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Nucleate Boiling Example Video Lecture - Heat Transfer - Mechanical Engineering
FAQs on Nucleate Boiling Example
| 1. What is nucleate boiling? |  |
Ans. Nucleate boiling is a phase change phenomenon that occurs when a liquid boils and forms bubbles on a heated surface. It is characterized by the rapid formation and collapse of small bubbles, known as nucleation sites, on the surface.
| 2. How does nucleate boiling differ from film boiling? | |
Ans. Nucleate boiling and film boiling are two different modes of boiling. In nucleate boiling, small bubbles are formed on the heated surface and provide efficient heat transfer. In film boiling, a continuous vapor film forms on the surface, reducing heat transfer efficiency.
| 3. What factors affect the nucleate boiling heat transfer coefficient? | |
Ans. Several factors influence the nucleate boiling heat transfer coefficient, including the surface roughness, temperature difference between the surface and the liquid, pressure, liquid properties (e.g., viscosity), and the presence of dissolved gases or impurities in the liquid.
| 4. How is nucleate boiling used in industrial applications? | |
Ans. Nucleate boiling is widely used in various industrial applications, such as heat exchangers, refrigeration systems, power plants, and chemical processes. It provides efficient heat transfer and helps in cooling or condensing processes.
| 5. What are the challenges in nucleate boiling research? | |
Ans. Nucleate boiling research faces challenges in understanding and predicting the critical heat flux (CHF), which is the point at which the heat transfer significantly deteriorates due to the formation of a vapor film. Additionally, studying nucleate boiling in complex geometries or with different fluids requires advanced experimental techniques and computational models.
Text Transcript from Video
This is a pool boiling problem where we have
ten copper tubes that are 15 millimeters in
diameter and each one is 0.5 meters long,
and they're used to boil saturated water at
1 atmosphere. And if the tubes are operated
at 80% of the critical heat flux, what we
want to know is what is the vapor production
rate with these ten tubes. In addition, we
want to know what is the corresponding
surface temperature of the tubes. So the heat
transfer rate is the surface flux times the
surface area, which is the surface flux times
the number of tubes times pi D L. And this
is equal to m dot h sub fg. And we should
realize that this is nucleate boiling because we
are only at 80% of the critical heat flux. So what
we're looking for is this vapor production
rate, or our m dot, which is our heat flux
times the number of tubes times pi D L, divided
by h sub fg, or the heat of vaporization.
So the first thing we're going to need to
do is find this heat flux, which is, again,
0.8 times q max. So let's do our properties
for our saturated water. So our rho sub l
is 957.9 kilograms per meter cubed, that's
the density of the liquid. The heat capacity
of the liquid is 4217 joules per kilogram
K. The viscosity of the liquid is 279 times
10 to the minus 6 newton seconds per meter
squared. The Prandtl number of the liquid
is 1.76, and the heat of vaporization is 2257
kilojoules per kilogram. And we're actually
going to change that to 2257 times ten to
the third joules per kilogram since everything
else is in joules. Finally, we have the surface
tension, which is 58.9 times 10 to the minus
3rd newtons per meter. And our density of
the vapor, which is 0.5955 kilograms per meter
cubed. So the equation for the critical heat
flux is 0.149 h sub fg rho of the vapor, the
surface temperature times gravity times the
difference in densities, over the density
of the vapor squared, and this is all raised
to the 1/4. And when we calculate that out,
we get 1.26 times 10 to the 6 watts per meter
squared. So our q, or our heat flux, is 1.008
times 10 to the 6 watts per meter squared.
And when we put that number times ten times
pi D L, and divide it by our heat of vaporization,
our vapor production rate, or m dot, is 0.105
kilograms per second. In order to find the
corresponding surface temperature of a tube,
we're going to need to find our difference
in excess temperature. And so we're going
to use the Rohsenow correlation, which equals
C sub sf h sub fg Prandtl to the n, divided
by our heat capacity of the liquid, and this
is multiplied by our surface heat flux divided
by our liquid viscosity h sub fg, all raised
to the 1/3, and this is multiplied by the
surface tension divided by gravity and the
difference in the densities, and this whole
thing is raised to the 1/6. Now the C sub
sf and our n are dependent on the solid and
liquid combination, and since this is copper
and water, C sub sf is 0.0128 and our n is
equal to one. So when we calculate our delta
T e, we come up with 19.2 degrees Celsius.
Therefore, our surface temperature is T sat
plus delta T e, or 100 plus 19.2, or that's
going to equal 119.2 degrees Celsius.
ten copper tubes that are 15 millimeters in
diameter and each one is 0.5 meters long,
and they're used to boil saturated water at
1 atmosphere. And if the tubes are operated
at 80% of the critical heat flux, what we
want to know is what is the vapor production
rate with these ten tubes. In addition, we
want to know what is the corresponding
surface temperature of the tubes. So the heat
transfer rate is the surface flux times the
surface area, which is the surface flux times
the number of tubes times pi D L. And this
is equal to m dot h sub fg. And we should
realize that this is nucleate boiling because we
are only at 80% of the critical heat flux. So what
we're looking for is this vapor production
rate, or our m dot, which is our heat flux
times the number of tubes times pi D L, divided
by h sub fg, or the heat of vaporization.
So the first thing we're going to need to
do is find this heat flux, which is, again,
0.8 times q max. So let's do our properties
for our saturated water. So our rho sub l
is 957.9 kilograms per meter cubed, that's
the density of the liquid. The heat capacity
of the liquid is 4217 joules per kilogram
K. The viscosity of the liquid is 279 times
10 to the minus 6 newton seconds per meter
squared. The Prandtl number of the liquid
is 1.76, and the heat of vaporization is 2257
kilojoules per kilogram. And we're actually
going to change that to 2257 times ten to
the third joules per kilogram since everything
else is in joules. Finally, we have the surface
tension, which is 58.9 times 10 to the minus
3rd newtons per meter. And our density of
the vapor, which is 0.5955 kilograms per meter
cubed. So the equation for the critical heat
flux is 0.149 h sub fg rho of the vapor, the
surface temperature times gravity times the
difference in densities, over the density
of the vapor squared, and this is all raised
to the 1/4. And when we calculate that out,
we get 1.26 times 10 to the 6 watts per meter
squared. So our q, or our heat flux, is 1.008
times 10 to the 6 watts per meter squared.
And when we put that number times ten times
pi D L, and divide it by our heat of vaporization,
our vapor production rate, or m dot, is 0.105
kilograms per second. In order to find the
corresponding surface temperature of a tube,
we're going to need to find our difference
in excess temperature. And so we're going
to use the Rohsenow correlation, which equals
C sub sf h sub fg Prandtl to the n, divided
by our heat capacity of the liquid, and this
is multiplied by our surface heat flux divided
by our liquid viscosity h sub fg, all raised
to the 1/3, and this is multiplied by the
surface tension divided by gravity and the
difference in the densities, and this whole
thing is raised to the 1/6. Now the C sub
sf and our n are dependent on the solid and
liquid combination, and since this is copper
and water, C sub sf is 0.0128 and our n is
equal to one. So when we calculate our delta
T e, we come up with 19.2 degrees Celsius.
Therefore, our surface temperature is T sat
plus delta T e, or 100 plus 19.2, or that's
going to equal 119.2 degrees Celsius.
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Apr 19, 2026 Last updated
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