228k views
5 votes
An object of irregular shape has a characteristic length of � = 1.00 [�] and is maintained at a uniform surface temperature of �. = 400 [�]. When placed in atmospheric air at a temperature of �- = 300 [�] and moving with a velocity of � = 100 [ 3 . ], the average heat flux from the surface to the air is 13,000[ 2 3&]. If a second object of the same shape, but with a characteristic length of � = 5.00 [�], is maintained at a surface temperature of �. = 400 [�] and is placed in atmospheric air at �- = 300 [�], what will the value of the average convection coefficient be if the air velocity is � = 20 [ 3 . ]?

User Bhupen
by
3.1k points

1 Answer

4 votes

The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Step-by-step explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

Thus,

h'_2 = [20,000/(400 - 300)]•(1/5)

h'_2 = 40 W/K.m²

User Danivovich
by
2.9k points