Help me deepen my system design skills + pipe sizing knowledge, Thanks!

[Mike Monstrosity]

I would like to broaden my theoretical understanding by using an simplified example:

There is a building containing one room with a floor surface of 40 m2 which needs exactly 2.8 kW’s to maintain its room temperature while outside temp is -10 celcius (The heat-loss of the house is 70 Watt / m2 at -10 degrees celcius).

· The room is heated by a floor-heating-system which has double the capacity to heat the room. It has 5.6 kWs of heating capacity.
· The distance between the heat source(a traditional naturalgasburing heatsource) and the floor heating dispenser is 20 meters.
· For this exercise lets assume there is 0% friction in the piping. The piping is straight and leveled.
· The pressure that the heat-source can exert on the water by means of a pump is limitless.
· The speed of the water may never exceed 1 m/s
· The water temperature is 35 degrees Celsius.
· The goal is to maintain a delta T of 5 degrees Celcius. This the temperature of the returning water should be 30 degrees.

Given the above stated facts what should be the minimum diameter of the pipes? Can you please show the formula by which this can be calculated?

Thanks in advance.

___

I'll give it a try myself.

To determine the diameter we first need to determine the flow that is needed to meet our quota of 2.8 kW's. The formula to determine that is:

Q = kW / 4.18 x Delta T

2.8 kw / ( 4.18 x 5 degrees celcius) = 0.1339 kg /s

0.1339 x 3.6 = 0.4823 m3/h

Q = 0.48 m3/h

If this formula is correct. Than where does the 4.18 come from?

Now based on Q and the maxium allowed speed of 1 m/s i can calculate:

Q / 3600 / maximum allowed speed*4) / PI^0.5 = 0,0131 m x 1000 = 13,1 mm.

 

[bernardgreen]

The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C which is higher than any other common substance.

 

[Mike Monstrosity]

The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C which is higher than any other common substance.

Thanks for replying Bernardgreen. Help me understand please. How does this help me awnser my question?

 

[Mike Monstrosity]

Sorry now i know!
So the formula is correct? And my awnser of 13,1 mm aswell?

 

[bernardgreen]

(The heat-loss of the house is 70 Watt / m2 at -10 degrees celcius)

If that is -10°C diferential ( from inside to outside ) then at -10°C external your temperature inside would be 0°C

If external temperature is -10°C and required internal temperature is to be 20°C then the differential is 30°C

(The heat-loss of the house is 70 Watt / m2 at -10 degrees celcius).
.The heat-loss of the house is 210 Watt / m2 at 30°C differential

 

[Mike Monstrosity]

If that is -10°C diferential ( from inside to outside ) then at -10°C external your temperature inside would be 0°C

If external temperature is -10°C and required internal temperature is to be 20°C then the differential is 30°C

(The heat-loss of the house is 70 Watt / m2 at -10 degrees celcius).
.The heat-loss of the house is 210 Watt / m2 at 30°C differential

I see. I received the 70 w / m2 as a standard number based on buildinglaws in my country. It means that it can maintain its temperature at minus -10 celcius outside. Nevertheless your sharp and i'll reread the information.

For now lets assume i'll just need the 2.8 kW's in capacity. And my formula to get the Q is correct.

How can i now determine if my heatsource operating modus complies with my installation. My hydraulic capacity is:

Max volumeflow (dT = 4K) 1620 l/h
min volumeflow (dT = 8 K) 810 l/h
Nominal available pressure at Outside temp 7 degree's (this is the countries avarage temp ) A7/ and water temp 35 degree C = 36 kPa

 

[Mike Monstrosity]

The pumpcurve starts at 7 MWK and has 0,5 m3/h and goes down to 1.8 mWK and has 2 m3/h