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PUSH NOTE : ARC2047H-Building_Science_Materials_and_Construction_3.md
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ThinkWithPbody committed Oct 16, 2024
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@@ -105,7 +105,7 @@ Thermal comfort directly affects productivity
- metrics and terminology
- air temperature
- humidity and dewpoint
- Relative humidity: the actual water vapour in the air / the total water vapour that could be in the air
- Relative humidity: the actual water vapour in the air ÷ the total water vapour that could be in the air
- metabolism & clothing: met & clo
- dry bulb & wet bulb temperature
- surface temperatures
@@ -129,7 +129,7 @@ Thermal comfort directly affects productivity
- Draw straight line between two air states representing linear interpolation of every possible mix of two air streams
- Use mixed air stream T>dry to determine point on line
- T1 \* R1 + T2 \* R2 = T3, R1 + R2 = 1
- Dry air mass (kg) = Room volume (m^3) / SV (m^3/kg)
- Dry air mass (kg) = Room volume (m^3) ÷ SV (m^3/kg)
- Water (g) = Dry air mass (kg) × HR (g/kg)
- Energy (kJ) = Dry air mass (kg) × Enthalpy (kJ/kg)
- Sensible Heat (): H>s
@@ -140,8 +140,8 @@ For a well-sealed and highly insulated room with a wall air conditioner, how muc
h1=27.2kJ/kg
h2=65.5kJ/kg
V=450m^3
Dry air mass1 (kg) = Room volume (m^3) / SV1 0.825 (m^3/kg) = 545.4545
Dry air mass2 (kg) = Room volume (m^3) / SV2 0.868 (m^3/kg) = 518.4332
Dry air mass1 (kg) = Room volume (m^3) ÷ SV1 0.825 (m^3/kg) = 545.4545
Dry air mass2 (kg) = Room volume (m^3) ÷ SV2 0.868 (m^3/kg) = 518.4332
Energy1 (kJ) = Dry air mass1 (kg) × Enthalpy (kJ/kg) = 14,836.3624
Energy2 (kJ) = Dry air mass2 (kg) × Enthalpy (kJ/kg) = 33,957.3746
Excess = 19,121.0122 kJ
@@ -179,9 +179,9 @@ Excess = 19,121.0122 kJ
- HSA = azimuth
- **VSA = arctan(tan(altitude)/cos(HSA))** (Calculator in degrees)
- tan(VSA) = tan(altitude)/cos(HSA)
- **Horizontal Projection = Window Height / tan(VSA)**
- **Horizontal Projection = Window Height ÷ tan(VSA)**
- **Lateral Projection = tan(HSA) × Horizontal Projection**
- **Vertical Projection = Window Depth-Width / tan(HSA)**
- **Vertical Projection = Window Depth-Width ÷ tan(HSA)**
### Q2

#### Module 04
@@ -192,26 +192,26 @@ Excess = 19,121.0122 kJ

- Balance Point Temperature, BPT ^ba60bb
- When Q_i = q_total
- BPT = T_indoor - Heat gain rate / Heat losses rate per T_delta
- BPT = T_indoor - Heat gain rate ÷ Heat losses rate per T_delta
- T_delta = (T_indoor - T_outdoor)
- q_total = q_skin + q_infiltration = UA_total × T_delta
- BPT = T_indoor - Q_i / UA_total
- BPT = T_indoor - Q_i ÷ UA_total
- Q_i: Internal Heat Gain (Btu/h) = q_solar + q_internal
- q_solar = Solar Insolation (BTU/(day-ft^2)) \* Surface Area South (ft^2) × 1d/24h
- q_internal = A: Floor Area (ft2) × (q_people + q_equipment + q_lights) (Btu/(h·ft2))
- UA_total: Heat Loss (Btu/(h\*oF)) = UA_envelope + UA_infiltration
- UA_envelope = U-value × Area (ft^2)
- U (Btu/(h\*ft^2\*^oF)) (W/m^2 ^oK) = 1 / R
- U = k: Conductivity / Thickness
- U-Value = 1 / R-Value
- U (Btu/(h\*ft^2\*^oF)) (W/m^2 ^oK) = 1 ÷ R
- U = k: Conductivity ÷ Thickness
- U-Value = 1 ÷ R-Value
- 0 ^oC = 273.15 \^oK
- UA_infiltration = ACH (h^-1) × Indoor Air Volume (ft^3) × Heat Capacity of Air (Btu / (ft^3 \* ^oF)) (J/(m^3 \* ^oK))
- UA_infiltration = ACH (h^-1) × Indoor Air Volume (ft^3) × Heat Capacity of Air (Btu ÷ (ft^3 \* ^oF)) (J/(m^3 \* ^oK))
- ACH: Air Changes per Hour ?= 0.44 (h^-1)
- Heat Capacity of Air (Btu / (ft^3 × ^oF)) = Density of Air × Specific Heat of Air = 0.075 lb/ft^3 × 0.24 Btu/(lb × ^oF) = 0.018 Btu/(ft^3·°F)
- Heat Capacity of Air (Btu ÷ (ft^3 × ^oF)) = Density of Air × Specific Heat of Air = 0.075 lb/ft^3 × 0.24 Btu/(lb × ^oF) = 0.018 Btu/(ft^3·°F)

Units of Power: Btu/h, W
Units of Energy: Btu, kWh
W = J / s
W = J ÷ s
1 kWh = 3600000 J

##### Lab
@@ -265,7 +265,7 @@ Biogenic Carbon?

- Cooling Loads
- **q_total: Total Heat Gain = q_sensible + q_latent**
- **Tons of cooling** = q_total / 12000 ^caf758
- **Tons of cooling** = q_total ÷ 12000 ^caf758
- Ton of cooling is the rate of heat transfer (power)
- 12000 Btu/h needed to melt one ton of ice in 24 hours.
- **Sensible Heat Gain**
@@ -309,7 +309,7 @@ Infiltration Factor = 0.7 - 1.5* ((0.7-1.1)/-5) = 0.58

Ventilation Factor = 11 - 1.5* ((11-16)/-5) = 9.5

(3.4 + 5.1)* 29426* 12 / 24
(3.4 + 5.1)* 29426* 12 ÷ 24

132765 × 1.25

@@ -358,7 +358,7 @@ Ventilation Factor = 11 - 1.5* ((11-16)/-5) = 9.5
**Cost of Thermal Comfort (Heating and Cooling)**
- Heating
- Annual Heat Needed (Btu)= UA_total (Btu/h\*^o𝐹) × HDD × 24 hours ^359eb9
- Annual Fuel Needed, E (ft3) = Annual Heat (Btu) / (AFUE × heat value of fuel (Btu/ft3)) ^591155
- Annual Fuel Needed, E (ft3) = Annual Heat (Btu) ÷ (AFUE × heat value of fuel (Btu/ft3)) ^591155
- Specific to the fuel source and equipment
- AFUE is the Annual Fuel Utilization Efficiency
- Assume 95%
@@ -372,13 +372,13 @@ Ventilation Factor = 11 - 1.5* ((11-16)/-5) = 9.5
- Historically 50°F was used as the base temperature for calculating annual cooling.
- If we can provide shading to some of the south glass we can improve the Summer BPT.
- assume 80% effective shading, thus allowing 20% of the solar gain.
- T_balance = T_desired: 78°F (25.6°C) - Q_i: Heat gain / UA_total: Heat loss
- T_balance = T_desired: 78°F (25.6°C) - Q_i: Heat gain ÷ UA_total: Heat loss
- **Annual Cost of Cooling** (\$) = Cooling Load (Btu/h) × CLH (h) × 1/SEER (W/Btu/h) × cost of electricity ($/Wh) ^8fe61e
- Cooling Load (Btu/h) From W6
- July Solar insolation for vertical surface, south facing glazing is 1,035 Btu/(day·ft2) in Toronto
- q_insolation cube summer@80%shade = 3,996 ft2 × 0.20 × 1,035 Btu/(day·ft2) × (day/24 h) = 34,687 Btu/h
- Only change solar insulation q_solar_summer_shaded, q_internal remains same
- CLH: Cooling Load Hours (h) = CDD × 24 (h/day) / (Design Temperature - Summer BPT) ^f100e8
- CLH: Cooling Load Hours (h) = CDD × 24 (h/day) ÷ (Design Temperature - Summer BPT) ^f100e8
- SEER: Seasonal Energy Efficiency Ratio ((Btu/h)/W)
- Btu/h heat rejected per Watt of electricity used
- For example AHU: Air Handling Unit could be 14
@@ -436,7 +436,7 @@ HDD = 13 + 11 = 24 °F·days
| High | 60F | 79F | 80F |
| Low | 51F | 62F | 68F |
| **Ave** | 55.5 | 70.5 | 74 |
CDD = 5.7 + 14.9 = 20.6 °F·days
CDD = 5.7 + 9.2 = 14.9 °F·days

6. For a given building that has a UA_total = 3,382 Btu/(h × F) and an HDD of 414 F*days determine the **Annual Heat Needed** in Btu.  Provide your answer as an integer.  Do not write units in your answer.

@@ -450,7 +450,8 @@ Annual Heat Needed = 3382 × 414 × 24 = 33,603,552 Btu

![[ARC2047H-Building_Science_Materials_and_Construction_3#^423ef7|^423ef7]]

Annual Cost = (139818553 / (98% × 1050)) × 11.38 / 1000 = 135.88
E = 139818553 ÷ (0.98 × 1050) = 135,878
Annual Cost = E × (11.38 ÷ 1000) = 1546.29

8. A building located in Vancouver, British Columbia with solar insolation data found in the link below has the following:
1) Heat gains from people, equipment, and lights = 20,982 Btu/h.
@@ -464,7 +465,7 @@ Find the approximate **Summer Balance Point Temperature** for the building in
![[ARC2047H-Building_Science_Materials_and_Construction_3#^ba60bb|^ba60bb]]

Q_i = (1239 × 3991 /24 × (1-0.76)) + 20982 = 70,430.49
BPT = 72 - (Q_i / 4185) = 55.17 oF
BPT = 72 - (Q_i ÷ 4185) = 55.17 oF

9. For a given building located in Ottawa, Ontario find the **cooling load hours (CLH)** using Table B.1 (linked below) if:
1) CDD = 1,513 F*days 
@@ -474,7 +475,7 @@ Provide your answer in hours as an integer.  Do not write units in your answer.

![[ARC2047H-Building_Science_Materials_and_Construction_3#^f100e8|^f100e8]]

CLH = 1513 × 24 / (80.6 - 56.7) = 1519 h
CLH = 1513 × 24 ÷ (80.6 - 56.7) = 1519 h

10. For a given building what is the **annual cost of cooling** if:
1) Cooling Load = 222,343 Btu/h

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