Lecture 9 Domestic Hot Water Supply

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

[email protected]


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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


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


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:


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


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


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)



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Plate heat exchanger in

LTDH


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)


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)


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


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


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


Parallel flow PHECounter flow PHE

In a condenser

In an evaporator


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PHE heat

transfer

analysis


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


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


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


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

qq

)( ,,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


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


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


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



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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.



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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



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1‐D pipe heat loss


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


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


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


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


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


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



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Flat station



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Legionella


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


Legionella treatment


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Heat pump


Condenser

Evaporator

Expansionvalve

Compressor

House heatingdemand

Coldenvironment

, /


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Heat pump as thermal booster



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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


Lecture 9 Domestic Hot Water Supply

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