Water at 120oc boils inside a channel with a flat surface measuring 45 cm x 45 cm. air at 62 m/s and 20oc flows over the channel parallel to the surface. determine the heat transfer rate to the air. neglect wall thermal resistance.

Q = 2.532 x 10³ W

Explanation:

The properties of air at the film temperature of

T(f) = (T(s) + T()∞)/2

T(f) = (120 + 20)/2 = 70 °C

From the table of properties of air at T = 20 °C we know that ρ = 1.028 kg/m³ , k = 0.02881W/mK, Pr = 0.7177 , v = 1.995 x 10⁻⁵ m²/s

Calculate the Reynold’s Number

Re(l) = V x L/ν, where V is the speed of air flowing, L is the length of the surface in meters and ν is the kinematic viscosity

Re(l) = 62 x 0.45/(1.995 x 10⁻⁵) = 1.398 x 10⁶

Re(l) is greater than the critical Reynold Number. Thus, we have combined laminar and turbulent flow and the average Nusselt number for the entire plate is determined by

Nu = (0.037 x (Re(l)^0.8) - 871) x Pr^1/3, where Pr is the Prandtl’s Number

Nu = (0.037 x (1.398 x 10⁶)^0.8) – 871) x (0.7177)^1/3

Nu = 1952.85

We also know that

Nu = h x L/k, where h is the heat transfer coefficient, L is the length of the surface and  k is the thermal conductivity

Rearranging to solve for h

h = (Nu x k)/l

h = 1952.85 x 0.02881/0.45 = 125.03 W/m²C

The heat transfer rate Q, may be determined by

Q = h x A x (T(s) - T(∞))

Q = 125.03 x 0.45 x 0.45 x (120 - 20)

Q = 2.532 x 10³ W

The heat transfer rate is 2.532 x 10³ W to the air.

twist boundary

explanation: