Lecture Objectives:
Discuss HW4
• Chiller modeling
• Cooling towers and modeling
• Project 1
• HW 4:
• Solve the problem 5.6 (water – ammonia)
• from the textbook
• Based on example 5.5.
Modeling of Water Cooled Chiller
Chiller model:
Chiller data:
QNOMINAL nominal cooling power,
PNOMINAL electric consumption for QNOMINAL
Available capacity as function of evaporator and condenser temperature
Cooling water supply
Cooling tower supply
2
2
CPATF  a1  b1  TCW S  c1  TCW
S  d1  TCTS  e1  TCTS  f1  TCW S  TCTS
Full load efficiency as function of condenser and evaporator temperature
2
2
EIRFT  a2  b2  TCW S  c2  TCW

d

T

e

T
S
2
CTS
2
CTS  f 2  TCW S  TCTS
Efficiency as function of percentage of load
EIRFPLR  a3  b3  PLR  c3  PLR
Part load: PLR 
Q( )
QNOMINAL  CAPFT
The consumed electric power [KW] under any condition of load
P  PNOMINAL  CPFT  EIRFT  EIRFPL
The coefiecnt of performance under any condition
COP( ) 
Q( )
P( )
Reading: http://apps1.eere.energy.gov/buildings/energyplus/pdfs/engineeringreference.pdf page 597.
Example of a chiller model
http://www.comnet.org/mgp/content/chillers?purpose=0
Cooling Towers
Power plant type
Major difference: NO FAN
Combining Chiller and Cooling
Tower Models
P  PNOMINAL  CPFT  EIRFT  EIRFPL
Function of TCTS
3 equations from previous slide
Add your equation for TCTS
TCTS  a4  b4 WBT  c4 WBT 2  [d4  e4 WBT  f 4 WBT 2 ]  R  [ g 4  h4 WBT  i4 WBT 2 ]  R 2
→ 4 equation with 4 unknowns
(you will need to calculate R based on water flow in the cooling tower loop)
Merging Two Models
Temperature difference:
R= TCTR -TCTS
Model:
TCTS  a4  b4 WBT  c4 WBT 2  [d4  e4 WBT  f 4 WBT 2 ]  R  [ g 4  h4 WBT  i4 WBT 2 ]  R 2
Link between the chiller and tower models is the Q released on the condenser:
Q condenser = Qcooling + Pcompressor ) - First law of Thermodynamics
Q condenser = (mcp)water form tower (TCTR-TCTS)
m cooling tower is given - property of a tower
TCTR= TCTS - Q condenser / (mcp)water
Finally: Find P() or COP( ) 
Q( )
P( )
The only fixed variable is TCWS = 5C (38F) and Pnominal and Qnominal for a chiller (defined
in nominal operation condition: TCST and TCSW);
Based on Q() and WBT you can find P() and COP().
Cooling Tower Performance Curve
R
TCTR
Outdoor
WBT
TCTS
from chiller
to chiller
Temperature difference:
TCTS
R= TCTR -TCTS
Most important variable is
wet bulb temperature
TCTS = f( WBToutdoor air , TCTR , cooling tower properties)
or for a specific cooling tower type
T
= f( WBT
, R)
WBT
Cooling Tower Model
Model which predict tower-leaving water temperature (TCTS) for arbitrary
entering water temperature (TCTR) and outdoor air wet bulb temperature (WBT)
Temperature difference:
R= TCTR -TCTS
Model:
TCTS  a4  b4 WBT  c4 WBT 2  [d4  e4 WBT  f 4 WBT 2 ]  R  [ g 4  h4 WBT  i4 WBT 2 ]  R 2
For HW 3b:
You will need to find coefficient a4, b4, c4, d4, e4, f4, g4, h4, and i4 based on the
graph from the previous slide and two variable function fitting procedure
Two variable function fitting
(example for a variable sped pump)
Function fitting for a chiller
q = f (condensing and evaporating T)
200
q[kW]
25 C
35 C
45 C
150
100
50
0
0
11
2
4
6
Tevaporator [C]
8
10
Merging Two Models
Temperature difference:
R= TCTR -TCTS
Model:
TCTS  a4  b4 WBT  c4 WBT 2  [d4  e4 WBT  f 4 WBT 2 ]  R  [ g 4  h4 WBT  i4 WBT 2 ]  R 2
Link between the chiller and tower models is the Q released on the condenser:
Q condenser = Qcooling + Pcompressor ) - First law of Thermodynamics
Q condenser = (mcp)water form tower (TCTR-TCTS)
m cooling tower is given - property of a tower
TCTR= TCTS - Q condenser / (mcp)water
Finally: Find P() or COP( ) 
Q( )
P( )
The only fixed variable is TCWS = 5C (38F) and Pnominal and Qnominal for a chiller (defined
in nominal operation condition: TCST and TCSW);
Based on Q() and WBT you can find P() and COP().
Low Order Building Modeling
Measured data
or
Detailed modeling
Find Q() = f (DBT)
For Austin’s Office Building
Model: (Area = 125,000sf)
1200
Hours in a year
kW
1000
800
600
Used for component
capacity analysis
400
200
800
Model
600
400
200
1
285
569
853
1137
1421
1705
1989
2273
2557
2841
3125
3409
3693
3977
4261
4545
4829
5113
5397
5681
5965
6249
6533
6817
7101
7385
7669
7953
8237
8521
0
Cooling water demand [kW]
1000
Number of hours
=0 when building is off
0
0
10
20
Outdoor temeprature [C]
30
40
For project 1 you will need Q()
for each hour
Yearly based analysis:
You will need Q() for one week in July
Use simple molded below and the Syracuse TMY2 weather file posted in the
course handout section
20
Q [ton]
16
12
8
Q=--27.48+0.5152*t
4
Q=-0.45 +0.0448*t
0
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
t [F]