# Modeling a kongming lantern (sky lantern)

• FranzDiCoccio
In summary, the problem involves finding the amount of fuel needed for a hot air balloon to lift a candle, as well as estimating the thermal conductivity of the balloon material. The given data includes external and internal air temperatures, lantern volume, surface area, and mass, convective heat transfer coefficients, balloon thickness and thermal conductivity, fuel calorific value, and atmospheric pressure. The solution involves comparing the total mass of the hot air-lantern system to the mass of the displaced cold air to ensure that the lantern will lift. The estimate for the amount of fuel needed is calculated using the convective heat transfer coefficients and an efficiency coefficient. However, there is some confusion about the meaning of the convective heat transfer coefficients
FranzDiCoccio
Homework Statement
make an estimate the fuel needed to keep an air lantern airborne for one minute. Check that the size of the balloon is enough to lift the candle.
Assume that the air pressure and temperature are constant.
Relevant Equations
Buoyancy equations
Heat transfer equations
The problem gives these data
• external (cold) air temperature: ##T_c = 20^\circ{\rm C}##
• internal (hot) air temperature: ##T_h = 80^\circ{\rm C}##
• lantern volume ##V=0.2 {\rm m^3}##
• lantern surface ##A=2.0 {\rm m^2}##
• lantern mass ##m_{\rm L}=30 {\rm g}##
• external convective heat transfer coefficient: ##C_{\rm e}=60##
• internal convective heat transfer coefficient: ##C_{\rm i}=25##
• balloon thickness: ## \tau = 0.1 {\rm mm}##
• balloon thermal conductivity: ## k = 12 {\rm W\,m^{-1}\,K^{-1}}##
• superior calorific value of the fuel ## h= 48 {\rm MJ/kg}##
• atmospheric pressure: ## p = 1.013 \cdot 10^5 {\rm Pa}##
Checking that the balloon is able to lift the candle is easy. One should check that the total weight is less than the buoyancy.
Since weight is proportional to mass, this is the same as comparing the total mass of the hot air-lantern system and the mass of the displaced cold air. Treating air as an ideal gas with molar mass 29, the mass of a volume #V# of gas at temperature ##T## and pressure ##p## is
$$m = 29 \cdot 10^{-3} {\rm \frac{kg}{mol}} n = 29 \cdot 10^{-3} {\rm \frac{kg}{mol}} \frac{p V}{R T}$$
Thus the mass of hot air is ##m_{\rm h} \approx 0.2003 {\rm kg} ##, while the mass of the cold air displaced by the balloon is ##m_{\rm c} \approx 0.2413 {\rm kg} ##, or perhaps something more than that, if the thickness of the balloon is taken into account. Now, the fact that ##m_{\rm c} > m_{\rm h}+m_{\rm L} ## ensures that the lantern will "fly". Also, however small, there's an extra buoyancy from the cold air displaced by the candle and other "solid" parts.

Estimating the fuel needed is trickier for me, because I'm not really clear on the meaning of the two provided convective heat transfer coefficients.

I'd say that the heat loss from the hot air inside the balloon in a time interval ##\Delta t## is
$$Q= \frac{k A \Delta T\, \Delta t}{\tau} \approx 864 {\rm MJ}$$
which seems like a lot. If the process was perfectly efficient it would require a mass of fuel
$$m_{\rm f} = \frac{Q}{h} \approx 18 {\rm kg}$$
which is way more than the mass of lantern itself.

There is definitely something wrong at this point. It could be me.
However, the thermal conductivity of the lantern seems too large to me.
I looked up the value for paper, and I found ## k = 0.05 {\rm W\,m^{-1}\,K^{-1}}##, more than two orders of magnitude less than the provided value...
Using this value the mass of fuel would be $$m_{\rm f} \approx 75 {\rm g}$$. This is probably still too much, but realistic. I think that the real ##k## should be even less.

Indeed, the heat transfer from the fuel to the air would not be perfect. I think I should use at least one of the heat transfer coefficients. My guess is the internal one, so that the mass of fuel would be
$$m_{\rm f} = \frac{Q}{h C_{\rm i}}$$
where I think the coefficient should be a percentage perhaps... 0.25.

I'm not sure whether this is correct, nor whether one needs to use both of the provided thermal transfer coefficients.

At this point I realize I might have to include the fuel mass in the buoyancy problem. I originally assumed that "lantern mass" meant "everything except the hot air", but perhaps it could be "everything except the hot air and the fuel".

Thanks a lot for any insight

Francesco

PS I used latex, but the preview does not appear to work. I hope math displays correctly after I posted my question.

Last edited:
Me again.
I am still not clear about the convective heat transfer coefficients, which in the original problem are dimensionless numbers.

I think one could work out an upper limit for the thermal conductivity coefficient of the balloon "skin". I'm going to introduce an efficiency coefficient ##e##, so that the heat lost throught the balloon is only partly compensated by the heat produced by the fuel burning.
$$\frac{k A \Delta T \, \Delta t}{\tau} =e\, m_f h$$
For the lantern to be airborne, the fuel mass equals at most the payload
$$m_f<m_c-m_h-m_L\approx 11 {\rm g}$$
thus
$$k_{\max}=\frac{e m_f h\tau }{ A \Delta T \, \Delta t} <\frac{(m_c-m_h-m_L) h\tau }{ A \Delta T \, \Delta t} \approx 7.33\cdot 10^{-3} {\rm \frac{W}{mK}}$$
The real figure should be even lower ##k=e k_\max ##, where ##e## should be worked out from the convective heat transfer coefficients.

Anyway, it seems to me that ##k## should be 4 orders of magnitude less than the given value.
Does this make sense?

FranzDiCoccio said:
the meaning of the two provided convective heat transfer coefficients.
Assuming those numbers are for units ##Wm^{-2}K^{-1}##, you need to consider three temperature drops: inside surface, balloon thickness, outside surface. Each carries the same heat flow.
On the numbers given, there's barely any drop through the balloon thickness.
Changing k to the value you found makes the drops all the same order of magnitude.

Hi haruspex,

I'm not sure I understand your suggestion, though. You're saying that I should assume that ##T_h## is the temperature "at the flame" and it decreases through the hot air inside the balloon, so that it reaches ##T_1<T_h## at the internal side of the balloon skin?
Also, the flame heats up the air just outside the skin. Am I understanding correctly?

If that is case, does that mean that I should change my buoyancy analysis, since it is based on the assumption that the outside air is cold? I am really confused.

I guess I have to make an assumption on the balloon shape. In the case of a cylinder with a single "lid" on the top part, I get a radius ##r=0,216 {\rm m}## and a length ##\ell=1,36 {\rm m}##, which are kind of reasonable figures. If I assume two "lids" I get ##r=0,248 {\rm m}## and a length ##\ell=1,04 {\rm m}##. I also get other solution where the lids are much wider and the cylinder is much shorter.
But what now? How do I use those two coefficients?
Do I assume that heat flows through the bottom lid, the air inside the balloon and the top lid towards the cold air? And what temperatures shoud I use?
Now that I read the assignment more carefully, ##80^\circ {\rm C}## is the average temperature of the air inside the balloon. So probably I'll have to figure out a system of equations that allows me to work out the temperatures at the interfaces and the transferred heat.
Not sure how to do that.
I'm thinking maybe I could assume a very "squat" lantern, ##r=0,398 {\rm m}## and a length ##\ell=0,402 {\rm m}## and ignore the heat transfer from the lateral area, which maybe simplifies the equations...I have to say this is a homework that was assigned to a friend of mine's daughter. She has to complete the proposed analysis in order to get some credits for a university course she is attending. According to her, the lectures were given online, there were a lot of connection problems, and the teacher refused to provide the handouts fot his lectures and did not give his permission to record his lectures.

My friend asked if I could help her daughter, since I have a degree in physics. However I am not expert at all in this subject. I do not remember analyzing anything like this in the standard courses I attended in my time.
Maybe I'm wrong, this is a simple problem, and I'm making it harder than it is.
However, it does not look like an "introductory physics" homework to me. That's why I have originally posted in "advanced physics homeworks".

Thanks a lot for any further insight.

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FranzDiCoccio said:
assume that ##T_h## is the temperature "at the flame" and it decreases through the hot air inside the balloon,
No, not quite. At each surface of the balloon's skin there is effectively an insulating boundary layer. We don't know how thick it is but we are told how effectively it insulates. So there are three consecutive temperature drops, one through the inside boundary, one through the skin, and one through the outer boundary.
You have to use the fact that in steady state the rate of heat flow is the same through each. That tells you how the overall drop partitions, and from that you can find the heat fliw.

FranzDiCoccio
Uhm... I see. I tried representing what you're suggesting in a sketch.

Since the dimensions for the coefficients should be ##\rm W m^{-2} K^{-1}## , I guess that the three relevant equations are
$$\left\{ \begin{array}{l} P = C_{\rm i} A (T_h-T_1) \\ P = k \frac{A}{\tau} (T_1-T_2) \\ P = C_{\rm e} A (T_2-T_c) \end{array} \right.$$
Solving for the transferred power I get ##P = C_{\rm eff} A (T_h-T_c)\approx 2.12 \rm kW## where
$$C_{\rm eff} =\frac{1}{C_{\rm i}^{-1}+C_{\rm e}^{-1}+\tau/k}\approx 17.6 \rm W m^{-2} K^{-1}$$

The heat transferred in one minute through the whole surface should then be ##Q = P\cdot 60 {\rm s} \approx 0.127 \rm MJ## and the fuel needed to achieve that has a mass
$$m_{\rm f} = \frac{Q}{h} \approx 2.65\,\rm g$$

This mass is well within the payload of the lamp.

Nice... I guess this is correct, but I'll appreciate any feedback.

I was thrown off by the meaning of the "convective heat transfer coefficients" (which I never encountered in my life), and by them being dimensionless in the original assignment.
If what I did is correct, then I agree that that the problem is not that hard, with a proper background.I have another question. As you mention, the temperature drop at the lantern skin is very small. I get ##T_1\approx 37.8^\circ \rm C## and ##T_2\approx 37.6^\circ \rm C##.

In assessing the buoyancy of the lantern, should I take into account the external layer at ##T_2##?
Is there a way to asses the thickness of the convective layers?
If these are for some reason negligible, I guess that my previous calculation is allright.Thanks a lot, and Merry Christmas again!

Last edited:
Yes, that all looks good.
FranzDiCoccio said:
In assessing the buoyancy of the lantern, should I take into account the external layer at ##T_2##?
Is there a way to asses the thickness of the convective layers?
I would think they're taken to be of negligible thickness. In fact, my explanation might be a bit off. It may have more to do with an inefficiency in heat transfer between a solid and a fluid, a bit like waves encountering a change of medium.

FranzDiCoccio

## 1. What is a kongming lantern?

A kongming lantern, also known as a sky lantern, is a small hot air balloon made of paper and fueled by a small flame, typically a candle. It is traditionally made in the shape of a sphere or cylinder and is released into the sky for cultural or celebratory purposes.

## 2. How do you model a kongming lantern?

To model a kongming lantern, you will need to gather materials such as paper, a lightweight frame, and a heat source. The paper can be cut into a cylinder shape and attached to the frame, leaving an opening at the bottom for the heat source. The heat source can be a small candle or a fuel cell. Once the heat source is lit, the hot air will fill the lantern and cause it to rise.

## 3. What are the physics behind a kongming lantern?

The physics behind a kongming lantern involves the principles of buoyancy and convection. The hot air produced by the heat source is less dense than the surrounding air, causing the lantern to rise. As the hot air cools, it becomes more dense and the lantern will eventually descend. The shape of the lantern also plays a role in its flight, as the curved surface creates lift.

## 4. Are there any safety concerns when modeling a kongming lantern?

Yes, there are safety concerns when modeling a kongming lantern. The heat source can pose a fire hazard, so it is important to take precautions and monitor the lantern while it is in flight. It is also important to check the weather conditions before releasing the lantern, as strong winds can cause it to fly off course and potentially cause harm.

## 5. Are there any environmental concerns with using kongming lanterns?

Yes, there are environmental concerns with using kongming lanterns. The paper and frame used to make the lantern can contribute to litter and pollution if not disposed of properly. In addition, there have been reports of lanterns causing fires and harm to wildlife. It is important to consider these factors and use caution when using kongming lanterns.

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