Renovation of St. John
Neumann High School
Rocco D’Uva
Introduction/Project Review
Architectural Analysis & Design
HVAC Analysis & Design
Miguel Armijos
Structural Analysis & Design
Rick Howley
Environmental Analysis & Design
Budget
Project Review

Location:
2600 Moore St.
Philadelphia, Pa

Size:
3 story building
Aerial Photo
Existing Footprint

Age:
Built in 1955
Problem Statement

Owner wants to change use from private
school to residential apartments

Specific problems to solve:
–
–
–
–
What building systems are affected
How are they affected
What is their current state
How should they be redesigned
ARCHITECTURAL
SYSTEM
OUTLINE:
 Existing Site & Floor Plans
 New Site Plan
 New Floor Plans
1. Entry Moved
2. Structural Walls Used
3. 11,000 SF Removed for Parking
4. 5,600 SF Added for Ballroom/Natatorium
EXISTING SITE PLAN
NEW SITE PLAN
RESIDENTIAL SECTION – FLOOR PLANS 1, 2, 3
COMMERCIAL SECTION
1ST FLOOR PLAN – BALLROOM
COMMERCIAL SECTION
2ND FLOOR PLAN – NATATORIUM
Architectural Program
Building
Section
Residential
Residential
Residential
Residential
Residential
Commercial
Commercial
Commercial
Commercial
Floor
1,
1,
1,
1,
1,
2,
2,
2,
2,
2,
3
3
3
3
3
1, 2
1
2
2
Space
Apartment Units
Studio Apartment
One Bedroom Apartment
Deluxe One Bedroom Apartment
Common Areas
Total Residential Space
Common Areas
Ballroom
Indoor Pool
Pool Deck
Total Commercial Space
Total Building Space
Approx Ceiling Window
Quantity Area (sf) Height (ft) Area (sf) Comments
78
47,640
10
5,616 Assume an average of 4 windows
27
460
10
1,944 per apartment.
30
600
10
2,160
21
820
10
1,512 Assume each window = 18 sf
n/a
47,343
10
648
94,983
6,264 See floor plans for window locations
n/a
1
1
1
24,144
22,746
9,490
13,310
69,690
164,673
13
16
13
13
2,556
n/a
n/a
828
3,384
9,648
Assume each window = 18 sf
Max Pool Depth = 6'-0"
Assume each window = 18 sf
See floor plans for window locations
HVAC
SYSTEM
OUTLINE:
 Specific Design Focus
 Residential Area Problems/Solutions
 Natatorium Design Problems/Solutions
 Optimal Solutions
 Equipment Example
Design Focus –
What To Design?

Residential vs. Commercial Areas

Wide Array of Problems

Residential = More Common

Focus will be on the Natatorium
Residential Areas

Design Parameters include:
–
–
–
–

Heating & Cooling during winter/summer
Low initial cost
Low operating cost
High efficiency
Assumptions:


Basic options include: Split VAV Rooftop units or
PTAC Units
Alternative with the highest efficiency and lowest
cost is the best solution
Natatorium Design
PROBLEMS
 High Evaporation Rates
–
–
–
–


Humidity Control
Condensation
Loss of Pool Water (costs to refill)
Thermal Comfort
Air Quality/Exchange
Heating of Pool Water
Evaporation Rate Formula

Evaporation Rate
= ERF x AF x Pool Water Surface Area
–
–
–
Where: ERF = Evaporation Rate Factor (table)
AF = Activity Factor (assumed to be 0.65)
Pool Water Surface Area = Approx. 9500 ft2
Evaporation Rate Factor Table
Evaporation Rate Calculation
NATATORIUM PARAMETERS/ASSUMPTIONS
Design Conditions
Condition
Number
Air
Water
o
Relative
o
Activity
Temp ( F) Temp ( F) Humidity (%)
1
80
82
50
60
Factor
0.65
0.65
86
0.65
0.65
2
85
50
60
Water Surface
Area (SF)
9,490
9,490
9,490
9,490
ERF
(lb/h sf)
0.06
0.048
0.068
0.052
Evaporation
Rate (lb/h)
Comments
370
At minimum temperatures
296
419
321
At maximum temperatures
Evaporation Rate Trend
Evaporation Rate (lb/h)
500
450
400
350
300
250
Evaporation Rate
200
150
100
50
0
Increasing Air/Water Temperature --->
Decreasing Relative Humidity --->
Condensation

Windows
–
–

Walls, Ceilings, Other Thermal Bridges
–

Recommended 3-5 CFM per SF of exterior glass
Therefore, natatorium requires 2,400-4,200 CFM
Vapor Retarder
Proper Air Distribution is Key to Minimizing Damage
–
–
Minimize Eddy Effects Over Window (Mullions)
Minimize Air Flow Over Pool Surface
Loss of Pool Water

The amount of condensate recovered in a year
by the HVAC system is approximately equal to
one entire pool fill.
Condensate Calculations
Design Conditions
Active
Inactive
Condition Evaporation Evaporation
Number
Active
Rate (gal/h) Rate (gal/h) Hours/day
1
44
20
10
35
16
10
2
50
23
10
38
18
10
Inactive
Space
Pool Water
Avg Pool
Depth (ft)
4
Inactive
Evaporation
Total
Total
Evaporation Evaporation
Hours/day Rate (gal/day) Rate (gal/day) Rate (gal/day)
14
444
287
730
14
355
229
584
14
503
325
827
14
384
248
633
Average =
694
Pool Dimensions
Approx
Area (sf)
9,490
Active
Evaporation
Approx
Approx
Volume (cf) Volume (gal)
37,960
283,979
(gal/year)
266,482
213,186
302,013
230,951
253,158
Air Quality & Air Exchange

ASHRAE Recommendations:
–
0.5 CFM of Outside air per FT2 of pool and wet deck Area… (For
us, approx 4,800 CFM)
–
15 CFM of Outside air per spectator/user… (For us, assume
above is larger)
–
4-6 Air Changes per Hour… (vs. 6-8 for spectator facilities)
–
13 – 37 Pa of Negative Pressure (We will use multiple exhaust
fans)
–
Must be able to purge if necessary
Pool Water Heating
Heating Pool Water Using Recovered Heat
Symbol Variable
a
b
c
d
e
f
g
h
I
j
k
Pool Water Temp
Air Temp
ERF - Active
ERF - Inactive
Active hrs/day
Inactive hrs/day
Activity Factor
Average ERF
Pool Water Surface Area
Pool Evaporation Rate
Energy Consumption to Heat Pool Water
Value
80
82
0.036
0.048
10
14
0.650
0.024
9,490
228
2,194,695,360
Unit
Formula
o
n/a
n/a
Table
Table
n/a
n/a
Table
(e x c x I x f x d x .5)/ 24
n/a
hxI
j x 8,760 h/yr x 1,100 Btu/lb
F
F
lb/hr ft 2
lb/hr ft 2
hr
hr
o
lb/hr ft 2
ft2
lb/hr
Btu/yr
Assume $.017 kWh for Gas Heat
Assume $.06 kWh for Electric Heat
14,631 $/yr
38,582 $/yr
Cost to Heat without Recovery
Cost to Heat without Recovery
Assume 80% Efficiency with Recovery - Gas
Assume 80% Efficiency with Recovery - Electric
11,705 $/yr
30,866 $/yr
Cost Savings with Recovery
Cost Savings with Recovery
Basic Equipment Example
Pool Design
Design of a Swimming Pool


Swimming pool on the 2nd floor
Dimensions:
- 30ft by 90ft
- 6ft deep
Considerations

Water in pool become heavy loads
62.4 lb/ft x 30ft wide x 90ft long x 6ft deep =
A total of 1,010,880 lbs

Architectural Considerations
-
Sizes of structural members
Column spacing
-
Structural System
Pick a system that can handle heavy loads
and is efficient:
-
Structural System : Option 1

One-way Joist Slab
-
Joists act as t- beams to distribute the loads to the
girders
Span 15 to 36 ft
Economical system for heavy
loads or long spans
-
Structural System : Option 2
1. Concrete Waffle Slab
Because there are joists in both directions, this
floor system is the strongest and will have the least
deflection
20 to 50 ft spans
Good for high gravity loads
High stiffness
Small deflections
Expensive due to formwork
Structural System –
One-Way Joist Slab
Slab Design

ACI requires for the height of slab to be at
least
Height = (Length of span)/20
In our case ----- Minimum Height = 3.6 “
*But Code requires a 6” slab for in our design for
fire requirements*
Slab Design





Live Loads: Water
Dead Load: Weight of concrete slab
6” slab
Reinforcement
# 4 @ 12” O.C
ACI Requirement –
Ends of slab
(length of span)/4 &
Interior Spans (length of span)*(.3)
Beam Design - Joist Layout
Skip Joists
- Every 6 ft

Span
- 30ft

Trial Structural System Sizing
Joists
Girders
Width
Width
Depth
Depth
Weight/ft
Weight/ft
16
38
0.61
18
36
0.675
20
34
0.7
22
32
0.73
0.375
26
30
0.81
0.408
30
28
0.875
11
38
0.43
10
36
0.37
10
34
0.35
11
32
0.367
12
30
14
28
Total Weight of Structural System
Depth Vs. Weight
150
140
PSF
130
120
110
100
90
27
28
29
30
31
32
33
34
Depth of Beam
35
36
37
38
39
Joists




Mu = 515.7 ft-kips
Span = 30 ft
Dimensions
B = 11”
D = 32”
Reinforcement
- 6 #7’s
- # 3 Stirrups
Girder Design

Max Moment = 1092 Ft-Kips
Girder Design

B = 22”
D = 32”
Compression Rebar
( ACI required in this case)

-2#8

Tension Rebar
- 8 # 10’s

# 3 stirrups
Column Design

Column layout
Column Design



Dimension
18” x 18”
11 ft high
Reinforcement
8 # 9’s
# 3 Ties
According to code
Spacing = 18”
Column Interaction Diagram

Evaluation of the
strength of the column
subjected to combined
bending and axial loads
Balanced Failure limit @
P = 400 kips
M = 250 ft-kips
Interaction Diagram
1000
800
600
Phi Load

400
200
0
-200
0
50
100
150
-400
-600
Phi M
200
250
300
Stormwater Detention/Retention
Why retain stormwater?






Impervious surfaces
High runoff flow
High river flow
Erosion of stream banks
Altered groundwater
tables
Contamination of streams
– Carried from streets
– Turbidity
Schuylkill River bank on Kelly Drive
Drainage



Crest of the field
Roof and lot
drainage to site
storm drains
Swales direct
current around field
to basin
Regulations
Philadelphia City Codes
 1 inch must be infiltrated
 Detention design for 100 year storm
 Only storm water may enter drainage pipes
 Design so that post-development infiltration
equals pre-development infiltration
Specs
Reference: 100 year storm with duration of 1 hour (duration assumption
based on size and slope of parcel)
Area=379146.24ft2
-Half of the area is the impervious parking lot and roof, the other half is the football and
baseball fields
Soil Class C-Soils having slow infiltration rates if thoroughly wetted and consisting chiefly of soils
with a layer that impedes the downward movement of water, or soils with moderately fine to fine
texture. They have a slow rate of water transmission.
Storm
Analysis
Design for 100 year storm



Duration 1 hour (based on size and slope of parcel)
Rainfall Depth 3.25 inches from chart
Infiltration

Pre-development
–
–
–
–
Runoff Curve Number
CN=79 (open space fair
condition grass cover 5075%)
Potential Maximum Retention
S=1000/79-10=2.66
Qrunoff=(P-0.2S)2/(P+0.85)
Q=(3.25-.2*2.66)2/(3.25+0.85)
Q=1.8in=0.15ft*A
Q=56871.93ft3/hr
I=3.25-1.8=1.45in

Post-development
–
–
–
–
CN=79 (field)
CN=98 (Parking lots, roofs,
etc.)
f=0.5
CNw=CNp(1-f)+f(98)
CNw=79(0.5)+0.5(98)=88.5
S=1000/88.5-10=1.3
Qrunoff=(3.250.2*1.3)2/(3.25+0.85)
Q=2.18in=0.1817ft*A
Q=68878.2ft3/hr
I=3.25-2.18=1.07in
Underground Basin Design









Qpost-Qpre=68878.2-56871.9=12006.3ft3/hr
Must re-infiltrate this volume.
Design a detention basin with two tanks with weir flow from storage
tank to discharge tank
Weir 1ft wide by 10ft tall by 60ft long=600ft3
Storage Tank 20ft wide by 10ft tall by 60ft long=12000ft3
Entire Tank 77ft wide by 15ft tall by 60ft long=69300ft3
Maximum volume of basin=69300-600=68700ft3
Storage Tank will supply sprinkler system
Discharge to storm sewer at Mifflin Street
Underground Basin
5ft
10ft
60ft
57ft
20ft
Re-infiltration




Storage tank supplies sprinkler system when it contains water
Rotary sprinklers will apply about 0.7in/hr to fields over a radius
of 50ft
The area of the field divided by the area covered by sprinklers
indicates that 25 rotary head sprinklers will be needed to
sufficiently reapply the contained storm water
Sprinklers run for a very short time so that runoff is minimized
during operation
Budget
Budget