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When evaporation takes place at the liquid-vapor interface, the heat transfer is solely due to free convection and the film coefficient follows the relation
The functions f 1 and f 2 depend upon the geometry of the heating surface.
Fritz criterion is given by
Fritz formulated the following formula for water boiling at atmospheric pressure in free convection in a vertical tube headed from outside.
A 0.10 cm diameter and 15 cm long wire has been laid horizontally and submerged in water at atmospheric pressure. The wire has a steady state voltage drop of 14.5 V and a current of 42.5 A. Determine the heat flux of the wire.
The following equation applies for water boiling on a horizontal submerged surface
H = 1.54 (Q/A) 0.75 = 5.58 (d t) 3 W/m2 K where Q/A is the heat flux rate in W/m2 and d t is the temperature difference between surface and saturation
Q = E I = 616.25 W and A = 4.71 * 10 -4 m2.
Consider the above problem, find the excess temperature of the wire
1.54 (1.308 * 10 6) 2 = 5.58 (d t) 3.
Natural convection heat transfer coefficients over surface of a vertical pipe and a vertical flat plate for same height. What is/are the possible reasons for this?
(i) Same height
(ii) Both vertical
(iii) Same fluid
(iv) Same fluid flow pattern
Select the correct answer
The fluids must be same so their flow pattern.
The heat flux in nucleate boiling varies in accordance with
Q/A = δ f h f g [(p f – p g) g/σ] 0.5 [C f d t/h f g p C s f] 3.
In nucleate pool boiling, the heat flux depends on
The heat flux must depends on liquid properties material and condition of the surface.
Identify the wrong statement with respect to boiling heat transfer?
For boiling to occur, the heated surface must be exposed to a liquid and maintained at a temperature higher than the saturation temperature of the liquid.
Estimate the peak heat flux for water boiling at normal atmospheric pressure. The relevant thermo-physical properties are
p f (liquid) = 958.45 kg/m3
p g (vapor) = 0.61 kg/m3
h f g = 2.25 * 10 6 J/kg
σ = 0.0585 N/m
(Q/A) = 0.18 p g h f g [σ (p f – p g)/p g 2] 0.25.
A 1.0 mm diameter and 300 mm long nickel wire is submerged horizontal in water at atmospheric pressure. At burnout, the wire has a current of 195 A. Calculate the voltage at burnout. The relevant thermos-physical properties are
p f (fluid) = 959.52 kg/m3
p g (vapor) = 0.597 kg/m3
h f g = 2257000 J/kg
σ = 0.0533 N/m
(Q/A) MAX = 1480000 W/m2. Let E b be the voltage at burnout. Then electric energy input to wire is E b I = 195 E b W.