HW1-ARE 346N-BES-F20-Solutions-PDF (2)
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10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
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Homewo
rk 1
ARE 346N Building Environmental Systems
Professor Dr. Zoltan Nagy
Due: September 29th, 2020
Solve each of the following problems and show your work. Make sure to label each answer with the
same label as the problem.
You will be graded on both the correct method and solution. Partial credit is awarded for use of the
correct equation or method, correct values chosen from tables and charts when required, and the
correct answer with appropriate units.
This assignment includes TWO
problems.
PROBLEM
POINTS
1.a
3 pts
1.b
2 pts
1.c
3 pts
1.d
3 pts
1.e
3 pts
1.f
3 pts
2.a
18 pts
TOTAL
35 pts
Instructions
For this Homework, the following resources might be useful for you:
1. Psychrometric Chart
1. Online Interactive Chart
2. Downloadable Charts
2. Table 3.6 from your textbook- Chapter 3
3. Table 3.7 from your textbook- Chapter 3
Resources
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
2/12
Problems
Outdoor air with dry bulb temperature of 40
o
F and relative humidity of 60% is heated using an
electric heater to DBT= 95
o
F, and after that humidi±ed to relative humidity of 35% by injection of dry
steam.
Problem 1 - Psychrometrics
Draw the heating and humidifying process in psychrometric chart. Show all points.
Problem-1(a)
# YOUR CODE HERE
#
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
3/12
Using the psychrometric chart, determine the amount of energy supplied to 1 lb of dry air in this
heating and humidifying processes
Problem-1(b)
# YOUR CODE HERE
#
23.5 #enthalpy of external air
h1=13.1
# Btu/lbda
#enthalpy of air after humidifier
h3=36.6
#Btu/lbda
#Energy per unit of mass:
q13=h3-h1
print(q13)
Problem-1(c)
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10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
4/12
Find the amount of water while steam humidi±er is using for humidi±cation of 1 lb dry air. Assume
that all water is converted into steam.
# YOUR CODE HERE
E
0.0092 w3= 0.0124
#lbw/lbda
w2=0.0032
#lbw/lbda
Delta_w23= w3-w2
print(Delta_w23)
w
If the volume ²ow rate of air is 30 ft3/min (CFM), what is the amount of energy per hour (energy
rate) [Btu/h] needed for heating and humidi±cation? What is the ²ow rate of water (lbw/hour)
needed for humidi±cation?
Problem-1(d)
1815.0 1335.84 3172.5 3150.84 0.11536363636363636 # YOUR CODE HERE
#Volume flow rate
V=30 #ft3/min
Qsensible
= 1.1 * V * (95-40)
Q_latent = 4840*V*(Delta_w23)
Q_total= 4.5*V*q13
print(Qsensible)
print(Q_latent)
print(Q_total)
print(Qsensible+Q_latent)
GPM = Q_total / (500 * (95-40))
print(GPM)
#
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
5/12
3214.7999999999997 1.25856 #ALternative Solution : I accepted the solutions above because of the book's reference
#Volume flow rate
V=30 #ft3/min
Air_density=0.076 #lb/ft3
m = V*Air_density * 60 #Air mass flow rate lb/hr
#Total heat rate:
Qtotal = m*q13 #[Btu/h]
#Water flow rate:
G= m*Delta_w23
print(Qtotal)
print(G)
#
What is the sensible energy that electric heater provides per unit of air? What should be the power
of electric heater for heating of 30 ft3/min of air ±ow? Assume that there is no heat loss in electric
heater.
Problem-1(e)
# YOUR CODE HERE
E
545.25273408 1860.4799999999998 h2=26.7 #Btu/lbda
qsensible= h2 - h1 #Btu/lb
Qsensible= m*qsensible
#Power of electric heater is equal of sensible heat rate:
P= Qsensible # [Btu/h]
#Wattconversion
Power_Watt_electricHeater= P*0.293071
print(Power_Watt_electricHeater)
print(Qsensible)
h
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
6/12
What is the latent energy delivered to 1 lb of air by humidi±cation? If the humidi±er uses electric
energy to transform water into steam and there is no losses of electric energy, what should be the
power of humidi±er in Btu/h for volume ²ow rate of air of 30 ft3/min?
Problem-1(f)
# YOUR CODE HERE
E
396.9119167200001 #Power of humidifier is equal of latent heat rate Qlatent = m × qlatent = 136.8 × 9.9
qlatent= h3 - h2 #Btu/lb
Qlatent= m*qlatent
#Power of electric heater is equal of sensible heat rate:
P= Qlatent # [Btu/h]
#Wattconversion
Power_Watt_humidification= P*0.293071
print(Power_Watt_humidification)
Problem 2 - Cooling Loads
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HW2-ARE 346N-BES-F20-Solutions - Colaboratory
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7/12
A one story o±ce building is located in New York City is exposed to outdoor air at all four sides (no
shading). The building has ²at roof exposed to sun and the ²oor to ceiling height of 10 ft. It has 2
windows on south side and 4 on north side (see Figure above). On south side there is a double side
glass door with dimension 6 ft x 8 ft.
The wall assembly for all four sides is the same: 1” stucco + 2” insulation + 4” concrete block,
U value = 0.11 Btu/hft2F, Color –Light
Medium construction roof: 1” insulation + 2” concrete, U = 0.206 Btu/hft2F , Color –Dark
The windows are all with same dimensions: width is 3 ft and height is 5 ft. Win- dows
manufacturer is Atlantic All-Weather Windows, type is Horizontal slider (CDP number is ATL-K-
18-00052-00001). Hint: to ³nd U and SC factors for this type of windows, go to the website of
National Fenestration Rating Council http://search.nfrc.org/search/Searchdefault.aspx
.
The south glass door has the same glass properties for the solar radiation and conduction like
the windows.
The ²oor is well insulated. (Neglect the heat ²ow through the ²oor.) In³ltration provides
maximum 1 air changes per hour. Indoor design condition are de³ned by: DBT=78
o
F and
RH=50% for summer and DBT=72
o
F for winter. There are 50 o±ce workers from 8 AM to 5 PM
and there are 45 computers with the energy consumption of 120 W (each) during the working
hours. There are 80 light ³xtures (each with the power of 120 W) which are ON from 8 AM to 5
PM.
Outdoor conditions for New York City are : DBT=90
o
F and WBT=73
o
F for summer and
DBT=12
o
F for winter.
Calculate the cooling load (latent and sensible)
Problem-2(b)
#YOUR CODE HERE
E
Heating Load Calculation
Volume = 80 * 50 *10
Area_Window_South= 2 *(3*5) + (6*8)
Area_Window_North = 4 * (3*5)
Area_Wall_East= 80*10
Area_Wall_West= 80*10
Area_Wall_South= 50*10- Area_Window_South
Area Wall North= 50*10
Area Window North
V
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
8/12
112114.8Btu/hr Area_Wall_North= 50*10- Area_Window_North
Area_roof = 50*80
#Outdoor Conditions
winter_dbt_outdoor = 12 #F
summer_dbt_outdoor= 90 #F
summer_wbt_outdoor = 73 #F
#Indoor Conditions
winter_dbt_indoor = 72
summer_dbt_indoor = 78
summer_RH_indoor = 0.50
U_Wall = 0.11
#[btu/hr*ft2*F]
U_roof= 0.206
#[btu/hr*ft2*F]
U_window = 0.32#[btu/hr*ft2*F]
Roof_conduction= U_roof *Area_roof
#[btu/hr*F]
Walls_conduction= U_Wall * (Area_Wall_East+Area_Wall_West+Area_Wall_South+Area_Wall_No
Windows_conduction = U_window* (Area_Window_South+Area_Window_North) #[btu/hr*F]
Qconduction= (Roof_conduction+Walls_conduction+Windows_conduction) *(winter_dbt_indoor
ACH = 1
Vinf = (ACH *Volume)/ 60
qAir = 0.076 # lb/ft3
m = Volume * qAir
Cp = 0.24
Delta_T = winter_dbt_indoor-winter_dbt_outdoor
Qinfiltration =
m*Cp*Delta_T
Total_Heating_Load = (Qinfiltration +Qconduction)
print(str(Total_Heating_Load)+"Btu/hr")
# Import modules
import pandas as pd
# Create an example dataframe
data = {'Load': ["Q conduction Heating Load", "Q infiltration Heating Load", "Q total
'Btu/hr': [Qconduction,
Qinfiltration, Total_Heating_Load],
'kBtu/hr': [(Qconduction/1000), (Qinfiltration/1000), (Total_Heating_Load/1000
df = pd.DataFrame(data)
df
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
9/12
Load
Btu/hr
kBtu/hr
0
Q conduction Heating Load
68338.8
68.3388
1
Q infiltration Heating Load
43776.0
43.7760
2
Q total Heating Load
112114.8
112.1148
Cooling Load
Volume = 80 * 50 *10
Area_Window_South= 2 *(3*5) + (6*8)
Area_Window_North = 4 * (3*5)
Area_Wall_East= 80*10
Area_Wall_West= 80*10
Area_Wall_South= 50*10- Area_Window_South
Area_Wall_North= 50*10- Area_Window_North
Area_roof = 50*80
#Outdoor Conditions
winter_dbt_outdoor = 12 #F
summer_dbt_outdoor= 90 #F
summer_wbt_outdoor = 73 #F
#Indoor Conditions
winter_dbt_indoor = 72
summer_dbt_indoor = 78
summer_RH_indoor = 0.50
U_Wall = 0.11
#[btu/hr*ft2*F]
U_roof= 0.206
#[btu/hr*ft2*F]
U_window = 0.32#[btu/hr*ft2*F]
SHGC= 0.57
SC = SHGC /0.86
# Shading Coefficient=SC
# SHGC is the ratio of energy transmitted to energy indicent on the glazing material.
# SHGC is equal to 86% of shading coefficient.
V
#Heat Gains to Space
SHGF_North = 28 # Table 3.6
SHGF_South = 29
#Table 3.7
TETD_roof = 81
TETD North facade = 15
#
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HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
10/12
TETD_South_facade =26
TETD_East_facade = 37
TETD_West_facade = 19
TETD_South_door =29 # has same properties as windows
6860.92 #Heat Gains to Space
#roof
Q_roof = U_roof * Area_roof *TETD_roof
#Wall (Heavy Thermal Mass)
Q_wall_East = Area_Wall_East *U_Wall*TETD_East_facade
Q_wall_West =
Area_Wall_West*U_Wall*TETD_West_facade
Q_wall_South = Area_Wall_South*U_Wall*TETD_South_facade
Q_wall_North =
Area_Wall_North*U_Wall*TETD_North_facade
Q_Walls = Q_wall_North +Q_wall_South +Q_wall_East +Q_wall_West
print(Q_Walls)
#
3142.640930232558 #windows+door ( door area was included in the window area calculation. So we don't nee
Q_window_North = U_window * Area_Window_North *(summer_dbt_outdoor-summer_dbt_indoor)
Q_window_South = U_window * Area_Window_South*(summer_dbt_outdoor-summer_dbt_indoor) +
Q_windows = Q_window_North +Q_window_South
print(Q_windows )
#
60 78 12 print(Area_Window_North)
print(Area_Window_South)
print(summer_dbt_outdoor-summer_dbt_indoor)
p
#Internal cooling load
Q_occupant_sensible= 250 * 50
Q_occupant_latent= 200 * 50
Q li hti
80 *120 *3 4 #[3 4 i
f
i
f
KW t
kbt /h
]
#
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
11/12
Q_lighting = 80 *120 *3.4 #[3.4 is for conversion from KW to kbtu/hr ]
Q_computer = 45*120*3.4
# infiltration Sensible & Latent
# Q sensible= m *Cp * Delta T
# Q latent = m* hfg * (Wroom -Woa).
## Wroom and Woa have been found by using psychro
m = 3040 #lb/hr
Q_infiltration_sensible = m* 0.24 *(summer_dbt_outdoor-summer_dbt_indoor)
Q_infiltration_latent = m *1000 *(0.0138-0.0103)
Q_total_sensible = (Q_roof +Q_Walls+Q_windows+Q_occupant_sensible+Q_lighting +Q_comput
Q_total_latent = (Q_occupant_latent+Q_infiltration_latent )/1000
DataFrame with default index Heat Gains to Space Heat Gains_Sensible [Btu/hr] Heat Gains_Latent [Btu/hr] 0 Roof 66744.00000 0.0 1 Walls 6860.92000 0.0 2 Windows 3142.64093 0.0 3 Lighting 32640.00000 0.0 4 Computer 18360.00000 0.0 5 Occupant 12500.00000 10000.0 6 Infiltration 8755.20000 10640.0 #initialize a dataframe
import pandas as pd
df = pd.DataFrame(
[['Roof', Q_roof, 0],
['Walls', Q_Walls, 0],
['Windows', Q_windows, 0],
['Lighting', Q_lighting, 0],
['Computer', Q_computer, 0],
['Occupant', Q_occupant_sensible, Q_occupant_latent],
['Infiltration', Q_infiltration_sensible, Q_infiltration_latent]],
columns=['Heat Gains to Space', 'Heat Gains_Sensible [Btu/hr]', 'Heat Gains_Latent [
print('DataFrame with default index\n', df)
#
#initialize a dataframe
import pandas as pd
#
10/20/2020
HW2-ARE 346N-BES-F20-Solutions - Colaboratory
https://colab.research.google.com/drive/1ggGirkwmJoZ-jcb8BemptYpUNLPLhgES#scrollTo=-pBSsflz_JyZ&printMode=true
12/12
DataFrame with default index Heat Gains to Space Heat Gains_Sensible [Btu/hr] Heat Gains_Latent [Btu/hr] 0 Roof 66.744000 0.00 1 Walls 6.860920 0.00 2 Windows 3.142641 0.00 3 Lighting 32.640000 0.00 4 Computer 18.360000 0.00 5 Occupant 12.500000 10.00 6 Infiltration 8.755200 10.64 7 Total 149.002761 20.64 df = pd.DataFrame(
[['Roof', Q_roof/1000, 0],
['Walls', Q_Walls/1000, 0],
['Windows', Q_windows/1000, 0],
['Lighting', Q_lighting/1000, 0],
['Computer', Q_computer/1000, 0],
['Occupant', Q_occupant_sensible/1000, Q_occupant_latent/1000],
['Infiltration', Q_infiltration_sensible/1000, Q_infiltration_latent/1000],
['Total', Q_total_sensible, Q_total_latent]
],
columns=['Heat Gains to Space', 'Heat Gains_Sensible [Btu/hr]', 'Heat Gains_Latent [
print('DataFrame with default index\n', df)
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