Help!How to calculate heat load with resistivity values?

Hi there i am currently doing an assignment, im trying to work out the heat load of the structure of my cold room. I have been given the following data to work with.

* The room is 15m x 10m x 5m
*The walls & ceiling are 150mm thick & are typical rollbound construction, 150mm polystyrene sandwhiched between 1mm sheet steel
*The floor is 200 mm thick , again of rollbound construction, it is standing on a 200mm concrete slab, the concrete slab is standing on the ground
*The ambient temp surrounding the room is 25 degrees celcius, The RH is 60 %

R1, Thermal resistance ,air film on outside of walls & ceiling is 0.03 W/m2/K

R2, Thermal resistance ,air film on inside of walls & ceiling is 0.12W/m2/K

R3, Thermal resistance, sheet steel 0.021W/m thick/m2/K

R4, Thermal resistance , polystyrene 28.9 W/m thick /m2/K

R5, Thermal resistance, concrete 1.07 W/m thick/m2/K

Now how would i go about working out the gain using a resistivity formula , i seem to recall it was similar to working out the resistance of an electrical circut in paralell.

Is this the formula here? resistivetey/m thick/m2/K with K being the out side ambient? then once all the equations have been solved i go R1+R2+R3+R4+r5=??

Thanks in advance if anyone can help me

Hi JBomber
I think what you are looking for is -
U = 1/(R1 + R2 + R3 + R4 + R5)
Q = A*U*deltaT
K = deltaT
K will equal 25 if your room temp is 0degC.
Then it's not going to be 25degC under that concrete slab - maybe around 15degC in NZ, and yes it is similar to the electrical formula.

Quote:

---

Originally Posted by Tesla

Hi JBomber
I think what you are looking for is -
U = 1/(R1 + R2 + R3 + R4 + R5)
Q = A*U*deltaT
K = deltaT
K will equal 25 if your room temp is 0degC.
Then it's not going to be 25degC under that concrete slab - maybe around 15degC in NZ, and yes it is similar to the electrical formula.

---

thanks for the help. i think the resistivity values given are for 1m thickness, how would i go about getting the correct value at the thickness stated above? Or would those resistivity values be at the thicknesses stated? Its a bit unclear when you have noone to ask, thanks alot for the help.

We need to convert those values into metres, so if we have 200mm - that would be 0.2 metres and 1mm would be 0.001 metres.

Quote:

---

Originally Posted by Tesla

We need to convert those values into metres, so if we have 200mm - that would be 0.2 metres and 1mm would be 0.001 metres.

---

ok so id divide each resistivity value by the desired thicknes then go 1/r1+r2+r3+r4+r5=?

Yes jbomber
Don't forget there are two sheets of steel and two calcs - roof and walls then floor. Please post you answers with a little working incase you or I made a mistake and others can learn from the exersise.

Quote:

---

Originally Posted by jbomba

Hi there i am currently doing an assignment, im trying to work out the heat load of the structure of my cold room. I have been given the following data to work with.

* The room is 15m x 10m x 5m
*The walls & ceiling are 150mm thick & are typical rollbound construction, 150mm polystyrene sandwhiched between 1mm sheet steel
*The floor is 200 mm thick , again of rollbound construction, it is standing on a 200mm concrete slab, the concrete slab is standing on the ground
*The ambient temp surrounding the room is 25 degrees celcius, The RH is 60 %

R1, Thermal resistance ,air film on outside of walls & ceiling is 0.03 W/m2/K

R2, Thermal resistance ,air film on inside of walls & ceiling is 0.12W/m2/K

R3, Thermal resistance, sheet steel 0.021W/m thick/m2/K

R4, Thermal resistance , polystyrene 28.9 W/m thick /m2/K

R5, Thermal resistance, concrete 1.07 W/m thick/m2/K

Now how would i go about working out the gain using a resistivity formula , i seem to recall it was similar to working out the resistance of an electrical circut in paralell.

Is this the formula here? resistivetey/m thick/m2/K with K being the out side ambient? then once all the equations have been solved i go R1+R2+R3+R4+r5=??

Thanks in advance if anyone can help me

---

If you look at the formulas that they have provided, you will see that the first 2 are for the air film and they are per m^2, so just work out the m^2 and multiply.

in the 3rd, 4th and 5th, they are per M thick x M^2, so again, just work out the M^2 and multiply by the M thick.(for 150mm read 0.15m)

The only thing missing from the calculation - or your post - is the Delta C (K)

You state that the outside temperature is 25C but you do not give an inside box temperature

Also, you need a ground temperature for the floor. I would use 10C.

ok i finally had time to sit down and do this. the temp inside the freezer is -35. which gives me a Delta T of 10 degrees. underneath my floor is 10 degrees. here are my workings. Can somebody please have a look through these and let me no if ive messed up at all. You guys have been very helpful.

Areas of roof , walls & ceilings

15x10=150m^2
5x10=50x2=100m^2
15x5=75x2=150m^2

R1 resistivety of 0.03W - airfilm on outside of walls & ceiling
R2 resistivety - air film on inside walls & ceiling of 0.12W
Total area of walls & ceiling = 310m^2

r1= 0.03x310x10=93
r2=0.12x310x10=372

Sheet Steel 0.021x0.01x310x10=0.651x2=1.302
Poly 28.9x0.15x310x10=13439

walls & ceilings = 93+372+1.302+13439=13.905kW

(i think this next bit may be incorrect?)

Floor:

Concrete - 1.07x0.20x150x15=481.5
Sheet steel - 0.021x0.01x150x15= 0.4725x2= 0.945
Poly - 28.9x0.20x150x15= 13005

481.5+0.945+13005=13.488kw+13.905kW=27.393kW

27.393kW Total.

If somebody could please take the time to check my calculations it would be great. I have a feeling ive done something wrong with the floor section

Quote:

---

ok i finally had time to sit down and do this. the temp inside the freezer is -35. which gives me a Delta T of 10 degrees.

---

Outside temperature of 25°C and inside temperature of -35°C gives you delta T of 60°C and not 10°C.
Also, ground below floor temperature of 10°C and indoor temperature of -35°C gives you delta T of 45°C.

You cannot simply add two numbers. You need to find interval between these two temperatures.

HI jbomba
A couple mistakes here they are
K roof and walls = 60K
K floor = 45K
roof area =150msq - You forgot this one
You need two areas 1=150msqfloor
2=400msqroof+walls
Remember that U value - it's important because the basic formula is Q=A*U*deltaT, then transpose it
All you do with the R values is thickness then add like this
Rroof+walls=R1+R2+R3+R3+R4. Then take the reciprocal (1/R Total)and you should get an answer around Uroof+walls = 0.229
As above for floor R2+R3+R3+R4+R5...Ufloor = 0.163
Then Q = A*U*deltaT(K) add floor and (roof+walls), your answer should be around 6455 watts or 6.455Kilowatts depending on your calculator
Note R4 will be differant for floor as to roof+walls
Thanks for the first effort please repost with workings - even I made a few mistakes in my first workings on paper.
There are more than one way of skinning the cat as there are for working out this problem

Quote:

---

Originally Posted by Tesla

HI jbomba
A couple mistakes here they are
K roof and walls = 60K
K floor = 45K
roof area =150msq - You forgot this one
You need two areas 1=150msqfloor
2=400msqroof+walls
Remember that U value - it's important because the basic formula is Q=A*U*deltaT, then transpose it
All you do with the R values is thickness then add like this
Rroof+walls=R1+R2+R3+R3+R4. Then take the reciprocal (1/R Total)and you should get an answer around Uroof+walls = 0.229
As above for floor R2+R3+R3+R4+R5...Ufloor = 0.163
Then Q = A*U*deltaT(K) add floor and (roof+walls), your answer should be around 6455 watts or 6.455Kilowatts depending on your calculator
Note R4 will be differant for floor as to roof+walls
Thanks for the first effort please repost with workings - even I made a few mistakes in my first workings on paper.
There are more than one way of skinning the cat as there are for working out this problem

---

ok so do i get my resistivety value , multiply it by the desired thickness. Then go 1/answer to get my u value. Then do i add those answers together to get a final U value then do Q=U*A*T of the entire area M^2

any chance of an example. As you can prob tell maths is my downfall :D

Quote:

---

Originally Posted by jbomba

ok so do i get my resistivety value , multiply it by the desired thickness. Then go 1/answer to get my u value. Then do i add those answers together to get a final U value then do Q=U*A*T of the entire area M^2

any chance of an example. As you can prob tell maths is my downfall :D

---

No, you need to calculate u-value for every different composition of materials, then calculate separate Q for total areas of same compositions at same delta T and finally sum all calculated Q-s.

HI jbomba
almost right - get your resistivity value by multiplying by the thickness of the material. Add all R values for roof and walls, then take the reciprocle - this will give you the U value for the roof and walls, then apply to the formula Q=AUdT. Here is an example using the same R values as you have and material thicknesses as they are standard.

A freezer room of 20m*6m*10m
AOT(aotside air temp)=28C
Space(room) Temp = -24C (icecream)
Resistivity values same as yours
R1 sir outside wall 0.03 W/m2/K
R2 air inside wall 0.12 W/m2/K
R3 sheet steel 0.021 W/m/m2/K
R4 poly 28.9 W/m/m2/K
R5 cocnrete 1.07 W/m/m2/K

Formulae
Q=A*U*dT
U=1/R1+R2+R3...

1 calculate A of (a) walls + roof
1(a) = roof 10m*20m=200m2
wall 20m*6m =120m2
wall 20m*6m =120m2
wall 10m*6m =60m2
wall 10m*6m =60m2
Area(a) = 560m2

1 calculate A of (b) floor
1(b) = 10m*20m=200m2
Area(b) = 200m2

2
R1(a) = 0.03 W/m2/K
R2(a)(b)= 0.12 W/m2/K
R3(a)(b) = 0.000021 W/m2/K (.021*.001)
R4(a) = 4.335 W/m2/K (28.9*.15)
R4(b) = 5.78 W/m2/K (28.9*.2)
R5(b) = 0.214 W/m2/K (1.07*.2)

3
Find U for roof + walls (a)
U(a) = 1/(.03+.12+.000021+.000021+4.335)
U(a) = 0.223

Find U for floor (b)
U(b) = 1/(.12+.000021+.000021+5.78+.214)
U(b) = 0.163

4
dT(a) = 28C - -24C
=52K
dT(b) = 10C - -24C
=34K

5
Apply Q=A*U*dT
5(a) = 560m2*.223*52K
=6494W

5(b) = 200m2*.163*34K
= 1108W

6 Add (a) + (b)
= 6494 + 1108

Q = 7602W

Q = 7.602kiloWatts

Your panels are not made for such temperature.
You need polyurethane panels 200 mm thick and you must insulate the floor as well with the thickness as the walls.

Other wise, you will be waisting a lot of energy and you will see very quick green condensation lines on the outside.

Quote:

---

Originally Posted by chemi-cool

Your panels are not made for such temperature.
You need polyurethane panels 200 mm thick and you must insulate the floor as well with the thickness as the walls.

Other wise, you will be waisting a lot of energy and you will see very quick green condensation lines on the outside.

---

this is not an actual design its just an exercise

Quite correct jbomba
We (experienced) know that that 150mm pannel would not hold straight after a period of time along with the thickness in insulation issue. How did you go with the calcs? did you get it? I will be visiting wgtn and pnorth next week

Quote:

---

Originally Posted by Tesla

Quite correct jbomba
We (experienced) know that that 150mm pannel would not hold straight after a period of time along with the thickness in insulation issue. How did you go with the calcs? did you get it? I will be visiting wgtn and pnorth next week

---

im just about to sit down and do them now , ive been putting it off for a while.