Answer:
5945.27 W per meter of tube length.
Explanation:
Let's assume that:
- Steady operations exist;
- The heat transfer coefficient (h) is uniform over the entire fin surfaces;
- Thermal conductivity (k) is constant;
- Heat transfer by radiation is negligible.
First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:
Q = A*h*(Tb - T∞)
A is the area of the section of the tube,
A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.
A = π*0.05*1 = 0.1571 m²
Q = 0.1571*40*(130 - 25)
Q = 659.73 W
Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:
Qfin = nf*Afin*h*(Tb - T∞)
Afin = 2π*(r2² - r1²) + 2π*r2*t
r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).
Afin = 0.006 m²
Qfin = 0.97*0.006*40*(130 - 25)
Qfin = 24.44 W
The heat transferred at the space between the fin and the tube will be:
Qspace = Aspace*h*(Tb - T∞)
Aspace = π*D*S, where D is the tube diameter and S is the space between then,
Aspace = π*0.05*0.003 = 0.0005
Qspace = 0.0005*40*(130 - 25) = 1.98 W
The total heat is the sum of them multiplied by the total number of fins,
Qtotal = 250*(24.44 + 1.98) = 6605 W
So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.
