Calculated cooling loads difference

[Anonymous]

Hello,

I've have a question about the results of the Heating and Cooling Loads calculation.

I have a space with the function where people eat. It also can be seen from the occupancy schedule in my attachment. (schedule and results in attachments) But I have a question about this, I only changed my occupancy schedule from the original schedule to the new schedule with zero occupancy.

But an odd result appears, with the ventilation load for cooling. This value increases, in my opinion for no reason because I haven't changed anything further. I thought this might happen because I put all values in the schedule to zero. But when I change the occupancy to 1% the ventilation cooling load still remains 651 [W]. 

I hope someone can explain this.

Thanks in advance!

[LjubanL]

Hi,

Probably because is selected (but Iam not sure you can test) :

If I remembered posts from autodesk forums that means : outdoor Air per person + outdoor air per area?

And when you choose occupancy schedule without people  you have some impact because of this sum (but why this value is greater-is not clear to me).

But if you want force Revit to use just once you can choose :

outdoor air method: max(by people, by area) to force revit to use just one parameter and if you want to be sure that this parameter will be all time used put outdoor air per area: 0

Also you can find a lots of discussion about this OA rates and what is the proper way to calculate this

https://aws.state.ak.us/OnlinePublicNotices/Notices/Attachment.aspx?id=113385

Also its not clear to me how the revit calculate this ventilation outdoor rate, because I didnt find any detailed instructions (except some 10 years old pdf) how it works (with formulas)

Maybe someday this post read some @Anonymous revit specialist and give to us reasonable explanation.

Kind Regards

Ljuban

[Anonymous]

Hi,

Thanks for your response. 

For the forcing of Revit to use just one outdoor air method.. I've tried switching between them but there is still the same increase of the ventilation cooling load. 

The behavior of the people load is explainable. But I also searched in ASHRAE and the formula for ventilation/infiltration load ("simplified"): 

Sensible heat gain: qs = 1.23 * Qs * (t1-t0)

Latent heat gain: ql = 1.20 * 2500 * Qs * (W1-W0) = 3010 * Qs * (Wi-Wo)

Here comes for Wi = Indoor air humidity ratio, kg/kg; the indoor air humidity also depends on the presence of people.

Maybe this plays a role?

Kind Regards

[LjubanL]

Hi,

I can do some test: and maybe someone makes some conclusions

Space :100m2 (10mx10m)

Height =5m

Without widows, doors…

I TEST (SCHEDULE ON)

-space type : TEST ROOM

Also we will set (zone settings) :

Humidification in space will be calculated by REVIT – recommendation by autodesk (because if you fix this value revit in lot of case cannot perform calculation)

(and this is a part of your comment, humidification have impact in latent part of qtotal: qtotal=qlatent+qsensible)

Revit calculation:

Ventilation is : 4575 W (schedule on) also relative humidity in space is 56%

And for this case we will try use formulas

(http://albadr.org/www/pdf/library/499.pdf):

SENSIBLE:

qs=1210* Air flow(m3/s)*(toutdoor-tindoor)

toutdoor =33

tindoor=23

air flow =3l/s*50people=150l/s=0.150m3/s

qs=1210* Air flow(m3/s)*(toutdoor-tindoor)=1210*0.150*(33-23)=1815W

LATENT:

ql=3010*Air Flow(m3/s)*(Wo-Wi)

Wo=? Outdoor condition (we have toutdoor wet=26 and t outdoor dry=33) we can find Wo=18.4g/kg

https://www.kwangu.com/work/psychrometric.htm

Wi=? (indoor condition is tindoor=23 and relative humidity is =56%

And for tindoor=23 and relative humidity=56% Wi=9.804g/kg

(https://afim-luchtdroging.com/en/mollier-chart-air-calculator/)

And now we can calculate

ql=3010*Air Flow(m3/s)*(Wo-Wi)=3010*0.150*(18.4-9.804)=3881W

ql- capacity that we need to remove humidity from space

qs-capacity that we need to keep desire temperature

qtot=3881+1815=5696W

II TEST (SCHEDULE OFF)

Ventilation is : 5167 W (schedule on) also relative humidity in space is 48%

SENSIBLE:

Qs=1210* Air flow(m3/s)*(toutdoor-tindoor)

toutdoor =33

tindoor=23

Qs=1210* Air flow(m3/s)*(toutdoor-tindoor)=1210*0.150*(33-23)=1815W

LATENT:

Ql=3010*Air Flow(m3/s)*(Wo-Wi)

Wo=? Outdoor condition (we have Toutdoor wet=26 and t outdoor dry=33) we can find Wo=18.4g/kg

https://www.kwangu.com/work/psychrometric.htm

Wi=? (indoor condition is tindoor=23 and relative humidity is =48

And for tindoor=23 and relative humidity=48% Wi=8.385g/kg

(https://afim-luchtdroging.com/en/mollier-chart-air-calculator/)

And now we can calculate

Ql=3010*Air Flow(m3/s)*(Wo-Wi)=3010*0.150*(18.4-8.385)=4520W

Qtot=4520+1815=6335W

COMPARE REVIT CALCULATION:

I TEST - qtot=4575

II TEST – qtot=5167W

5167-4575=592W

COMPARE MANUAL CALCULATION

I TEST - qtot=5696

II TEST – qtot=6335W

6335-5696=639W

Very similar difference, probably you have right, also for qlat, qtot we used some ashrae formulas, we don’t have information which formulas revit use (they probably differ in terms of value 1210, 3010)...

Kind Regards

Ljuban

[Anonymous]

Hi,

Thank you for the detailed explanation and test. 

I think we have it at the right end. Because it takes some energy to adjust the humidity in a room, so the components in the installation have to work harder. 

But I ran also some tests, I have simulated three scenarios. (And didn't select humidification control..)

a. Basic scenario

b. ONLY the occupant schedule with 1 % up on each hour 

c. ONLY the occupant schedule with 2 % up on each hour 

The difference between scenario basic and the scenario where the scedule is zero is indeed a difference in the humidity factor. 

But when I change my occupant schedule, and put 1 % on each hour. In the first from basic scenario vs +1% its again the humidity. But when I put it +2% there isn't any difference/odd results anymore..

So why is the basic vs the +1% a significant difference? And with +1% and +2% there isn't any difference anymore..

Below the results of each scenario a. b. c.

Why is the ventilation in a. 117 [W] for cooling load and for b. and c. 19 [W]?

The increase of cooling load for people is logic. An increase of 58 [W] per percentage. 

It still doesn't seem to make quite sense.. Maybe is in Revit some sort of constant step graph..?

Thanks again.

[LjubanL]

Hi,

Very interestingly test .

We have conclusion above that this formula have impact on ventilation (humidity) in space:

Ql=3010*Air Flow(m3/s)*(Wo-Wi)

Because only fluctuated value is Wi because relative humidity isn’t fix on one value in space

Wi- you can calculate this value because revit calculate relative humidity in space and you have temperature in space (like in previous reply)

Your case :

1. Basic scenario

 relative humidity is 68%, ventilation =117

1. ONLY the occupant schedule with 1 % up on each hour 

relative humidity 72%, ventilation =19

1. ONLY the occupant schedule with 2 % up on each hour

relative humidity is; 72%, ventilation is=19

in case b. and c. revit wasn’t change relative humidity and there is no reason to change ventilation (for b and c Wi is the same).

And about People :

1. Basic scenario

4779W

1. ONLY the occupant schedule with 1 % up on each hour 

4837W

1. ONLY the occupant schedule with 2 % up on each hour

4895W

If you look trane in pdf (http://albadr.org/www/pdf/library/499.pdf);

Qpeople=Qsensiblepeople+Qlatentpeople

Qlatentpeople=number of people * latent heat gain per person

Qsensiblepeople= number of peole* sensible heat gain per person x CLF

This CLF factor depends on occupant schedule. It’s a function of total hour in space and hours after people enter in space ( above table it’s a short version but in old books I have detailed table )

But If I look this formula

Qpeople=Qpeoplesensible + Qpeople latent

Only Q peoplesensible depends on schedule but this value have no impact on humidity in room only have impact on temperature

Qpeople latent have impact on humidity on space

But Qpeople latent must be increase depend on occupancy schedule (like people part in W)

Maybe when you change occupant from 1 on 2% a small impact to change humidity in space, did you try to rise on 50%?

Kind Regards

Ljuban

[Anonymous]

Hi,

I've tried testing your advice to increase the schedule with 50%.

The results are shown below.

Still here the results are shown as;

a. Basis scenario

b. ONLY the occupant schedule is increases by 50%.

[LjubanL]

Hi again 🙂

I was try to replicate your procedure to make some conclusion

same space as above (10x10=100m2)

1. case schedule of:

2. case schedule 1% increase:

3. case schedule 2% increase:

and finally i will put much more in schedule to be sure that is increase enough to change humidity in space 

4. case schedule 50% occupant:

and we can see that is relative humidity was change on 52%-but we need to put much more occupant in space.

because we have humidity on 50%, and when I increase a occupant on 50% in every hour revit was recalculate the value and its a very small change only 2%, in other case when I put 2% increase revit was calculate some 50.25564525 etc (just as example) but in report we can see as 50.00%.

Kind Regards

Ljuban

[Anonymous]

Hi,

Also an interesting test. 

So it's important to know which standard outside humidity percentages it takes, to see the difference between outside and inside humidity. 

But as we see, the more people, the less ventilation load. This still doesn't make quite sense to me.

Because you selected outdoor air per person..

There is probably more airflow given, but less load.. quite odd appearance. 

[LjubanL]

Hi,

"So it's important to know which standard outside humidity percentages it takes, to see the difference between outside and inside humidity."

agree with you

if you look this formula :

Ql=3010*Air Flow(m3/s)*(Wo-Wi)

only value which need to determinate is Wi because Wo is outdoor condition and we can not influence on oudoor temperature and humidity.

"But as we see, the more people, the less ventilation load. This still doesn't make quite sense to me.

Because you selected outdoor air per person.."

There is probably more airflow given, but less load.. quite odd appearance."

also agree 

but the the reason why is less ventilation load is because we allow revit to calculate indoor humidity

if we see diagram:

 

revit it increases the humidity and approaching the external conditions and (wo-Wi) is less...

and the real question is which formula revit uses to get the humidity inside the room because if  I put fix humidity in space as we do in manual calculation ( 50%as example) revit cant perform calculation in lot of case... I dont want cooling load determine for humidity in space 72% as some example from above tests...

Kind Regards

Ljuban

[Anonymous]

Hi again,

And thanks again. 😉

So the real question is which formula revit uses to get the humidity inside the room.. 

For the humidity, I found some parameters involving humidity in ASHRAE.

It writes; "Basic Parameters

Humidity ratio W (or mixing ratio) of a given moist air sample
is defined as the ratio of the mass of water vapor to the mass of dry
air in the sample:

W = Mw/Mda (7)

W equals the mole fraction ratio xw/xda multiplied by the ratio of molecular masses (18.015 268/28.966 = 0.621 945):

W = 0.621 945xw/xda (8)

Specific humidity y is the ratio of the mass of water vapor to total mass of the moist air sample:
y = Mw/(Mw+ Mda) (9a)

In terms of the humidity ratio,
y = W/(1 + W) (9b)

Absolute humidity (alternatively, water vapor density) dv is the
ratio of the mass of water vapor to total volume of the sample:

dv = Mw/V (10)

Density p of a moist air mixture is the ratio of total mass to total volume:

p = (Mda + Mw)/V = (1/v)(1 + W) (11)

where v is the moist air specific volume, m3/kgda, as defined by Equation (24).

Dew-point temperature td is the temperature of moist air saturated at pressure p, with the same humidity ratio W as that of the given sample of moist air. It is defined as the solution td (p, W) of the following equation:
Ws (p, td) = W (13)

Thermodynamic wet-bulb temperature t* is the temperature at which water (liquid or solid), by evaporating into moist air at drybulb temperature t and humidity ratio W, can bring air to saturation adiabatically at the same temperature t* while total pressure p is constant. This parameter is considered separately in the section on Thermodynamic Wet-Bulb and Dew-Point Temperature."

To calculate absolute humidity, you must first use the dewpoint temperature to calculate vapor pressure in millibars.

[LjubanL]

Hi,

This is the basic formulas from books, ashrae and that’s ok. But how revit use this formulas to determinate relative indoor humidity (probably some kind of iteration?) we don’t know because @Anonymous did not provide such kind of guidance.

Also when I analyze above formulas and try to determinate relative humidity in space I find that we have “one more unknown” in this formulas to determinate relative humidity in space (and I assume that revit use some kind of iteration).

And how it connect these formulas with number of people in space and qlatent which value has impact on humidity in space?

I am afraid that without the help of @Anonymous, which would share the formulas used for the analysis,

we will not be completely sure what the revit works.

Also if I fixed humidity in space why revit cannot perform calculation in many cases?

Also very interesting study;

https://www.tesisenred.net/bitstream/handle/10803/406007/THBG1de1.pdf?sequence=1&isAllowed=y

Kind Regards

Ljuban

[michel_martinss]

HI,
Could you forward me this pdf?
I'm seeing this same type of problem in Revit.
The correct thing would be to fix the relative humidity without problems, as in other thermal simulation programs.

http://albadr.org/www/pdf/library/499.pdf