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District heating engineering
- DH distribution
Energy Technology
Energiatekniikan laitos
Risto Lahdelma
Professor, Energy technology for communities
Department of energy technology
School of Engineering
Aalto University
Otakaari 4, 02150 ESPOO, Finland
Email: [email protected]
Tel: +358 40 503 1030
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28.10.2015
DH distribution
Energy Technology
• The two pipe system is most commonly used
• In some systems, three or more pipes are used to distribute heat and/or
cooling at different temperatures
• The water is pressurized to prevent it from boiling at
operating temperatures
• Max temperature in Finland is 120oC (115oC in effect)
• In other countries max temperature from 90oC to 180oC
– A high temperature:
• Increases transmission capacity as it allows greater temperature
difference
• Decreases pumping costs
• Allows long transfer distances
- But increases heat losses
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DH distribution
Energy Technology
• Controlling the network is in principle possible by
– changing temperature difference or
– changing the water flow (or both)
• The production plant controls only the supply
water temperature based on outdoor temperature
– Pumps in the network maintain sufficient pressure
difference between supply and return pipes
• Customer substations control the water flow
through heat exhangers with adjustable valves
– Return water temperature depends on how well heat
exchangers operate
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DH distribution
Energy Technology
• The transfer capacity of the network depends on
•
•
•
•
Pipe diameters
Allowed pressure level, pressure loss and difference
Size of pumps
Dimensioning of customer equipment
– Temperature levels of supply and return pipes
• Supply water temperature must exceed tap water temperature
• Heat loss during transfer decreases temperature for remote
customers
• To reduce heat losses, supply water temperature should be as
low as possible
• Temperature difference should be as large as possible to
improve efficiency of heat plant
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DH distribution
Energy Technology
• Transferred heat power in two pipe system
= cpm’ T = cp V’ T
–where
•
= heat power (W)
• cp = specific heat capacity of water ( 4200 J/kg,K)
• m’ = mass flow rate of water (kg/s)
• = density of water ( 1000 kg/m3)
• V’ = volume flow rate of water (m3/s)
• T = temperature difference between supply water and return
water (K)
–Both cp and
depend a little on water temperature
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DH distribution
Energy Technology
• Dependency between heat power, temperature
difference and mass flow
Heat power
Temp. diff (K)
Mass flow (kg/s)
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Examples
Energy Technology
• A heat plant operates at flow rate 1 m3/s with T =
40oC. What is the heat power?
Flow rate 1 m3/s of water 1000 kg/s
4200 · 1000 · 40 = 168 MW
• The peak heat power of a detached house is 57 kW
and T = 30oC. What is the water flow rate?
m’ 57000/4200/30 = 0.45 kg/s
V’ 0.45 litres/s
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DH distribution
Energy Technology
• Networks are built from
standardized components
– Rigid steel pipes designed for
• 16 bar pressure
• 120 oC temperature
• Components should last
– 30 years in 120 oC operating
temperature
– 50 years in 115 oC operating
temperature
– Elastic plastic pipes designed for
• 10 bar pressure
• 80 oC temperature
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DH distribution, pipes
Energy Technology
• Pipe elements are classified according to their
structure and insulation material
– Most common are polyurethane insulated, bonded steel
pipes with plastic jacket
• 2 MPuk = two single pipes, one for supply and
one for return water
• MPuk = one dual pipe for supply and return water
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DH distribution, pipes
Energy Technology
• Structural alternatives
– Coating
• M = plastic jacket (muovi)
• E/W/T/Y/P = different kinds of concrete channels
– Insulation
•
•
•
•
pu = polyurethane
pe = polyethylene foam
mv = mineral wool (mineraalivilla)
kb = aerated concrete (kevytbetoni)
– Fastening
• k = insulation is bonded to steel pipe and jacket (kiinteä)
• l = insulation not bonded (liikkuva).
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DH distribution – standard
dimensions of steel pipes
Energy Technology
Diameter
Outer diameter
Insulation
Diameter
Outer d.
Insulation
DN
Single
Dual
mm
DN
Single
mm
201)
125
125
46
200
400
86
25
125
140
43
250
450
83
321)
140
180
46
300
500
82
40
140
180
43
400
630
105
50
160
200
47
500
710
94
65
180
250
49
600
800
87
80
200
280
52
700
900
86
100
250
355
64
800
1000
84
125
280
400
66
900
1100
83
150
315
455
69
1000
1200
81
1200
1400
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1) Not
recommended
Energy Technology
DH distribution – dimensions of
welded steel pipes
Inner diamater
mm
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Outer diamater d
Wall thickness t
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DH distribution
Energy Technology
• Installation of single and dual bonded pipes
– Pipes must lay at least 40 cm below surface
– Not fastened – soil friction keeps pipes on place
– To reduce heat loss, dual pipes are installed so that
supply pipe is below the return pipe
Final filling
Initial filling
Compressed sand
0-16mm
Fine sand 0-20mm
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DH distribution – network design
Energy Technology
• Networks are expensive so they are built to last
– The networks are designed to anticipate current and
future needs of district heating in different areas
• Factors to consider
–
–
–
–
–
–
–
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Specific heating demand of buildings, W/m3
Heat demand of industrial processes
Demand of heat for heating tap water
Pressure losses in network, bar/km
Temp. difference in different operational situations
Coincidence of heat demand
Distance of customers/areas from heat plant(s)
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DH distribution – network design
Energy Technology
• The heat demand varies in different geographical
areas
Heat index
Specific
heating power
W/m3
kWh/m3
Building
type
old
new
old
New
Small house
22-30 18-20
55-70 40-50
Block of flats
22-28 15-20
55-75 45-55
Office building
20-34 20-30
45-80 35-45
Public buildings 28-38 25-32
50-80 35-45
Industrial
50-70 30-55
25-35 15-25
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DH distribution – network design
Energy Technology
• Network design implies considering three design
task
– Mechanical design
– Thermal desing
– Flow dynamic design
• Design is complicated, because each of these
affect the others
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Mechanical design
– Thermal expansion
Energy Technology
• Steel pipes extpand when their temperature rises
and contract when their temperature lowers
– The linear thermal expansion coefficient L relates the
change in a material's linear dimensions to a change in
temperature.
– It is the fractional change in length per degree of
temperature change.
= dL/(LdT)
dL = L L dT
• L is the length of the object
• dL is the (differential) change in the length
• dT is the (differential) change in temperature
L
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Mechanical design
– Thermal expansion
Energy Technology
• Example
– A steel pipe is 20 m long at 10oC
– L = 12.3 10-6 1/K
– How much longer does the pipe become at 50oC?
dL =
LL
dT = 12.3 10-6 20 40 m = 9.84 mm
– The new length is (1+
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L)
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times the old length
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Mechanical design
– Stress due to thermal expansion
Energy Technology
• If the pipe is not allowed to expand due to thermal
expansion, it causes stress in the material
– Stress due to compression is measured as force per area
= F/A
• F = compression force
• A = cross section area
– Suitable units 1 N/mm2 = 1 MPa
– The force due to deformation depends on the elastic modulus
(coefficient of elasticity) of the material
•
•
=
L/L
= elastic modulus
L/L = proportional compression
– This is valid in the elastic range below the materials yield strength
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Energy Technology
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Mechanical design
– Stress due to thermal expansion
• Example
– A 20m long steel pipe is compressed 10 mm
• = 208 kN/mm2
• L/L = 10 mm/20000 mm = 0.0005
• Stress =
L/L = 208000 0.0005 = 104 N/mm2
– The yield strength of steel is (depending on quality)
448 to 690 MPa
• 1 MPa = 1000000 N/m2 = 1 N/mm2
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Energy Technology
Mechanical design – Compensation of
thermal expansion
• Different techniques
– Natural compensation
• L-bends, U-bends and Z-bends allow pipes to expand and
contract
– Compensatory elements
• allow parts of the pipe to compress, expand or bend
– Uncompensated, stiff network
• Immobilized by friction with earth
• Anchoring elements
– Preheated installation to reduce the maximum stress
• The pipes are heated to middle between max and min operating
temperature to reduce the maximal stress
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Mechanical design – Earth friction
Energy Technology
• Earth friction
– An infinitely long pipe element is immobilized by earth
friction
– The free end of a finite length pipe in earth expands up
to a given length L depending on earth friction
– The first immobile point is called a natural fixpoint
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Mechanical design – Earth friction
Energy Technology
Friction force F
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Mechanical design – Earth friction
Energy Technology
– The frictional force per unit length can be computed as
•
•
•
•
•
•
•
F’ = ((1+K0)/2· gZ Dc + G)
= friction between pipe and earth 0.2 – 0.6
K0 = earth type specific pressure coefficient, 0.5 for sand
= density of earth
g = gravity acceleration (9.81 m/s2)
Z = average depth of pipe (depth of pipe axis)
Dc = outer diameter of the pipe
G = effective mass of pipe + water per unit length
– The expanding length L is determined so that F’L
equals the thermal expansion force
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Energy Technology
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Energy Technology
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Mechanical design – Compensation of
thermal expansion
• Compensatory elements
– Bellows compensator (a)
• Compresses, extends and bends
– Glide compensator (b)
• Axial compensator that extends and
compresses
– Angular compensator (c)
• Bends
– Single step compensator –
compresses once after installation
(e)
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Energy Technology
Mechanical design – Compensation of
thermal expansion
• Compensatory element between
anchor/fixed points (1)
• Natural compensation through
bends allow pipes to expand and
contract
• L-bend (2)
• Z-bend (3)
• U-bend (4)
– Sufficient space around the bend
allows compensation
– Shank length must be sufficient
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Energy Technology
• Necessary shank length
in L-bend
– Determine maximum
length expansion L
– Determine outer
diameter of pipe
– Read shank length B
from diagram
– Example:
• L = 25 mm
• = 42.4 mm
• B = 1.6 m
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Energy Technology
• Necessary shank length
in U-bend
– Determine maximum
length expansion L
– Determine outer
diameter of pipe
– Read shank length U
from diagram
– Example:
• L = 45 mm
• = 76.1 mm
• B = 1.9 m
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Energy Technology
• Necessary shank length
in Z-bend
– Determine maximum
length expansion L
– Determine outer
diameter of pipe
– Read shank length U
from diagram
– Example:
• L = 35 mm
• = 76.1 mm
• Z = 2.4 m
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Thermal and flow dynamic design
Energy Technology
• The pipes should be dimensioned so that
– sufficient amount of heat can be transferred during peak
demand
– pressure loss is not too large
– heat losses are not too large
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Thermal design
– Heat transfer capacity
Energy Technology
• Transferred heat power in two pipe system
= cpm’ T = cp V’ T
–where
•
= heat power (W)
• cp = specific heat capacity of water ( 4200 J/kg,K)
• m’ = mass flow of water (kg/s)
• = density of water ( 1000 kg/m3, 971.8 kg/m3 at 80oC)
• V’ = volume flow of water (m3/s)
• T = temperature difference between supply and return water
(K)
cp and
depend a little on water temperature
• cp
1.140 kWh/(m3K) in the summer; 1.125 kWh/(m3K) in
the winter
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Thermal design
– Heat transfer capacity
Energy Technology
• Examples
– What is summer flow rate to transfer 1000 kW heat
when T = 40K
•
= cp V’ T
V’ = /(cp
T) = 1000 kW/1.125 kWh/(m3K) /40 K
= 22.2 m3/h = 6.17 l/s
– What is corresponding winter flow rate?
V’ = 1000 kW/1.140 kWh/(m3K) /40 K
= 21.9 m3/h = 6.09 l/s
– Assuming supply/return temperatures 100oC and 60oC ?
V’ = 1000 kW/ 971.8 kg/m3 / 4.198 kJ/(kgK) /40 K
= 6.13 l/s
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Dimensioning the pipes
Energy Technology
• The pipes should be thick enough to supply the
necessary heat during peak demand
– Pressure loss P/L should be less than:
• 1 bar/km (0.1 kPa/m) at main lines
• 2 bar/km (0.2 kPa/m) at branches near consumer
– The following diagrams show how to dimension pipes
as function of allowed pressure loss and water flow
• At bottom, choose water flow (kg/s) or heat flow (kW) at
different temperature
– E.g. for steel pipes 300 kW at T=50 oC
• Choose necessary thickness from diagram that yields
acceptable pressure loss on the right
– E.g. 50 mm pipe gives pressure loss 0.2 kPa/m
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Pressure loss in steel pipes
Energy Technology
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Pressure loss in copper pipes
Energy Technology
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Pressure loss in plastic pipes
Energy Technology
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Determining the peak heat demand
Energy Technology
• The peak demand for
– space heating depends linearly on the number of
apartments
• Demand is coincident: when it is cold, every apartment is
likely to consume maximum amount of heat
– tap water heating depends non-linearly on the number
of apartments
• Demand is highly non-coincident: different apartments are
likely to use water at different times
• The following diagram can be used to determine
peak demand for heating space and water as
function of number of apartments
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Determining the peak heat demand
Energy Technology
For heating water
t = temperature
difference between
supply and return water
Number of apartments
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Heat demand for heating water
Energy Technology
kW
ET/SKY
450
400
350
300
250
ET/SKY
200
150
100
50
0
0
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20
40
60
80
40
100
120
nof apartments
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Manual dimensioning the pipe
network
Energy Technology
• Dimensioning starts from the far ends of the network
– First the peak demand for heating space and water is determined
for each building
• Previous chart can be used in lack of more accurate information
• Space heating demand can also be estimated based on specific heat
demand per building volume or area
– E.g. 20 W/m3
– The connection lines to buildings are dimensioned based on which
is greater, space or water heating power
• Normally water heating power dominates when number of apartments
is < 20 (see previous diagram)
• In particular for small detached houses the value 57 kW for water
heating can be used
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Energy Technology
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Manual dimensioning of the pipe
network
• Example
– A building with 15 apartments
• Water heating power from diagram 220 kW
• Space heating power from diagram 150 kW
• Necessary peak power is 220 kW
– Assuming 50 oC temperature difference the necessary
water flow is
• V’ = /(cp T) = 220 kW/1.125 kWh/(m3K) /50 K
= 3.9 m3/h = 1.1 l/s ( 1.1 kg/s)
– From diagram we get pressure loss
• 40 mm steel pipe
• 50 mm steel pipe
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dP/L = 3 bar/km (TOO LARGE)
dP/L = 1 bar/km (OK)
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Manual dimensioning the pipe
network
Energy Technology
• Dimensioning proceeds from buildings towards
the heating plant
– At each branch, the peak demand is added separately
for water and space heating
– Again, the greater of these is used to determine the
necessary pipe diameter
• At main lines supplying over 60 apartments the space heating
normally power dominates (see previous diagram)
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Manual dimensioning the pipe
network
Energy Technology
• Dimensioning proceeds from buildings towards
the heating plant
– Example
• Two branches with space heating power 120 kW and 20
apartment each are combined
–
–
–
–
Combined space heating power = 240 kW
Water heating for 40 apartments = 300 kW (from diagram)
Combined power = 300 kW
Pipe diameter 50 mm and T = 50 oC yields pressure loss of 2
bar/km (OK)
• After the network is dimensioned, the actual pressure
losses are computed, and if necessary pipes are changed
to thicker or thinner
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DH Network flow dynamics
Energy Technology
• When the water flows in the network, friction causes a
pressure loss
– Both the properties of the pipe and properties of the fluid affect the
flow
• Viscosity is a measure of the resistance of a fluid which is
being deformed by either shear or tensile stress
– viscosity is "thickness" or "internal friction“
•water is "thin", having a lower viscosity, while honey is "thick", having a higher
viscosity.
– The less viscous the fluid is, the greater its ease of
movement (fluidity)
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Viscosity
Energy Technology
– Consider fluid between a stationary plate and a plate
moving at speed u
• Fluid particles close to boundary plate are stationary and those
close to moving plate move at speed u
• Between plates speed is proportional
to distance from bottom
– The force F keeping top plate
moving is proportional to
plate area and velocity
gradient in fluid
F = A u/y
is the (dynamic) viscosity (Pa s)
Shear stress = du/dy
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Viscosity of water
Energy Technology
– The viscosity of water depends on temperature
Temperature
[°C]
Viscosity
[mPa·s]
10
1.308
20
1.002
30
0.7978
40
0.6531
50
0.5471
60
0.4668
70
0.4044
80
0.3550
90
0.3150
100
0.2822
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Network flow dynamics
Energy Technology
• When the water flows in the network, viscosity (internal
friction) causes pressure loss
pv = (L/d)
v2/2
5
= (8L/d )
V’2/ 2
= (8L/d5) m’2/( 2)
– Where
pv = pressure loss (Pa)
L = pipe length (m)
d = inner diameter of pipe (m)
= friction factor
= density of water (kg/m3)
v = flow speed (m/s)
V’ = volume flow (m3/s) = v d2/4
m’ = mass flow (kg/s) = V’
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Network flow dynamics
Energy Technology
• The friction factor is not a constant, but it depends on the
flow speed
• Fluid can flow in different modes
– Laminar flow (streamline flow) takes
place at low velocities
• Fluid particles move mostly in the
same direction without lateral mixing
– Turbulent flow takes place at higher
velocities
• Fluid particles move in a chaotic manner
forming crossflows, eddies and vortexes
– The mode of flow depends on the
• speed, viscosity & density of fluid
• the smoothness of the pipe
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Reynolds number – mode of flow
Energy Technology
• Reynolds number R is a dimensionless number that gives a
measure of the ratio of inertial forces to viscous forces
R = vd/
– Where
= density of fluid
v = average axial speed of fluid
d = diameter of pipe
= (dynamic) viscosity of fluid
– Flow type depends on Reynolds number
•Laminar flow: R < 2000
•Turbulent flow: R > 4000
•Transitional area when 2000 < R < 4000
– Flow may randomly change from laminar to turbulent and back
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Reynolds number – mode of flow
Energy Technology
• Example
– Consider a steel pipe with
• diameter d = 50 mm = 0.050 m
• water flow V’ = 1 liter/s = 0.001 m3/s
• water temperature 80oC
– We have
•
•
•
•
cross-section area A = d2/4 = 0.00196 m2
flow speed v = V’/A = 0.509 m/s
water density at 80oC = 971.8 kg/m3
viscosity at 80oC = 0.000355 Pa s
R = vd/ = 69700 (Turbulent flow)
– How large a flow remains laminar?
• vd/ = (V’/A)d/ = 2000 V’ = 0.029 liter/s
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Friction factor and Moody diagram
Energy Technology
• The Moody diagram determines the friction factor as a
function of Reynolds number and relative roughness of
pipe
– Relative roughness = e/d
– d = pipe diameter
– e = pipe roughness, mean height of inner surface roughness
• roughness is given by manufacturers
• for steel pipes e = 0.04 mm is typical
– Example: previous steel pipe
• steel pipe with diameter d = 50 mm
• relative roughness 0.04/50 = 0.0008
• Reynolds number R = 69800
– From Moody diagram (next page)
• Friction factor
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0.0224
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Moody diagram
Energy Technology
= (dynamic) viscosity of fluid
– Flow type depends on Reynolds number
• Laminar flow: Re < 2000
• Turbulent flow: Re > 4000
• Transitional area when 2000 < Re < 4000
– Flow may randomly change from laminar to turbulent and back
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Formulas for friction factor
Energy Technology
• For laminar flow the friction factor can be determined
analytically
= 64/R = 64 /( vd)
– Substituting this into the pressure drop formula gives
pv = 32L v/d2
– Pressure drop is proportional to length, viscosity, speed and
inversely proportional to squared diameter
• For turbulent flow is determined from experimental
curve fits, e.g. by Colebrook (1938)
1/
1/2
= -2 log10(e/d/3.7 + 2.51/R/
1/2)
– here e = roughness of pipe
– As appears on both sides, equation must be solved
iteratively
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Computing pressure loss
Energy Technology
1. Determine flow speed v = V’/A
2. Determine Reynolds number R = vd/
3. Determine friction factor from Moody diagram or
Colebrook formula
4. Compute pressure loss
pv = (L/d)
v2/2
• Example:
1 m length of previous pipe (d=50mm, v= 0.51m/s)
pv = (1/0.05) 0.0224 971.8 0.512 / 2 56.6 Pa
• Calculator available e.g. at
http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm
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Pressure loss pipe fittings
Energy Technology
• Pipe fittings, forks, bends etc also cause pressure
loss
– pk =
v2/2
– It is possible to
consider a typical
amount of such
components by
multiplying the
pipe lengths
e.g. by 1.1
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Pressure loss in a network
Energy Technology
• The circulation pump of the network must generate
sufficient pressure difference between the supply and
return pipes in the peak load situation
– At least 50 kPa pressure difference must be guaranteed to each
customer in the peak load situation
– Bottom-up computation
• The pressure loss of pipe elements is added up starting from each customer
along the network
• The maximum pressure loss at each branch is applied in the combined pipe
– The heat exchanger at the DH plant must also have
sufficient pressure difference (50 kPa)
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Pressure levels in a DH network
Energy Technology
• In addition to the pressure losses, the absolute
pressure at each point in the supply pipe must be
sufficient so that water does not start to boil
– At 2 bar the boiling point is 120oC (see diagram)
– The elevation of each node in the network affects the
overall pressure
•
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ph = - g h
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Absolute
pressure
Boiling point
(bar)
(°C)
1
99.63
104.81
Energy1.2
Technology
1.4
109.32
1.6
113.32
1.8
116.93
2
120.23
3
133.54
4
143.63
5
151.85
6
158.84
7
164.96
8
170.42
9
175.36
10
179.88
11
184.06
12
187.96
13
191.6
14
195.04
15
198.28
16
201.37
17
204.3
18
207.11
19
209.79
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20
212.37
Boiling point of water
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Heat loss in DH network
Energy Technology
• The heat loss is about
– 10-20% in small networks with average pipe size DN 50
– 4-10% in large networks with average pipe size DN 150
– Larger losses in small networks are caused by larger pipe surface
area in comparison to tranferred heat
• Heat loss is due to conduction of heat from the pipes into
the ground
• A part of the heat is conducted from the supply pipe into
the return pipe (especially in the dual pipes)
– This heat is not totally lost, but it reduces the transmission capacity
and worsens the efficiency of the DH plant somewhat
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Heat loss in DH network
Energy Technology
• Typical heat conductivity of insulation material is
– 0.463 W/m,K for mineral wool
– 0.033 W/m,K for polyurethane
• The heat conductivity of soil
and moisture
g
depends soil type
– Ranges from 0.5 to 3.5 W/m,K
– Typically g = 1.5 W/m,K (see diagram on next page)
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Heat conductivity of soil
Energy Technology
Gravel 2050 kg/m3
Moraine 1920 kg/m3
Sand 1670 kg/m3
Fine sand 1770 kg/m3
Silt 1600 kg/m3
Clay 1850 kg/m3
Clay 1235 kg/m3
Moisture (%)
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Heat loss in bonded PUR pipes
Energy Technology
• Heat loss per unit length depends linearly on
temperature difference between DH water and
ground
T T
2 K s r Tg
2
• K = combined heat transmission factor that takes into account
heat conductivity of insulation, ground, and from supply pipe
to return pipe (W/m,K)
• Ts = supply water temperature
• Tr = return water temperature
• Tg = ground temperature (5oC + annual average temperature)
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Heat loss in bonded PUR pipes
Energy Technology
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Heat loss in bonded PUR pipes
Energy Technology
• The combined heat resistance is computed as
RM
1
2
g
h'
ln 1 4
b
2
• g = heat conductivity of ground per unit length (W/m,K)
• h’ = h + g/ (corrected depth of pipes)
• h = actual depth of center of pipes from surface (m)
• b = distance between centers of pipes (m)
• = heat transfer coefficient from ground surface to air
(W/m2,K)
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R.Lahdelma
Supply temperature drop in the pipeline
Energy Technology
Temperature drop
Flow line temperature
in the pipeline
90
80
70
60
50
40
30
20
10
0
Temperature drop is biggest where
the flow is low ( house connections)
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