Week 6

2904ENV Principles of Sustainable Design

Tutorial / Workshop 6 – Week 6

Thermal comfort

Aim: To understand thermal comfort indicators and the meaning of PMV (Predicted Mean Vote)

WORKING IN GROUPS OF TWO CONSIDER THE FOLLOWING TWO climatic conditions on the Gold Coast:

•       Summer

•       Winter

Task:

-        Calculate the PMV for the three sample days in summer and winter (Data from table 1)

-        For the sample Day 2 in summer, calculate the air temperature, humidity and air velocity range in which people will be comfortable and PMV will be between -0.5 and 0.5*.  (Table 2)

-        Further discuss how other factors such as activity or clothing can improve the PMV for the studied sample days

* This means that for calculating the air temperature range in which PMV would be between -0.5 and 0.5, we would keep all the figures in the row of the Day 2 in summer the same and only change air temperature in a way that the red dot in the Psychrometric Chart falls in the Blue Zone. We repeat the procedure for humidity and air velocity.

Use CBE Thermal Comfort Tool available at http://comfort.cbe.berkeley.edu/

Get the climatic data from the table 1.

Fill in the blank spaces in Table 1 and Table 2.

Table 1: Climate Data

Summer	Air temperature (oC)	Globe Temperature (oC)	Air speed (m/s)	Humidity (%)	Activity	Clothing	Mean Radiant Temperature (oC)	PMV	Sensation
Day 1	22.8	24.1	0.1	60	Filing, seated	Knee-length skirt, short sleeve shirt, sandals	24.9	-0.13	Neutral
Day 2	30.0	32.0	2.2	85	Standing relaxed	Shorts, short sleeve shirt	38.8	0.82	Slightly Warm
Day 3	38.5	34.0	0.1	80	Sleeping	Shorts, short-sleeve shirt	31.6	2.44	Warm
Winter
Day 1	19.1	24.0	0.3	55	Sleeping	Trousers, short sleeve shirt, socks, shoes	29.5	-4.15	Cold
Day 2	13.2	20.0	0.1	50	Standing relaxed	Trousers, long sleeve shirt	24.1	-1.81	Cool
Day 3	3.8	5.6	5.3	45	Walking 2mph	Jacket, Trousers, long-sleeve shirt	18.7	-3.28	Cold

Table 2:

Comfort condition range in which PMV is between -0.5 and +0.5	Air temperature Range (oC)	Humidity Range (%)	Air Velocity Range (m/s)
Day 2 in Summer	26.8-29.2oC	<70%	3.5-21 m/s

Week 9

Exercise 1

1. Determine the window head height for each window. The window head height (WHH) is the vertical distance from the finished floor level to the top of the glazing.

2700 mm

2. Determine the depth of the zone. The zone depth is one window head height into the area adjacent to the window.

1800 mm

3. Calculate the width of the zone. The zone width is the window’s width added to half the window head height on each side of the window.

4. Subtract any area blocked from receiving daylight by a permanent obstruction that is six feet or taller, Modular furniture is not considered a permanent obstruction.



Exercise 2

Exercise 2
Desk 1	Daylight factor (%)	Adequacy		Eo	Ei
Brisbane	17.95936693	6	-27.4	8909	1600
Darwin	16.400164	6	-12.4	9756	1600
Perth	18.48428835	6	-31.9	8656	1600
Sydney	19.13189047	6	-33.8	8363	1600
Adelaide	19.321338	6	-24.9	8281	1600
Canberra	19.44343177	6	-35.2	8229	1600
Melbourne	20.1435226	6	-37.8	7943	1600
Hobart	21.46498524	6	-42.8	7454	1600
Desk 2	Daylight factor (%)	Adequacy		LUX
Brisbane	5.612302166	6	-27.4	8909	500
Darwin	5.125051251	6	-12.4	9756	500
Perth	5.776340111	6	-31.9	8656	500
Sydney	5.978715772	6	-33.8	8363	500
Adelaide	6.037918126	6	-24.9	8281	500
Canberra	6.076072427	6	-35.2	8229	500
Melbourne	6.294850812	6	-37.8	7943	500
Hobart	6.707807888	6	-42.8	7454	500
Desk 3	Daylight factor (%)	Adequacy	400
Brisbane	4.489841733	4	-27.4	8909	400
Darwin	4.100041	4	-12.4	9756	400
Perth	4.621072089	4	-31.9	8656	400
Sydney	4.782972617	4	-33.8	8363	400
Adelaide	4.830334501	4	-24.9	8281	400
Canberra	4.860857941	4	-35.2	8229	400
Melbourne	5.03588065	4	-37.8	7943	400
Hobart	5.366246311	4	-42.8	7454	400
Exercise 3
The lamp capacity	140
Room Area	15940
Lamp Power density	113.85

Exercise 3

Step one – determine lamp capacity by adding the lamp wattage of the three lamps

40+40+60= 140 Watts

Step two – determine room area

Height 600+200+210 + 135+135+120 = 1400
Width = 1500

140 X 150 = 21000

1+26.5+12.5 = 40
13.5+13.5+12 = 39

40 X 39 = 1560

120+200+380 = 700

50 X 70 = 3500

21000-1560-3500 = 15940


Step three – determine lamp power density by dividing the wattage by floor area

113.85