Heat transfer coefficient - Coolpack

[Graham]

What tabels are used to determine m_WALL (the thermal resistance of the wall)? I cannot believe the variances in the tables that I have come accross recently. I previously did my calculations using the ASHRAE handbooks from the 1970's or Modern refrigeration & Airconditioning 1975. These are not metric/SI and converting is cumbersome.

[nike123]

What tabels are used to determine m_WALL (the thermal resistance of the wall)? I cannot believe the variances in the tables that I have come accross recently. I previously did my calculations using the ASHRAE handbooks from the 1970's or Modern refrigeration & Airconditioning 1975. These are not metric/SI and converting is cumbersome.

This is from help in CoolPack:

As a guideline for the thermal conductivity of prefabricated walls (walls, cieling og floor) the following can be used (Ref.: W.F. Stoecker, "Industrial Refrigeration Handbook", page. 610, table 18.4. McGraw-Hill, 1998. ISBN 0-07-061623-X)

[nike123]

This is whole text:

Description of model


Contents:

1) Introduction
2) Heat transfer through building parts
3) Air change
4) Cooling down of goods and heat of respiration
5) Auxiliary loads
6) Total cooling demand

1) Introduction

The model is used for calculating the cooling demand for a refrigerated room in a steady state condition. The cooling demand is made up of several individual heat loads which will be described in the following sections.

The purpose of the model is to provide a general tool for calculating the cooling demand in refrigerated rooms. To some extent the user has to estimate the individual loads or parameters that describe them. In most cases guidlines for estimating the values can be found in this help file. The user should notice that the actual values can differ significantly from these guidelines. If a particular individual heat load makes up a large portion of the total cooling demand, a more detailed investigation and calculation should be made.

2) Heat transfer through building parts

The heat load from heat transfer through building parts is calculated for all 4 walls, roof and cieling individually. The heat transfer is calculated as follows

Q = k · A · (T - T_ROOM)

where Q is the heat transfer in [W], A is the area of the building part [m˛], k is heat transfer coefficient [W/(m˛·K)], T_ROOM is the temperature in the room in [°C] and T is the temperature on the hot side of the building part in [°C].

The heat transfer coefficient (k-value) of each building part depends on the insulation material (foam or mineral wool), how the building part is constructed and the thickness of the insulating layer of air on both sides of the building. In refrigerated rooms there is usually a significant air movment (due to the fans) and the thickness of the air layer close to the wall is very thin. This means that the air layer (and thermal resistance of the layer) on the inside of the refrigerated room can be ignored. In most cases this is also true for on the outside of the wall. In both cases the thermal conductivity of the air layer should be taken into account if there is no air movement.

If you want to include the thermal resistance of the air layer you should use the following equation for calculating the heat transfer coefficient:



where m_INNER is the thermal resistance [(m˛·K)/W] of the air layer on the inside of the the wall, m_WALL is the thermal resistance of the wall and m_OUT is the insulation property on the outside of the wall.

Typical values of m_INNER and m_OUTER are 0.04 (m˛·K)/W if there is a good air movment (thin layer) and 0.08 (m˛·K)/W with none or low air movment (thick layer). The insulation property of the wall can be calculated as follows

Small refrigerated rooms are typically constructed of prefabricated walls. Larger rooms are typically constructed individually taking into account the temperature levels and the purpose of the room.

As a guideline for the thermal conductivity of prefabricated walls (walls, cieling og floor) the following can be used (Ref.: W.F. Stoecker, "Industrial Refrigeration Handbook", page. 610, table 18.4. McGraw-Hill, 1998. ISBN 0-07-061623-X).

Thickness k-value Recommended temperature band

[mm] [W/(m˛·K)]

50 0.34 Chilled room
75 0.27 Above 0 °C
100 0.17 Above -29 °C
125 0.14 Above -45 °C
150 0.11 Above -57 °C

For large rooms that are build individually (not prefabricated) the following values can be used.

Room with temperatures above 0°C
Building part k-value Typical values of k
[W/(m˛·K)] [W/(m˛·K)]

Walls 0.23 From 0.18 to 0.35

Cieling 0.23 From 0.18 to 0.35
Floor 0.23 From 0.18 to 0.35

Room with temperatures below 0°C
Building part k-value Typical values of k
[W/(m˛·K)] [W/(m˛·K)]

Walls 0.12 From 0.10 to 0.20
Cieling 0.12 From 0.10 to 0.20
Floor 0.17 From 0.12 to 0.25

In a room with temperatures below 0°C floor heating should be installed to prevent the ground from freezing. For further details about this please go to section 5.3.

3) Air change (infiltration)
Due to opening of doors and leaks in the building parts warm and humid air flows into the room. This air has to be cooled and maybe also dehumidified. The flow of air infiltrating the room can be specified either directly by the flow in [mł/h] or indirectly by the Air Change Factor (ACF) specifying the flow of air as number of times the total room volume is changed per 24 hours.

4) Cooling of goods
The cooling down of goods generates an additional heat load on the refrigeration system. The calculation of this load is based on the temperature of the goods when they enter the room and the quantity of goods. The user can specify two individual quantities of goods and for each of these choose between 6 different groups of goods (Vegetables, Fruit, Beef, Pork, Fish og Diary products).

If the temperature of the room is below the freezing temperature of the goods in the room the extra heat load from freezing in the goods is taken into account. The load is calculated as a mean value of the heat load for the first hour of cooling. The average cooling load seen over the entire cool down period is also calculated.

If no cooling down of goods is to be included in the calculations, the quantity should be set to 0 [kg].


5) Auxiliary loads

A part of the heat load in a refrigerated room is caused by persons working in the room or from heat developing equipment in the room. These loads can be specified individually.

5.1) Persons

The heat load from persons working in the room is based on the maximal number of persons that are in the room at the same time.

There is three different types of work.
Light - typically inspection of goods
Medium - typically handling of goods using equipment
Heavy - typically manual handling of goods

The calculations are based on the assumption that there is a linear relation between the heat developed by a person and the room temperature. See H. Drees "Kühlanlagen", Table 12.18, page 376, 14. edition, VEB Verlag Technik, Berlin 1987.

5.2) Lights
The heat load from lighting equipment corresponds to the electrical power consumption of the lighting eqiupment .The heat load can be specified in [W/m˛] or as a total heat load in [W].

5.3) Fans
All of the electrical power consumption for evaporator fans is converted to heat. The heat load from this equipment is specified as the power consumption in [kW].

5.4) Other equipment
If other power consuming equipment in placed in the room the heat load of this equipment has to be estimated. Normally, the heat developed corresponds to the electrical power consumption of the equipment.

5.5) Heat of respiration
If the cold room contains a large quantity of fruits and vegetables or cheese the heat load caused by respiration of these products may be significant in relation to the total cooling demand. A constant load from respiration can be entered.

5.6) Hours of operation
To allow for adequate defrost of evaporators the hours of operation pr. 24 hours can be specified. Setting the hours of operation to a value lower than 24 h will result in a higher demand for cooling. This calculation is based on a simpel scaling of the calculated individual demands with the ratio between 24 h and the number of hours entered for hours of operation.

6) Total cooling demand
The total cooling demand is the sum of all the individual heat loads described. Both the maximum value (based on the maximum load from cooling of goods) and the average value are calculated. The SHR for the evaporator is calculated based on the humidity of the infiltrating air.

[Graham]

Yes. I can read the help page. Where the question started to form was from a technical article in a magazine where ASHRAE tables were quoted and the resultant calc supplied the answer of 0.55 meter squared K/W for a 50mm wall (expanded polystrene). If you visit engineers toolbox the value for polystyrene .03 W/m K and no matter how I play with the 50mm thickness I cannot get anywhere near the .34 that W.F. Stoecker achieves. I also have some hand me down tables that alledgedly originate from CIBSE and AIRAH the one has a value of 25.68 W/m K for which if multiplied by 0.05 (the wall thickness) = 1.284 which then divided into 1 = 0.77 as the coeficient which is way different to the 0.34. Who's tables do I use when I want to have a 230mm Brick wall plastered with 15mm perlite and then 50mm Polystrene ?

[nike123]

Where is gone delete message option?

[nike123]

Maybe this table could help!
http://en.wikipedia.org/wiki/U-value

[Peter_1]

... question started from a technical article in a magazine where ASHRAE tables ...the resultant calc supplied the answer of 0.55 meter squared K/W for a 50mm wall (expanded polystrene). ...engineers toolbox value polystyrene .03 W/m K and no matter how I play with the 50mm thickness I cannot get anywhere near the .34 that W.F. Stoecker achieves. I also have some hand me down tables that alledgedly originate from CIBSE and AIRAH the one has a value of 25.68 W/m K for which if multiplied by 0.05 (the wall thickness) = 1.284 which then divided into 1 = 0.77 as the coeficient which is way different to the 0.34. ...I want to have a 230mm Brick wall plastered with 15mm perlite and then 50mm Polystrene ?

Graham, 0.03W/m.K seems for me a a little bit low value, but this figures doesn't change if you have a thin or a thick wall, This because you said you played with a thickness of 50 mm to achive some result.
Perhaps I understood it wrong.
There's also a difference between expanded and extruded polystyrene. Perhaps Eng. toolbox gave another sort of polystyrene.
I haven't looked in tables but I can remember out of my head that you need +/- double thickness for expanded polystyrene if you want the same U-value for a PU wall.

http://en.wikipedia.org/wiki/Polystyrene
http://www.texasfoam.com/technical-data.htm

Don't confuse conductivity with isolation factor (don't know if these are the right English expressions) because the first is the opposite of the second.
The 25.68 you mention seems to be 1/25.68 = 0.038 W/m.K. I think you or the article didn't used the right units.

[US Iceman]

Don't confuse conductivity with isolation factor (don't know if these are the right English expressions) because the first is the opposite of the second.

Hi Peter. I think the word/phrase you are looking for is Thermal Resistance, which is shown by R. Therefore R= 1/C where C equals the Conductivity of one material of the wall section. In other words, R equals the inverse of C.

It's important to remember the difference between K & C. K = conductivity and C = conductance.

C is defined as the heat transfer factor for a specific building material at a specified thickness.

K is the heat transfer factor for a specific building material at an actual thickness.

So....

R (total) = (X1/K1) + (X2/K2) + (X3/K3) + (1/fi) + (1/fo)


X = actual thickness of material
fi = inside wall film coefficient
fo = outside wall coefficient

and the numbers indicate the different building materials.

So, 1/R (total) = U, or the overall heat transfer coefficient of the entire wall cross-section.

I think this is what you were hinting at.

[Graham]

Peter,thanks but Wikipedea links to The engineering toolbox, Texas foam is Imperial, I have been using similar numbers for 20 years now. Iceman, thanks you have all the letters in the right place, they make sense and I can use them. My problem is the different pieces of paper that I have, contain column headings that are the same however the numbers opposite the various materials differ greatly. I do not know what chart/table to use. Consider this a poll.

[US Iceman]

Graham,

Look at the fine print on the pages you have. They may list the values at different specific temperatures or they may be based on new material versus aged material.

Or, they may reference different standards (ASTM, etc)

Or..... they could be wrong! (just kidding). There must be a reason, but off the cuff I don't know what it would be without doing some homework.

[botrous]

Who's tables do I use when I want to have a 230mm Brick wall plastered with 15mm perlite and then 50mm Polystrene ?

Look at the fine print on the pages you have. They may list the values at different specific temperatures or they may be based on new material versus aged material.

Hi . . . how are you all ???

For the polysterene you should use the manufacturer value for the convinient range of tempreture , many densities of polysterene boards exists and each one with a different value.(maybe that's why your not getting good resolution results).

Respects and regards