I'm having a party.
Suppose I'd like to have a fridge full of cold ($6~^\circ\text{C}$ or below) beer bottles, in as short a time frame as possible. The fridge indicates that it is targeting (and presumably currently at) $4~^\circ\text{C}$. All the bottles are currently at $30~^\circ\text{C}$, which is the temperature outside the fridge, and together the bottles will fill the fridge completely (as in: no further bottles could be fitted in).
What is the best strategy to achieve this aim?
(Should I put in all bottles at once? Should I put the bottles in at different times? And, if so, should I make them touch, or keep them initially apart as much as possible? Should I even consider taking out some bottles at some time and putting them back in later? (That would be really weird.) If so, should I have those bottles cool other bottles that weren't in the fridge yet?)
NB 1: I do not have a freezer available.
NB 2: Assume that I know how to get as much bottles into the fridge as possible (without breaking any).
Measurements: 1 bottle of beer: $0.61~\text{kg}$, 1 empty bottle (with cap): $0.28~\text{kg}$, 1 empty bottle (without cap): $0.28~\text{kg}$, contents of 1 bottle: $0.33~\text{l}$. My guesstimate is that I can get 72 bottles into the fridge.
Answer
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$.
No comments:
Post a Comment