Answer:
a) the rate of heat transfer from the pipe to the air is 23.866 watts
b) YES, the rate of heat transfer changes to 3518.61 watt
Explanation:
Given that:
steam is saturated at 17.90 bar.
the pipe is stainless steel and has an outside diameter of 6.75 cm
length = 34.7 m
Air flows over the pipe at 7.6 m/s
Bulk fluid temperature of 27°C
we know that
hD/k = 0.028 (Re)^0.8 (Pr)^0.33
Outside diameter of pipe = 6.75 cm
length of the pipe = 34.7 m
velocity of air = 7.6 m/s
Cp of air = 1.005 kJ/Kgk
viscosity of air = 1.81 × 10⁻⁵ kg/(m.sec)
thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k
so as
hD/k = 0.028 (Re)^0.8 (Pr)^0.33
hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33
(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))
h = 0.0414 w/m².k
a)
Now to find the rate of heat transfer Q
Q = hAΔT
Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)
Q = 23.866 watts
therefore the rate of heat transfer from the pipe to the air is 23.866 watts
b)
Now the flow direction changes to parallel flow, then
(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))
h = 6.1036 w/m².k
so from the steam table, saturated steam at 17.70 bar, temperature of steam will be 105.383°C
so to find the rate of heat transfer Q
Q = hAΔT
Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)
Q = 3518.61 watt
Therefore the rate of heat transfer changes to 3518.61 watt
