What you need to apply is the engineering equations for a heat exchanger. In the below equation, $\dot{Q}$ is the heat transfer rate, $UA$ is the heat transfer coefficient times the area and $\Delta T_m$ is the log mean temperature difference (LMTD).
$$\dot{Q}=U A \Delta T_m$$
For this problem there are 3 barriers to the heat transfer in series. You have the convective heat transfer coefficient of the syrup $h_c$, the convective heat transfer of the steam $h_s$, and that of the pipe metal $h_R$. Take $n$ to be the number of pipes, $r_i$ to be the inner radius of the pipe, and $r_0$ to be the outer radius of the pipe, and $R$ the heat transfer coefficient of the Copper.
$$UA = \frac{2\pi n}{\frac{1}{h_s r_i}+\frac{1}{R L} ln \frac{r_o}{r_i} +\frac{1}{h_c r_0} }$$
The LMTD is the average temperature difference, but for the specific case where one part of the heat exchanger is saturated, and thus constant temperature, just know that it is the following, where $T_{s}$ is the saturation temperature of the steam, or 100 degrees C. Then $T_h$ and $T_c$ are the temperatures after and before preheating respectively.
$$\Delta T_m = \frac{T_h-T_c}{ln\frac{T_h-T_s}{T_h-T_s} }$$
The hard part of the above equations is the $h_c$ and the $h_s$. You will probably use things like the Dittus-Boelter correlation. But it's more important that for the moment we address $\dot{Q}$ itself. For the described hood, I see two possibilities.
- The steam is being provided at a faster rate than it is being condensed
- The steam is being provided at a slower rate that it is being condensed
In the first case, you will see steam leaking out of the hood. In this case, the equilibrium operation see the air mostly evacuated because it is pushed out. In the other case, steam is present with air and the air reduces the heat transfer. In case #2 the given is the rate of steam condensation, which directly determines $\dot{Q}$ and $h_c$ adjusts to compensate. In case #1 $h_c$ is a given based on the assumption of your geometry and the atmospheric steam heat transfer properties.
For $h_s$ use Wikipedia:
$$h_s={{k_w}\over{D_H}}Nu$$
where
$k_w$ - thermal conductivity of the liquid
$D_H$ - $D_i$ - Hydraulic diameter (inner), Nu - Nusselt number
$Nu = {0.023} \cdot Re^{0.8} \cdot Pr^{n}$ (Dittus-Boelter correlation)
Pr - Prandtl number, Re - Reynolds number, n = 0.4 for heating (wall hotter than the bulk fluid) and 0.33 for cooling (wall cooler than the bulk fluid).
What I don't have right now are the numbers for applying the above for Maple Syrup and the approach for the steam side of the tubes. This is all I have time for right now, but I think this amount will still be helpful. I can try to look up these things later, maybe you can specify what you understand the least first.
Without doing the analysis, I would think that a cooling system
is more effective at extracting heat from a warm container than from a
cold one.
For a fridge, the effectiveness (or coefficient of performance) is
$Eff=Q_c/W$ is the ratio of the heat removed from the cold source (the fridge) to the energy used for
the purpose. It increases with the temperature of the cold source. This is actually the prime factor to be considered in the analysis.
If we assume that, for a given current temperature of its contents, and thus for a given coefficient of performance, the cooling capacity of the fridge (heat removed per second) is limited only by the
power of its cooling engine (I do not know whether that is the case),
then the heat pump will pump more heat per second when the fridge is warm.
Hence it is better to put all bottles at once and get the fridge
warmer to have a maximum heat pumpimg capacity from the heat pump.
Heat sharing rate within the fridge may also be an important issue, but there are no data available to measure how important. If it is really low, thus leaving an important temperature gradient in the fridge, it may be useful to exchange the position of bottles, so as to have the warmer part of the load
near the heat pump and have work with highest possible coefficient of performance.
Precise figures about the load do not matter very much. However, a load with large heat capacity will take longer to cool and will thus allow more time for heat sharing.
In the second part below, we prove formally that all bottles should be cooled at once, and we use the understanding to discuss the heat sharing issue in some more depth. The variability of the coefficient of performance with temperature is central to this analysis.
FORMAL STATEMENT AND PROOF
A refrigerator is a Carnot machine functionning as a heat exchanger,
where we are interested in removing heat from the low-temperature
reservoir, using work from an engine that provides compression.
The effectiveness, or coefficient of performance, noted here $Z$, is
defined as $Z=Q_c/W$ where $W$ is the work provided and $Q_c$ is the
amount of heat extracted from the cold source (the refrigerator) with
that work. If we note $Q_h$ the amount of heat delivered to the hot
source (outside the refrigerator), we have the equality $Q_h=W+Q_c$.
For an ideal Carnot cycle, we have
$Z_{ideal}=Q_c/W=Q_c/(Q_h-Q_c)=T_c/(T_h-T_c)$ where $T_h$ and $T_c$ are the
temperatures of the hot and cold source. (see http://en.wikipedia.org/wiki/Coefficient_of_performance).
Of course, the actual coefficient of performance $Z$ is less that the
Carnot ideal. Short of knowing its specific, we will only assume that,
like the ideal coefficient, it depends monotonically on the
temperature $T_c$ of the cold source, the hot source (outside the
refrigerator) being considered at constant temperature. Hence we only
assume that the coefficient of performance $Z$ is a strictly
increasing function of (cold source) temperature, i.e., such that
$T_1< T_2\ \Rightarrow\ Z(T_1) < Z(T_2)$
We also assume that the mechanical power available for compression is
invariant, i.e. does not depend on the temperature of the sources, at
least within the range of temperatures considered.
Finally, we also assume that the heat capacity of the refrigerator
itself is negligible, and that heat sharing within the refrigerator
takes negligible time compared to cooling time so that the
content may be considered to have uniform temperature. These later two
assumptions will be discussed afterwards.
With the above assumptions, given two masses $m_1$ and $m_2$ to be
cooled in the cold source, it is faster to cool booth simultaneously
than to try to cool one first and later add the second one. It also
consumes less energy.
PROOF
Cooling a mass $m$
Actually, the formulae above are about heat and work increments. This
is necessary since $Z$ is temperature dependant, and
temperature may vary. Also, since we intend to analyze the system from
the point of view of the cold source, the heat increments are actually
removed from that source and must be countd as negative.
So we can write $Z= -dQ_c/dW$, or $dQ_c/dW=-Z$.
The power provided by the compressor is a constant $P=dW/dt$.
Hence $dQ_c/dt= (dQ_c/dW)\times(dW/dt)=-Z\times P$.
On the other hand we know that removing heat reduces the temperature
according to the formula $\Delta Q=-cm \Delta T$, where $m$ is the mass
being colled and $c$ is the specific heat for the substance
constituting that mass.
Hence we have $dQ_c/dt= cm(dT_c/dt)$.
Combining the two formulae, we get $dT_c/dt=-ZP/cm$.
But we cannot resolve this equation since $Z$ is an unknown function of
$T_c$.
What we know is that $Z(T_c)$, $P$, $c$ and $m$ are strictly positive
values. So $dT_c/dt$ is strictly negative. Hence $T_c$ will
decrease with time. Since the function $Z(T_c)$ is a strictly
increasing function, its value will also decrease with time, and hence
the absolute value of the derivative $dT_c/dt$ will also decrease with
time.
Hence the graphical representation of the evolution of the temperature
will look like the red curve in the figure, where $T_0$ is the initial
temperature at time $t_0$.
Cooling the same mass $m$ in two steps
If we consider cooling independently (in an identical refrigerator) another mass $m_1$, smaller than
$m$, with initial temperature $T_0$ we get another curve, like the curve in blue in the left part of
the figure up to point C, corresponding to the equation
$dT_c/dt=-ZP/cm_1$. It is below the red curve because the smaller mass
$m_1$ cools faster than $m$. Formally, if we draw a horizontal line
like the line cutting both curves in A and B, this corresponds to the
same temperature for both curves, hence to a common value of the
coefficient of performance $Z$. Then $m_1<
m\ \Rightarrow\ (dT_c/dt)_A<(dT_c/dt)_B$. Since this is true for any
value of the temperature $T_c$, it confirms that the blue curve for
$m_1$ decreases faster than the red curve for $m$.
Suppose now that at time $t_2$ the mass $m_1$ has been cooled to
temperature $T_2$ corresponding to point C below the red curve.
We add to $m_1$ another mass $m_2$ such that
$m=m_1+m_2$, the mass $m_2$ being at the initial temperature $T_0$.
The mass $m_2$ being warmer than $m_1$ will share its heat with $m_1$
(in negligible time according to our hypothesis) so that both reach
the temperature $T_1$ corresponding to point D, and pursue cooling.
At any time between $t_0$ and $t_2$, the temperature of $m_1$ (blue)
is less than the temperature of $m$ (red). Hence the refrigerator
works with a lower coefficient of performance $Z$ for $m_1$ than for
$m$, and less heat has been removed from the refrigerator containing
$m_1$ than from the refrigerator containing
$m$ at time $t_2$. When we introduce the mass $m_2$ with $m_1$, the
total heat introduced in the refrigerator is that of $m_1+m_2$ at
temperature $T_0$. This is exactly the same as the heat introduced in
the refrigerator containing $m$. Since less heat was removed from the
$m_1+m_2$ refrigerator at time $t_2$, it is at a higher temperature
that the $m$ refrigerator. Hence the point D is above the red curve.
The $m_1+m_2$ refrigerator now contains the same mass as the $m$
refrigerator. Hence it will follow an identical curve. But it is at
temperature $T_1$ that was attained earlier, at time $t_1$ by the $m$
refrigerator. So the right part of the blue cooling curve for $m_1+m_2$,
starting at point D is the same as the right part of the red coling curve
for $m$ starting at point B, translated by a duration $\Delta
t=t_2-t_1$.
Conclusion
The masses $m_1+m_2$ will always reach any temperature with a delay
$\Delta t$ after the mass $m$ has reached it.
Actual figures would be required to be more precise.
Given the problem, the cooling engine will be working at maximum power
to get the fastest possible cooling in both cases. Then it is obvious
that the faster solution is also the most economical energetically.
This assumes either that the cooling is started just at the right
time, or that the cooling power is reduced once the right temperature
has been attained.
These results are based exclusively on our assumptions, independently
of any actual figures.
We will now discuss some of these assumptions.
DISCUSSION
Heat capacity of the refrigerator
We have assumed that the heat capacity of the refrigerator itself is
negligible. We should however analyse its effect. We first note that
the objective is to extract heat from a given number of bootles to
bring them from $T_0=30^{\circ}C$ to $T_f=6^{\circ}C$. That
corresponds to a precise amount $Q$ of heat to be removed,
independently of the process used for that purpose.
If the refrigerator itself is initially at temperature $T_f$, it will at
the start share heat with the mass to be cooled, thus warming up and
cooling the bootles. But
then it will have to be cooled down to $T_f$ again, so that it net
contribution to the cooling process will be null, and the same amount $Q$
of heat has to be removed by the heat pump. However, by sharing heat
at the beginning, it induces an early cooling, thus making the whole
heat removal process operate at a lower temperature, hence with a
lower coefficient of performance $Z$.
The net effect of the heat capacity of the refrigerator is thus to
provide some early cooling, which can sometimes be considered an
advantage, but at the expense of a lower effective coefficient of
performance $Z$. These effects increase with the heat capacity of the
refrigerator.
Note that if the initial temperature of the refrigerator is below the
targeted final temperature $ T_f$, the difference multiplied by the
heat capacity is a net contribution to the refrigeration process,
though the loss on the coefficient of performance remains.
Hence it is better not to have anything else in the refrigerator, even already cooled, unless it is cooled to a much lower temperature than the final temperature $ T_f$ intended for the bottles.
Heat sharing rate
As we have seen from the previous discussion, the main objective is to
remove a given amount $Q$ of heat, and the effectiveness of the
removal decreases as temperature is lowered. If the rate of heat
sharing within the refrigerator is small, the volume near the cooling
system will cool faster, thus reducing the coefficient of performance,
i.e. the rate of heat removal.
Hence, ensuring the best possible heat sharing can help in all
circumstances. It should be noted that direct sharing between
cylindrical bottles will be reduced to a minimum: just one line of
contact. So, if space allows, it is probably preferable to help air
circulate between the bottles. Keeping the bottles in vertical
position will help if the refrigerator has grid shelves that let air
through, rather than glass shelves. And, of course, the bottles must
be unpacked.
Opening the door to quickly exchange bottles so that the warmer ones
are placed near the cooling system will improve the coefficient of
performance and reduce cooling time. It may have some
cost in warming the refrigerator, but that is less important if
the heat capacity of the load to be cooled is large (actual
measurements would be useful).
Exchanging bootles will also avoid having to cool those close to the
cooling system below the required temperature $ T_f$ in order to have
all the bottles temperature at least as low as $ T_f$.
Best Answer
Assuming you're using degrees Fahrenheit 160F is only (about) 70C so you won't be boiling the water. The increase in pressure will come partly from heating the air in the bottles and partly from the increased vapour pressure of the water.
The pressure of an ideal gas (air is near ideal in this temperature and pressure range) is:
$$ P = \frac{nRT}{V} $$
where $n$ is the amount of gas, $R$ is the gas constant, $T$ is the temperature (in Kelvin) and $V$ is the volume. For any particular bottle the amount of air and the volume is constant, so we can make life simpler by taking the ratio of the pressure after heating to temperature $T$, $P_T$, and the pressure we started off with $P_0$ (the initial pressure, $P_0$, is just atmospheric pressure). So:
$$ \frac{P_T}{P_0} = \frac{T}{T_0} $$
or:
$$ P_T = P_0 \frac{T}{T_0} $$
and we need to add on the vapour pressure of water $P_w(T)$. As the notation suggests the vapour pressure of water changes with temperature. At 70C it's about 30% of atmospheric pressure. Anyhow our equation for total pressure is:
$$ P_{total} = P_0 \frac{T}{T_0} + P_w(T) $$
Note that the volume of the air doesn't appear in this equation, so for the fully immersed bottles (where we can be sure of the air temperature) both bottles will burst at the same temperature.
When you only have the bottles immersed up to the level of the water the temperature of the air isn't well defined because it's heated by the water below it and cooled by the air above it. I would have guessed the air in the bottle with less air would be hotter simply because it sticks out of the water less. I'd expect these bottles to burst at a higher temperature than the fully immersed bottles, but how much higher is hard to say.
In the above I've assumed that you're raising the temperature gradually. Your question implies you're heating the water to 160F then throwing the bottles in. If you're doing that all bets are off because it depends on how quickly the contents of the bottle heat up. You can only do the experiment reproducibly if the put the bottles into cold water and raise the temperature slowly enough for the contents on the bottles to keep up with the temperature rise.
Finally note that this treatment only applies if you keep the temperature below the boiling point of water. Once the water starts to boil the pressure in the bottles will rise extremely rapidly.