case 3 with h = 0)
Thus, T(0,10) = ¼ (35 + 64 + 2 * 46) = 191/4 = 47.75 o C. (or ~ 48 C)
B. This node (T2) is a surface node exposed to convection. Solve using Case 3 equation from the handout.
(2 * 46 + 35 + 50) + 2 * 65 W/m2-K (0.1 m)/(28 W/m-K) * (25) = 2 * {(65*0.1)/28 + 2} * T2
(NOTE: you can use oC or K in this equation, just be consistent. ALSO- watch units on the cm/m conversion.)
Solve for T2.
T2 = 42.2 C (check- does it make sense? Yes- between 35 and 50).
C. To determine SS heat transfer, use Newton’s law of cooling, with the node temperatures representing the
temperatures for the area around each node; i.e., for T(0,20), this represents the temperature of the block from
T(0,20) to T(5,20). T(10,20) represents the temperature from (5,20) to (15,20) and T(20,20) is the temperature
from (15,20) to (20,20).
Thus, q = h * dz * {(0.05) (35-25) + (0.10) (42.2 – 25) + (0.05) (50-25)}
q = 65 W/m
2
-K (0.20 m) * {0.5 + 1.72 + 1.25} m-K
q = 45.10 W

3. The steady-state temperatures for selected nodes in a solid are marked below in o C. The solid has a thermal
conductivity of 1.5 W/m-K, and the grid spacing and boundary conditions are noted in the figure.
(A) Determine node temperatures T
1
, T
2
and T
3
.
(B) If the slab is 1 m thick (dz out of the page), what is the rate of heat transfer from the right surface to the
fluid?
[Don’t forget about the node at the bottom edge that is at 200
o
C]
(B)
Or, as the question asked, q
conv
for a 1 m thick slab would be 743 W.

4. A 2-dimensional steady state heat transfer problem is solved using finite
divided differences and the grid at right:
In the grid, Δx ≠ Δy, and the material has a depth Δz.