A long stainless steel rod (k =16 W/m°C) has a equilateral triangle cross section. It is
attached to a plane wall which is maintained at a temperature of 450 °C. The surrounding
environment is at 50 °C. Compute the heat dissipated by the rod, assume the value of
surface conductance h is 90 W/m²°C.
90 mm
S mm
(a) 3.12 kW
(b) 1.7 w
(c) 5.12 W
(d) 6.12 W
Answers
Answer:
c. I hope this will help you
Explanation:
. ...
Given info : A long stainless steel rod (k =16 W/m°C) has a equilateral triangle cross section. It is
attached to a plane wall which is maintained at a temperature of 450 °C. The surrounding
environment is at 50 °C.
To find : heat dissipated by the rod is ..
solution : side length of equilateral triangle, s = 5 mm = 5 × 10^-3 m
so cross sectional area of rod, A = √3/4 s²
= √3/4 × (5 × 10^-3)²
= 25√3/4 × 10^-6 m²
= 10.825 × 10^-6 m²
perimeter of triangle, P = 3a = 3 × 5 × 10^-3 = 1.5 × 10^-2
length of rod, l = 90 mm = 90 × 10^-3 m
change in temperature, ∆T = 450°C - 50°C = 400°C
value of surface conductance, h = 90 W/m²°C
thermal conductance of steel , k = 16 W/m°C
using formula,
heat dissipated by the rod, Q = {√(hkPA)} ∆T
= √(90 × 16 × 1.5 × 10^-2 × 10.825 × 10^-6) × 400
= 153 × 10^-4 × 400
= 61200 × 10^-4
= 6.12 W
Therefore the heat dissipated by the rod is 6.12 W i.e., option (d) 6.12 W
