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7.2 Heat Load Calculation
7.2.1 Load Calculation of a Single Floor
Room
Master Room
Bathroom
Total Load
7.2.2 Arrangement Design of the Underfloor System for A Single Floor
Assumed conditions: the floor is cement or ceramics, the normal external diameter of the heating pipe is
20mm, thickness of the stuffer is 50mm, thickness of PS foam insulation is 20mm, supply water temperature is
45°C , return water temperature is 35°C , indoor design temperature is 20°C .
7.2.3 Arrangement Design of the Underfloor System for the Bath Room
Heat load of the bath room is 900W, heat dissipation per unit area is 75W/m2, tube spacing of the heat pipe
is 30mm, and heat loss is 25.4 W/m2, then the total heat loss is:
Based on the heat load listed in the table above, the heating load for the bathroom is:
According to the formula Q=CρGΔT, the flow rate of the heating pipe for the bathroom is:
If the outer diameter of the heating pipe is 20mm and thickness is 2mm, then the minimum flow for the
heating pipe is:
G=
It can be see that the arranged piping system for the bathroom does not meet the technical requirement and
must be used in common for the master room.
7.2.4 Arrangement Design of the Underfloor System for the Master and Both Rooms
According to the calculation results, the total heat load for the master and bath rooms is 3196W, heat
dissipation per unit area is 82W/m2, tube spacing of the heating pipe is 300mm, and heat loss 25.4 W/m
then the total heat loss is:
According to the formula Q= CρGΔT, the flow rate is
G=
Loop quantity is 0.3622/0.18=2.012 and the round-off number is 2.
7.2.5 Check
◆ A. Check for the flow rate
Floor rate of each loop is within 0.25~0.5m/s and the system can run stably.
◆ B. Check for the tube length
When the average tube spacing is 30mm, the required length of the heating pipe per square meter is 3.5m,
length of total coils is 3.5×40=140 and length for each loop is 140/2=70.
It can be seen that the length for each loop is less than 120m and there it meets the design requirement.
◆ C. Check for the ground average temperature
Area
Long-term Dwelling Area
Short-term Dwelling Area
Nobody Area
Area
28
12
Average Temperature of the Heating Pipe=(45+35)/2=40°C
25.4×12=304.8W
900+304.8=1204.8W
Q
1.2048kJ/1/3600 h
G=
=
=3.14/4*(20-2*2)
3196+1016=4212W
Q
4.212kJ/(1/3600h)
=
4.186kJ/(kg·℃)*1000kg/m
0.3622/2
3.14*0.008
tp=tn+9.82×(qx/100) 0.969=20+9.82×(82/100) 0.969=28°C
Upper Limits and Average Floor Temperature
Average Floor Temperature
Average Temperature
24~26
28~30
35~40
2
Heat Index (W/m
82
72
2296+900=3196w
=0.104m
*10
*0.25*3600 =(0.18m
2
-6
=0.3622m
*(45-35)℃
3
=0.2503m/s
*3600
2
Maximum Temperature
9
Design & Selection
)
Heat Load (W)
2296
900
/h
3
)/h
3
/＞0.18m
/h
3
3
28
32
42
2
is,