Question

please show work

use chart to get correct length and fin efficiency

4. Straight metal fins (k = 45 W/m °C) of triangular profile installed on a l m 1 m wall. The wall is maintained at 200 °C and are exposed to a convection environment at 25 °C with h = 204 W/m² °C. The fins have a length of 8 mm and a thickness of 2 mm. The space between two adjacent fins is 3 mm. Calculate (a) the fin efficiency, (8 points) (b) the heat transfer from one (1) fin, (7 points) (c) the heat transfer from one open space between two adjacent fins (5 points) (d) the heat transfer rate from this finned wall, (5 points) (e) the fin effectiveness, (5 points) and (f) the device (wall) effectiveness. (5 points)

Answer 1

DATE: PAGE: Finned surface Tempentare To Q1 (from 1 fon (surso cuzin) d I Emply surface em emitting az NE YA++) Total fins On wall. Total emty suo faces on wall.

Fin Type		Triangular
Fin length, L	mm	8
	m	0.008
Fin Thickness, t	mm	2
	m	0.002
Width of the wall,W	m	1
Length of the wall,l	m	1
slant surface area of one fin, A1	m^2	0.016125
Cross section surface area of one fin, A2	m^2	0.000008
Total surface area of one fin, A=A1+2*A2	m^2	0.016141
Wall temperature, To	DegC	200
Surrounding temperature, T∞	DegC	25
Conductivity of Fin material, k	W/m-DegC	45
Coeficient of convective heat transfer, h	W/m^2-DegC	204
S=(L^1.5)(h/kA2)^.5		0.54
Efficiency, η (From graph, against S, Calculated above)		0.85
To-T∞	DegC	175
Heat transfer from 1 fin Q1=ηAh(To-T∞)	W	489.78
Open space length, d	mm	3
	m	0.003
Area of single open space A3= d*W	m^2	0.003
Heat transfer by open space Q2=h*A3*(To-T∞)	W	107.1
Total fins, n= l/(t+d)	No's	200
Total heat transfer through fins, n*Q1	W	97957
Total heat transfer empty surface, n*Q2	W	21420
Total heat transfer from wall, Q= n*Q1+n*Q2	W	119377
Effictiveness of a single fin
Base area of fin A0, t*w	m^2	0.002
Q min from base wall= h*t*W*(To-T∞)	W	71.4
Effective ness of fins Q1/Qmin		6.86
Effictiveness of whole surface
Q min from whole wall= h*l*W*(To-T∞)	W	35700
Effectiveness of fins Q/Qmin		3.34