Methodology by Mistral Associates
This publication is © Copyright Mistral Associates (1987) and is provided as a guide to the mathematical methods and processes within computer programs such as COLDWIND and similar. It does not represent a total account of all of the methods, procedures, factors, algorithms, constants, equations and empirical data sources employed by the computer programs themselves as this would require several large volumes of text. It can however be regarded as a reliable summary or precis. Mistral Associates grants permission for this publication to be freely used for reference purposes, for qualifying parts of calculations and for manual calculation purposes, however Mistral Associates expressly forbids reproduction or re-printing of this document or any part of it by mechanical or electronic means without prior written agreement from Mistral Associates and which may be granted with strict additional conditions and limitations. No part of this document may be used to damage the legitimate business or financial interests of Mistral Associates, including its staff, directors and shareholders, whether this was the intended aim or otherwise.
The Cold Room Refrigeration Load Calculation©
The refrigeration load calculation is used to determine the cooling duty of equipment required to maintain a coldroom or freezer, including its contents, at a controlled design temperature other than that of its ambient surroundings.
All possible sources of thermal energy that can find its way into the coldroom must be considered and quantified.
The calculation process takes into account all of the energy sources plus other relevant factors, such as for example the quality and thickness of insulating materials, that have an effect.
The complete list of possible heat sources and their accompanying formulae are shown below. The list is as comprehensive as it needs to be for all practical purposes and also includes items that may not be relevant in every case E.g. Miscellaneous loads (where one might include gains for fork lift trucks).
Note that there is no provision for ‘Solar Gains’ in this calculation process. The program is designed to cope with insulated coldrooms and freezers only and these are invariably built as structures within another structure, thereby rendering the insulated structure isolated from any direct solar radiation for all practical purposes. That the air temperature surrounding the insulated space may be increased to above outside ambient due to energy conducted and re-radiated from the external building fabric is another matter and the program does provide for this phenomenon.
For smaller coldrooms (those say under 100 cubic metres) with ‘standard’ food storage applications refrigeration loads can generally be predicted quite reliably using a ‘worst case’ set of parameters without incurring undue penalties as a result of ‘over engineering’. This is partially due to the fact that smaller coldrooms inevitably tend to be used to capacity anyway even if that was not the coldroom operator’s originally intended plan. For this reason a set of ‘grid size’ coldroom and freezer designs have been included in the program where normal coldroom operating practices have been applied. These include such calculation parameters as typical food product entry temperatures, product turnover, product temperature pull-down times etc. The laws of physics, when combined with the likelihood that larger coldrooms will tend to have more specialised usage, makes ‘quick selection’ methods unreliable.
1) U x A x TD = Watts
U = U value and is the coefficient of thermal conductivity for a specific thickness of material W/M² K (Watts/metre² Kelvin).
U =
1 _
Thickness of material in metres
k factor
Example:-
75mm Polyurethane Foam
1 _
0.075M = 0.24 U
0.018 k
A = Total surface area of material in metres²
TD = Difference between coldroom design temperature and outside temperature.
In the United Kingdom 32ºC is generally considered to be the highest summer ambient dry bulb temperature to be used for coldroom load calculation purposes. Care should be taken though when, for example, considering a (usually small) coldroom in say a hot hotel kitchen where summer temperatures can easily exceed 32ºC.
1 Ground floor temperatures
For practical purposes the constantly renewable source of energy contributing to a stable temperature in the sub-strata under a coldroom floor is considered to occur at a depth of around 2 metres. This results in a temperature that is (conveniently!) around 50% of the highest mean summer ambient air dry bulb temperature. Of course this is a mathematical expedient with no proven basis of fact but nonetheless as an empirically based simple formula it has been well proven as a reliable ‘worst case’ figure. Therefore examples would be 16ºC as the ground temperature chosen for a calculation in the southern part of the United Kingdom and 25ºC in parts of Saudi Arabia where ambient temperatures can reach 50ºC. These ground temperatures have been recorded and documented by water extraction authorities who have cause to measure such.
To calculate U for a floor composite material construction:-
1 _
U = D1/k1 + D2/k2 + D3/k3
Where D1, D2 and D3 etc. are thickness of various materials, not forgetting the temperature ground under the floor. For calculation purposes only this is taken to occur at 2.0 metres at which depth it is considered a constantly renewable source of energy resulting coincidentally in a temperature at around 50% of ambient air temperature.
Example:-
75mm of concrete over 100mm slab styrofoam IB on 50mm sand/cement screed laid over 150mm concrete. Ground is clay.
1 _2 Air Change
0.075 + 0.1 + 0.05 + 0.15 + 2.0
1.73 0.033 0.721 1.73 1.44
1 _
= 0.043 + 3.03 + 0.069 + 0.086 + 1.389
1 _
= 4.617 = 0.217 U
Air change load (W) =
Room volume (m³) Heat in Joules removed per Number of air changes
X m³ of air per day X per day _
86,400 seconds
3 Product Load
3a Product load above freezing temperature
3b Product load below freezing temperature
3c Product freezing
3d Product heat of respiration
3a and 3b
Product load (W) =
Weight of product Product temperature Product Specific
loaded (kg) X reduction (K) X Heat (J/kg K)
86,400 seconds per day
Notes
i) Ensure that the correct product specific heat figure is used, noting that the value for products above their freezing temperature is different to that below freezing.
ii) When a product is introduced into a freezer room above the product freezing temperature then the Latent Heat of that product must first be removed for the product temperature to drop below its freezing temperature.
iii) The time allowed for reducing product temperature is critical to the calculation. In the example calculation above the figure of 86,400 seconds in a 24 hour day means that, according to the calculation, it will take 24 hours for the product temperature to be reduced to design temperature. For faster product temperature reduction times adjust the divisor accordingly.
E.g. For 12 hours, 86,400 x 12 = 43,200
24
iv) Freezing temperatures of products vary and for food products these are invariably below the freezing temperature of pure water (0ºC).
v) Remember, there will come a point in temperature reduction times where the heat transfer coefficient of the product itself must be considered, especially for bulk packed products E.g. barrels of beer. As even though sufficient cooling duty might be provided, the depth of the product and its own energy conductivity characteristics may prevent the product from giving its energy up in the limited time available. For most practical purposes, average food products packed in their ‘point of retail sale’ design of packaging and with adequate air circulation can be expected to be reduced in temperature from say ambient (32ºC) to typical chill store temperature of say 2ºC in as little as 6 hours. Any time expectation faster than this and a ‘Rapid Chiller’ must be considered. Below 90 minutes and a ‘Blast Chiller’ will be essential.
3c Product freezing
Removal of Latent Heat of fusion
Load (W) =
Weight of product (kg) x Product Latent Heat (J/kg)3d Product heat of respiration
86,400 seconds per day
Fresh fruit and vegetables are alive, consequently they respire and in doing so release heat. This must be included chill room in calculations. Fruit and vegetables that have been cooked or that are frozen do not respire.
Respiration Load (W) =
Total weight of live
product in room (kg) J/kg x Heat of Respiration
86,400 seconds per day
4 Heat of Occupancy
People produce heat and emit it in increasing amounts as the air surrounding them becomes lower.
Refer to the table for heat emitted by people in coldrooms at different temperatures.
The number of hours people spend in coldrooms varies with each application and should be estimated with common sense. People do not spend more time than they have to in very low temperatures.Coldroom temperature ºC Heat emitted per person in Watts
15 180
10 210
5 240
0 270
-5 300
-10 330
-15 360
-20 390
-25 420
Occupancy load (W) =5 Lighting Load
Number of occupants x Number of hours x Heat Equivalent per occupant (W)
24 hours
Lighting loads vary but unless specified assume approximately 10 watts per metre² of floor area. Hours of operation of lights are normally the same as for occupancy although can be longer. Rarely can they be less than hours of occupancy as usually there are no windows in coldrooms.
Lighting Load (W) =Total Lighting Power (W) x Hours of operation
24 hours
6 Miscellaneous Loads
For calculation purposes the total rated power (Watts) of any electrical machinery E.g. conveyor belts, mixers, lifts etc. is assumed to be converted into thermal energy added to that in the coldroom air and must therefore be included in the refrigeration load.
Do not forget to include fork lift trucks although its ‘plated’ power rating is unlikely to ever all be converted to energy inside the coldroom. Firstly a fork truck only consumes electricity from its batteries at its maximum plated power rating when it is working under maximum load I.e. when lifting to its maximum rated weight capacity and probably travelling at the same time. Clearly it cannot by the very nature of its design do this as a continuous operation. Secondly, most fork trucks do not spend all of their time inside the coldroom (although some do!). Most are moving product in and out of the coldroom and logistically this must mean the fork truck spends less than 50% of its total working time outside the refrigerated space. One leading electric fork truck manufacturer recommends that a ‘power factor’ of between 20% and 25% is applied to the fork truck’s plated power rating..
Miscellaneous Load (W) =9 Freezer under floor heater matsPower Rating (W) x Power factor x Hours of operation
24 hours
Gains from frost heave prevention Heater Mats should not occur. Firstly they are (or should be) thermostatically controlled. Secondly, the calculation process must always consider 'worst case' situations if a potentially catastrophic failure is to be avoided. With 'comfort cooling' in air conditioning applications for example, perhaps the worst consequence of the failure of a cooling system is a staff walk-out, whereas a freezer failure with the loss maybe of hundreds of tonnes of food product, then no further justification for 'worst case' design is required.
So why again are heater mat gains NOT considered? Simply because the calculation must instead consider the far higher and thus more onerous gains from the ground beneath the floor. Whether or not the floor is insulated.
In conjunction with universities specialising in the subject such as South Bank, London and Grimsby College, plus unlikely but nonetheless useful sources for their valuable expertise, the water authorities, Mistral conducted considerable research during the mid 1980s as to what really happens with energy sources under cold room and freezer room floors. Conclusions drawn from the results of much empirical data arrived at a complicated set of rules for calculation of energy gains from beneath floors. Gains from heater mats were firmly ruled out from these.
8 Cooler Fan Load
Whatever of the above sources of heat, the final total heat to be removed (the Primary Refrigeration load) finds its way into the air within the coldroom. Subsequently this heat is removed from the coldroom air in the evaporator (or unit cooler as it may also be called). Coldroom air is forced by a fan or fans through the evaporator which usually comprises a series of tubes over which tightly fitting fins are attached. Fans are driven by electric motors and in all but the most unusual of designs these electric motors are also mounted inside the coldroom. As it is not possible to either create or destroy energy, only to move it or to change its state, then all of this electrical energy must be considered as being converted into thermal energy and thus added to that energy already in the coldroom. Unfortunately, for cooler fan load to be added to the primary refrigeration load the make and model of the cooler needs to be known and until the fan heat is known it is not possible to select the cooler. An iteration problem thus arises and one which can only be solved by first making an estimate then trying this to see if it is right and if it is not, modifying the first estimate until a closer match is found. Fortunately there are strong relationships between the amount of air needed to be forced through evaporators to exchange a given amount of energy and the amount of electrical energy needed to drive the fans. The computer model uses some quite sophisticated techniques to arrive at estimates but when calculating manually a figure of 7.5% of the primary load is generally considered a good estimate to start with.
The sum of the Primary Load and the Cooler Fan Load is known as the Preliminary Load.
9 Running Load
In order to allow sufficient time for defrosting the evaporator(s) and also as a contingency against peak load conditions being exceeded the refrigeration equipment duty is adjusted to meet the required Preliminary Load in less than 24 hours. This time adjustment is known as the Running Load.
Generally 6 hours per day is allowed for total defrosting time for a low temperature coldroom, that is a room operating below 2ºC, and this also provides for removing the energy released into the coldroom from stray radiation and conduction of defrost system heat. Where defrost is effected by the ‘off cycle’ method, I.e. where there is sufficient energy in the air in the coldroom to remove frost from the evaporator(s) simply by allowing the fans to continue running, but without the refrigeration compressor in operation, then more time is needed and generally 8 hours is thus provided.
Preliminary Load (W) x 24
Hours of operation = Total refrigeration Load
© Copyright Mistral Associates (1987)