- #1

FranzDiCoccio

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- 41

- 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

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.

- 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}##

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.

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