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8/19/2019 Lecture 9 Domestic Hot Water Supply
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Lecture 9. Domestic
Hot
Water
Supply
Hongwei LiCivil Engineering Department
Building 118, room 206Technical University of Denmark
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Lecture Content
DHW Supply
Plate heat exchanger(PHE) Heat exchanger concept and design By‐pass control
Pipe heat transfer calculation Storage tank Flat station
Legenella Heat pumps
Lecture11 DHW Supply
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DHW supply
Lecture11 DHW Supply
Fig. 1 DHW supply with reciruclation
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DHW
supply
ComfortDanish regulation: 10 seconds waiting time, supply at 45oC for kitchen and 40oC for shower and
hand wash Hygiene: LegionellaEHP requires the minimum DHW tempearture at 50oC, and never fall below 55oC for storagetank or DHW with recirculationGerman guideline: a DHW system with total volume less than 3 L allows temperature lowerthan 50oC without risk of Legionella (3 liters not includes HE)
Energy efficiencyWell insulated pipe, avoid recirculation
Low costoverall cost reduction both in ‐house substaiton and in network
Lecture 9 DHW Supply
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DHW load
profile
Lecture 9 DHW Supply
0
2
4
6
8
10
12
14
16
0 5 10 15 20 25 30 35 40 45 50 55 60 65
l / m
min
Flow rate ShowerKitchen
0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40 45 50 55 60 65
k W
min
Power Shower
Kitchen
Bath tub Shower Kitchen wash Hand wash
Cold water temperature [oC] 10 10 10 10
Temperature at tapping [oC] 40 40 45 40
Required power [kW] 26.4 17.6 14.7 7.0
Norminal flow rate [l/min] 12.57 8.38 6.00 3.33
Duration of draw-off [s] 600 300 150 180
Volume of tapped water [l] 126 42 15 10
Total number of tapping 2 4 2 4
Delay between each tapping [min] 30 20 20 20
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Energy demand
for
DHW
Energy for taping requirement
i is the i th unit. T s,i is the i th unit DHW supply temperature. V i,day is volume flow rate of i th unit m3 /day, n and V i,day depends on the type of buildings and activities, Q has the unit MJ/day
Heat loss along distribution pipe:
Lecture 9 DHW Supply
L
l
e f d u ploss dlT lT U T T cmQ0
)(182.4 ,1
, cis
n
i
dayitapping T T V Q
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Factors affect time
delay
Lecture 9 DHW Supply
q
Ld t i
4
2
Factors for time delay due to service pipe and heat exchanger(i) Service pipe diameter, length, thermal capacity and re ‐circulation.
(ii) HE volume, primary flow rate, thermal capacity of HE and the setting of thermal by ‐pass (without by ‐pass, external by ‐pass, and internal by ‐pass).
Transprtation delay (simplest condition)
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DHW Distribution Systems in Small Buildings
Anti ‐Legionella equipment in large buildings
DHW system
design
No DHW recirculation Reduce transportation time Reduce total volume
Lecture 13 Domestc hot water supply
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ITHE in
LTDH
Lecture11 DHW Supply
1. DH supply 2. DH return 3. SH return 4. SH supply
5. DHW return 6. DHW supply 7. Heat exchanger 8. By‐pass thermostatic valve 9. Thermostatic controller
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Recommended Design
Parameters
(EHP)
Lecture11 DHW Supply
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Plate heat exchanger in
LTDH
Lecture11 DHW Supply
For lower temperature difference Improved flow pattern
Enlarged surface area and increased heat transfer (10%) Improved control
Low pressere drop
T11=50oC, T12
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Plate heat exchanger (PHE)
Lecture11 DHW Supply
Thin and rectangular metallic sheet with corrugated surface.
Two fluids alternatively passes through the plate surface and exchanges heat.
Numbers of plate depend on required thermal ouput Corrugated surface increaes effective surface area, distrupting boundary layer, creates turbulent
mixing and decrease fouling resistance, thus enhance the heat transfer.
Large heat transfer surface, fast temperature change, and compact size.
PHE is more efficient to cool DH water than stroage tank
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Gasketed Plate heat exchanger (PHE)
Lecture11 DHW Supply
Gasketed plate heat exchanger
Gasketed PHE:are made of two end plates and of form ‐pressed plates with gaskets, tightened between the end plates.
use sealing gasket to prevent intermixingof media and leakage to outside.
Capable for high temperature, easy to clean and maintain, plates can take apartfor expansion or contraction
Start, left ‐hand, right ‐hand, and end plate
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Brazed PHE
Lecture11 DHW Supply
The units are connected together not through the use of endplates and gaskets. Instead, all corrugated plate are brazed together at hightemperature.
1.Reliability, lightweight, large capacity
2.Low internal water volume. Need fast regulatingsystem to reach desired temperature within short
timeTwo media flow through alternate channels,
always in opposite directions (counter current flow).
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Flow arrangement
in
PHE
Lecture11 DHW Supply
U and Z type port connection. The flow pass the same distance in the Z type connection
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Temperature distribution
in
PHE
Lecture11 DHW Supply
Parallel flow PHECounter flow PHE
In a condenser
In an evaporator
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PHE heat
transfer
analysis
Lecture11 DHW Supply
Overall energy balance in hot/cold fluid
micoc pcohih ph T UAT T cmT T cmQ
)()( ,,,,
)ln()ln(2
1
21
1
2
12
T T
T T
T T
T T T m
Log mean temperature difference (LMTD)
For parallel flow For counter flow
ocoh
icih
T T T
T T T
,,2
,,1
icoh
ocih
T T T
T T T
,,2
,,1
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PHE heat
transfer
analysis
Lecture11 DHW Supply
hh and hc are heat transfer coefficient of hot and cold stream, p is plate thickness, k
p is thermal conduction of plate material, R
f,h and
R f,c are fouling resistance on hot and cold side. It is an approximation as U vary along with plate.
Overall heat transfer coefficient U
c f h f p
p
ch
R Rk hhU ,,
111
for the same U and inlet/outlet temperature, the meantemperature difference is smaller for parallel flow than for counter flow, thus the required surface area is smaller than parallel flow
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‐NTU
method
Lecture11 DHW Supply
Two types of analysis: Performance calculation (determine heat transfer rate): with known flow rate, inlet/outlet temperature, U and A
Design calculation (determine heat transfer area):
1. With knowing Tm, use LMTD method
2. Without knowing Tm, use ‐NTU method (only inlet temperatures are known). : Heat exchanger effectiveness
NTU: the number of heat transfer unit (thermal length)
Maximum possible heat transfer rate • It can be achieved in a counter flow HE with infinite length• with maximum temperature difference in HE: Thi‐Tci
If C c
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‐NTU method
Lecture11 DHW Supply
Heat exchanger effectiveness e (ratio between actual HT rate to maximum possible HT rate )
)(
)(
)(
)(
,,min
,,
,,min
,,
max icih
icocc
icih
ohihh
T T C
T T C
T T C
T T C
)( ,,min icih T T C q thus
NTU is the number of heat transfer unit (thermal length):
),,( ement flowarrang R NTU f
minC
UA NTU
max
min
C C
R
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‐NTU method
Lecture11 DHW Supply
Parallel flow PHE:
Counter flow PHE:
R
NTU R
1)1(exp1
NTU R R NTU R )1(exp.1 )1(exp1
When C h or C c tend to be infinite as in evaporator or condenser, R=0, e becomes independent of flow direction
NTU exp1
Parallel (top) and counter (bottom) PHE
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Example
Lecture11 DHW Supply
An instantaneous heat exchanger is used to heat up the cold water to the desired DHW supply temperature. The overall heat transfer coefficient for the PHE is 4000 W/m 2.oC. The primary side water inlet temperature is 60oC, with flow rate 15 L/m. The secondary water inlet temperature is 10 oC, with flow rate as 11 L/m. To get the desired secondary water outlet temperature as 55 oC. 1) what will be the heat transfer area for the PHE? 2). What will be the primary return temperature?
NTU method
As =765 W/ oC
, , =3.8 E4 W
, , =3.4 E4 W
The effectiveness is
0.9 0.73
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Example
Lecture11 DHW Supply
The NTU can be found from the figure as
4.6
Thus 0.88 2
8.9
)ln( )
10275560
ln(
)1027()5560(
1
2
12
T T
T T T m
LMTD
W E T UA m 44.3
Thus 0.87 2
isos pso pi p p p T T cmT T cmq ,,,,
Thus ,
27
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Thermal by ‐pass
Lecture11 DHW Supply
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Without by‐pass: water cool down
Initial water temperature is 50oC and insulation is 15oC
Test between 0 ‐720 min. Three ground tempeature 3, 8, 14 From 50oC to 20oC requires 3 and 4
hours for ground temperature at 8 and 14.
At ground 14oC, 1.0 hours to cool down from 50 to 35oC.
Lecture11 DHW Supply
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Without by‐pass: transport delay
Transport delay due to thermal capacity.
Temperature at outlet of service pipe
(inlet of HE) Aluflex 20/20/110 , 10 m long Ground temeprature 8oC Different initial temperature and flow
rates.
For Ti=20oC,
17.3
l/min:
6.1
s inlet
hot
water (50oC) reach the outlet, but cooled to 39oC due to pipe thermal capacity, need additional 2 s to reach 45oC.
For Ti=35oC, 17.3l/min: 7 s to reach
45oC
Lecture11 DHW Supply
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1‐D pipe heat loss
Lecture11 DHW Supply
r 1, h f
h ar 2, k 2
0)(1
dr dT
kr dr d
r
1‐D pipe temperature distribution
Heat transfer across the cylinder
dr dT
rLk q r )2(
T f
T aT s,1
T s,2 T s,3T s,4
T f
T s,1 T s,3 T s,4 T a
f hr 121 2
12
2)/ln(
k
r r
ahr 421 3
23
2)/ln(
k
r r
T s,2
4
34
2)/ln(
k
r r
a f a f
r T T U R R R R R
T T q
54321
R2R1 R3 R4 R5
[W/m]
a f hr k r r
k r r
k r r
hr
U
44
34
3
23
2
12
1
1)/ln()/ln()/ln(12
[W/m.K]
R: [K.m/W]
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Example 1 ‐D pipe heat loss
Lecture11 DHW Supply
Calculate pipe line heat transfer coefficient based on the properties given in the table. Pipe is made of PEX, casing is HDPE and insulation is PUR foam
Terms Values
Pipe outer diameter [mm] 20
Pipe wall thickness [mm] 2
Casing outer diameter [mm] 80
Casing thickness [mm] 2.5Thermal conductivity of PEX [W/m.K] 0.38
Thermal conductivity of PUR[W/m.K] 0.023
Thermal conductivity of HDPE [W/m.K]
0.43
Water flow velocity [m/s] 2
Water temperature [oC] 47
Air temperature [oC] 20
Pipe length [m] 5
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Internal flow heat transfer coefficient
Lecture11 DHW Supply
For turbulent flow
Determine flow status
6932465771000/20*2*1000
Re E Du m
Flow is in turbulent region
7.29177.369324*023.0Pr Re023.0 4.08.05/4 n D D Nu
4.93302.064.0
*7.291 Dk
Nuh D
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External flow heat transfer coefficient
Lecture11 DHW Supply
External heat transfer coefficient is calculated based on natural convection:
Assume pipe surface tempeature is 40oC to start calculation,
resulting average temperature as 30oC (for thermal properties)
13933694.22*619.16
02.0)2040()30273/(8.9)( 33
E E
DT T g Ra as D
RaD is Rayleigh number in free convection, is vometric thermal expansion coefficient, v is kinematic viscosity, a is thermal diffusivity. Air properties are evaluated at average temperature
)(1K T a
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External flow heat transfer coefficient
Lecture11 DHW Supply
Calculate heat resistance
2
27/816/9
6/1
Pr)/559.0(1
387.06.0 D D
Ra Nu
27.6
)706.0/559.0(1
13933*387.06.0
3200265.0
2
27/816/9
6/1
e Nu
Dk
h D
R1=0.84; R2=3.68; R3=361; R4=0.94; R5=25.
mW R R R R R
T T U q a f r /06898.0)2047(
54321
Calculate back to the surface temperature. It is an iteration process
s f r T T U q ' 7.21)(06898.047 4321' R R R RU q
T T r f s
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Simplifcation
Lecture11 DHW Supply
ahr k
r r U
43
23 1)/ln(2
R1 ,R2 and R4 are orders of magnitude lower than R 3 and R5. Thus expression of U can be simplified as (k 3> ha ). The difference is only 1.4%
mW R R
T T U q a f r /06996.0)2047(
53
a f hr k
r r
k
r r
k
r r
hr
U
44
34
3
23
2
12
1
1)/ln()/ln()/ln(12
mW R R R R R
T T U q a f r /06898.0)2047(
54321
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Flat station
Lecture 8 DHW Supply
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Flat station
Lecture 8 DHW Supply
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Legionella
Lecture 9 DHW Supply
A fatal disease named as Legionellapneumophila.
First discovered in 1976, USA, due to a
outbreak of pneumonia caused 34 death. Infected by inhaling legionella bacteria
through aerosols (tiny water droplet), or droplet nuclei contaminated with Legionella, or with ingestion of contaminated water
Legionella proliferation/death rate vs. temperature
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Chemical treatment
Thermal treatment Member
filtration
UV sterilization
Lecture 8 DHW Supply
Legionella treatment
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Heat pump
Lecture 13 Domestc hot water supply
Condenser
Evaporator
Expansionvalve
Compressor
House heatingdemand
Coldenvironment
, /
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Heat pump as thermal booster
Lecture 13 Domestc hot water supply
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Lecture 13 Domestc hot water supply
Twin Pipe Triple Pipe
Heat Pump COP 3.3 3.3
System COP 5.7 10.2
Heat pump as thermal booster
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End of lecture
Lecture11 DHW Supply