Chapter 8: TRANSIENT ANALYSIS
METHODS FOR BUILDING ELEMENTS
Agami Reddy (July 2016)
1.
2.
3.
4.
5.
6.
7.
8.
9.
Why dynamic models
Terminology
Importance of storage effects in buildings
Types of transient solutions
Finite difference numerical methods
Linear models and concept of superposition
Transfer function models for building elements
Conduction time series model
Thermal network models
HCB 3- Chap 8: Transient Analysis
1
What are dynamic models and when are they used?
In the previous chapter, heat transfer has been
assumed to be steady-state
For the case of a simple wall, at any given
time:
Heat in = Heat out
In certain cases this is a very poor assumption
since the thermal heat capacity of the element
cannot be neglected
For a Trombe wall element, the more
appropriate expression to use is:
Heat stored = Heat in – Heat out
+ Heat generated inside the element
HCB 3- Chap 8: Transient Analysis
2
Other instances where dynamic models are necessary
Embedded piping in ceiling walls or floors
HCB 3- Chap 8: Transient Analysis
3
Building Elements Requiring Transient Analysis
(a) thermal mass of the building envelope
subject to changes in outdoor ambient
T ( x , t)
air conditions and in solar radiation
T ( x , t)
over the day,
T ( x, t)
(b) transmitted solar radiation which is
first absorbed in the internal mass of
the various elements within a
room/space (such as slabs, internal
walls and furniture) and released later in time,
(c) internal mass interacting with operational schedules causing diurnal
fluctuations in internal load (lights, equip. and occupants),
(d) internal mass interaction with control strategies of the indoor
environment (diurnal thermostat set-up or set-back or variations in
supply air volume).
wall
1
wall
2
wall
HCB 3- Chap 8: Transient Analysis
4
Terminology
• Static models
- Steady-state solution (or response)
- Quasi-steady state solution
• Dynamic models
- Transient solution
- Steady-periodic transient solution
HCB 3- Chap 8: Transient Analysis
5
HCB 3- Chap 8: Transient Analysis
6
Conduction Transfer Coefficients- Roofs and Walls
Decrement factor
Time lag
TABLE
7.78.3
Table
Coefficients of Conduction Transfer Function*
(Layer Sequence Left to Right = Inside to
Outside)
n=0
n=1
n=2
n=3
n=4
n=5
n=6
Σcn
U
0.05362
0.080
1.63
0.97
0.02684
0.055
2.43
0.94
0.00673
0.043
4.85
0.82
0.01652
0.138
5.00
0.56
0.00550
0.057
6.32
0.60
0.01477
0.090
7.16
0.16
0.00349
0.140
7.54
0.15
0.00053
0.034
10.44
0.30
δ
λ
a. Roofs
Layers E0 A3 B25 E3 E2 A0
bn
0.000487
0.03474
0.01365
0.00036
0.00000
0.00000
0.00000
Steel deck with 3.33-in insulation
dn
1.00000
−0.35451
0.02267
−0.00005
0.00000
0.00000
0.00000
Layers E0 A3 B14 E3 E2 A0
bn
0.00056
0.01202
0.01282
0.00143
0.00001
0.00000
0.00000
Steel deck with 5-in insulation
dn
1.00000
−0.60064
0.08602
−0.00135
0.00000
0.00000
0.00000
Layers E0 E1 B15 E4 B7 A0
bn
0.00000
0.00065
0.00339
0.00240
0.00029
0.00000
0.00000
Attic roof with 6-in insulation
dn
1.00000
−1.34658
0.59384
−0.09295
0.00296
−0.00001
0.00000
Layers E0 B22 C12 E3 E2 C12 A0
bn
0.00059
0.00867
0.00688
0.00037
0.00000
0.00000
0.00000
1.67-in insulation with 2-in h.w. concrete RTS
dn
1.00000
−1.11766
0.23731
−0.00008
0.00000
0.00000
0.00000
Layers E0 E5 E4 B12 C14 E3 E2 A0
bn
0.00000
0.00024
0.00217
0.00251
0.00055
0.00002
0.00000
3-in insulation w/4-in l.w. conc. deck and susp. clg.
dn
1.00000
−1.40605
0.58814
−0.09034
0.00444
−0.00006
0.00000
Layers E0 E5 E4 C5 B6 E3 E2 A0
bn
0.00001
0.00066
0.00163
0.00049
0.00002
0.00000
0.00000
1-in insul. w/4-in h.w. conc. deck and susp. clg.
dn
1.00000
−1.24348
0.28742
−0.01274
0.00009
0.00000
0.00000
Layers E0 E5 E4 C13 B20 E3 E2 A0
bn
0.00001
0.00060
0.00197
0.00086
0.00005
0.00000
0.00000
6-in h.w. deck w/0.76-in insul. and susp. clg.
dn
1.00000
−1.39181
0.46337
−0.04714
0.00058
0.00000
0.00000
Layers E0 E5 E4 B15 C15 E3 E2 A0
bn
0.00000
0.00000
0.00002
0.00014
0.00024
0.00011
0.00002
contd…
HCB 3- Chap 8: Transient Analysis
7
TABLE 8.4 Thermal properties and code numbers of layers used in roof and wall descriptions of Table 8.3- SI Units
Layer
ID
Description
Thickness
Conductivity
Density
mm
W/(m·K)
kg/m
3
Specific Heat
Resistance
kJ/(kg·K)
(m ·K)/W
2
A0
A1
Outside surface resistance
25 mm (1 in) Stucco
25
0.692
18.58
0.84
0.059
0.037
A2
100 mm (4 in) Face brick
100
1.333
2002
0.92
0.076
A3
Steel siding
2
44.998
7689
0.42
0.000
A6
Finish
13
0.415
1249
1.09
0.031
B1
Air space resistance
-
-
-
B3
50 mm (2 in) Insulation
51
0.043
32
0.84
1.173
B6
50 mm (2 in) Insulation
51
0.043
91
0.84
1.173
B7
25 mm (1 in) Wood
25
0.121
593
2.51
1.760
B12
75 mm (3 in) Insulation
76
0.043
91
0.84
1.760
B13
100 mm (4 in) Insulation
100
0.043
91
0.84
2.347
B14
125 mm (5 in) Insulation
125
0.043
91
0.84
2.933
B15
150 mm (6 in) Insulation
150
0.043
91
0.84
3.520
B20
20 mm (0.76 in) Insulation
20
0.043
91
0.84
0.440
B22
42 mm (1.67 in) Insulation
42
0.043
91
0.84
0.968
C2
100 mm (4 in) low density concrete block
100
0.381
609
0.84
0.266
C8
100 mm (4 in) high density concrete block
100
1.731
977
0.84
0.059
C8
200 mm (8 in) high density concrete block
200
1.038
977
0.84
0.196
C9
200 mm (8 in) Common brick
20
0.727
1922
0.84
0.279
C11
300 mm (12 in) high density concrete
300
1.731
2243
0.84
0.176
C12
50 mm (2 in) high density concrete
50
1.731
2243
0.84
0.029
C13
150 mm (6 in) high density concrete
150
1.731
2243
0.84
0.088
C14
100 mm (4 in) low density concrete
100
0.173
641
0.84
0.587
E0
Inside surface resistance
-
-
-
-
0.121
contd…
HCB 3- Chap 8: Transient Analysis
0.160
8
Temperature Profiles in a Wall Under Transient Conditions
Forward and
Backward trains
Santamouris, M. and D. Asimakopoulos (1996)
HCB 3- Chap 8: Transient Analysis
9
System model
x1
f(x)
y1
A system model is a description of the system. Empirical and mechanistic
x2
y2
models are made up of three components:
(i) input variables (also referred to as regressor, forcing, exciting, exogenous
or independent variables which
on the system, and :
c1 xact
1
c1 y1  c2 y2
- can be controlled by the
c2 xexperimenter,
2
- are uncontrollable or extraneous such as climatic variables;
(ii) system structure and parameters/properties provide the necessary physical
description of the systems in terms of physical and material constants ; for
example, thermal mass, overall heat transfer coefficients, mechanical
properties of the elements; and
(iii) output variables (also called response, state, endogenous or dependent
variables) which describe system response to the input variables.
HCB 3- Chap 8: Transient Analysis
10
Estimate of Static Heat Capacity for the House of Example 7.5, Not Counting the Wooden
Support Beams in Walls, Floor, and Roof. Material Properties from Appendices in CD-ROM
Area
A
Component
m
2
Thickness
ΔX
Volume V
= At
m
m
3
Specific
Heat Cp
Density
ρ
kJ/(kg • K)
kg/m
3
ρcp
Heat
Capacity C =
Vρcp
kJ/(m •
K)
3
kJ/K
Drywall, roof
144
0.013
1.87
1.20
800
960.0
1797
Drywall, walls
360
0.013
4.68
1.20
800
960.0
4493
Plywood, floor
144
0.013
1.87
1.09
545
594.1
1112
Furniture (same
as floor)
Air
1112
144
2.50
360.0
1.00
1.20
1.20
Total (without
brick floor)
432
8946
If we add effect of wooden support beams in walls, floor and roof, total = 11,000
Distinguish between dynamic and static heat capacity
HCB 3- Chap 8: Transient Analysis
11
A brick or concrete floor would increase the heat capacity significantly.
Concrete has a ρcp value around 1500 kJ/(m3.K).
If the concrete floor is 0.1 m thick, its static heat capacity is 144 m 2×0.1 m×1500 kJ/(m3.K) = 21,600
kJ/K.
Since it is fairly thick, we will assume its effective heat capacity to be around one-half that value. The
same house as before with the plywood floor replaced with a concrete one will have an effective heat capacity
of about 9000+21,600/2 = 20,688 kJ/K which is about twice that of the room with plywood floor.
To get an idea of the importance of the heat capacity, suppose that during the course of a summer afternoon
the average temperature of the mass increases by 2 K in 5 h, implying a rate of temperature rise of
2K
T
 0.4 k/h
5h
This can happen even when the thermostat is set to maintain the air at constant temperature: Under dynamic
conditions, air and room mass can have different temperatures. Assuming an effective heat capacity Ceff =
11,000 kJ/K, the storage term in the heat balance, is
Qstor  Ceff T  11, 000 kJ/K  0.4 K/h  1.2 kW
HCB 3- Chap 8: Transient Analysis
12
For summer design conditions the temperature difference Ti−To is around 25° C−35° C = −10° C, and the
corresponding steady-state conductive heat loss (with Ktot = 205 W/K from Example 7.5) is
K tot (Ti  To )  2.05 kW
with a minus sign because it a heat gain of the building.
If there were no storage, the cooling load would be 2.05 kW (for simplicity, we assume that there are no other
terms in the thermal balance). But 1.2 kW of this heat gain is soaked up by the storage, and thus the effective
cooling load at this moment is only 2.05−1.2 = 0.85 kW.
The storage effect reduces the cooling load by over one-half.
HCB 3- Chap 8: Transient Analysis
13
Different Transient Solutions to Dynamic
Analysis of Buildings
a) Distributed models
- Harmonic/admittance methods- closed form solutions
b) Numerical methods (finite differences)- most flexible
c) Thermal network models: lumped and 3C models
d) Transfer function models for both conduction and rooms
(used by eQuest, DOE-2, TRACE, HAP…)
- CLTD/CLF method is a simplified older version (now obsolete)
e) Transfer function models for conduction and then heat balance
method for room response (EnergyPlus)
- Simplified approach: Radiant Time Series method
HCB 3- Chap 8: Transient Analysis
14
Coverage
• Building elements- Chap 8
– Walls, roof, slabs
subject to outdoor ambient air and solar radiation changes
• Entire rooms/spaces- Chap 9
- treatment of transmitted solar
- treatment of internal loads
- combining various individual heat loads
- peak load calculation methods
HCB 3- Chap 8: Transient Analysis
15
Finite Difference Method
The finite difference method relies on
discretizing a function on a grid.
A very flexible and widely used numerical method for solving differential
equations with any type of boundary condition
It involves discretizing the derivatives of the temperature profile both
spatially and temporally into discrete quantity
The solid, or say the building wall assumed to be a homogeneous element
with constant conductivity, is divided into a number of equally spaced
finite layers of thickness HCB 3- Chap 8: Transient Analysis
16
T , h

T1
T2


Each of the layers is assumed to be at a uniform temperature i.e., all the mass of the wall
element is located at the node. An energy balance
on node m yields:
T
Heat flow in from node (m-1)- heat flow out to node (m+1) = change in heat capacity of node m, or
(Tm1  Tm )
(Tm  Tm1 )
(Tmt t  Tmt )
k
k
  .c p .x
x
x
t
which simplifies to: Tm1  2  Tm  Tm1 
where  
 .t
x
2
1

(Tmt t  Tmt )
and is called the mesh Fourier number.
with thermal diffusivity   k / (  c p )
HCB 3- Chap 8: Transient Analysis
17
Equation from previous slide:
Tm1  2  Tm  Tm1 
1

(Tmt t  Tmt )
We now need to solve this equation over time. The simplest is the explicit finite difference
formulation wherein the temperature terms on the left hand side of the previous are expressed at time
step t.
Tmt 1  2  Tmt  Tmt 1 
1

Explicit forward difference formulation
(Tmt t  Tmt )
From which Tmt t   (Tmt 1  Tmt 1 )  (1  2   )Tmt
Since the initial conditions Tmt 0 for all nodes m will be known, Tmt t can be determined explicitly
for each successive time increment t .
HCB 3- Chap 8: Transient Analysis
18
The smaller the time step, the more accurate the solution. However,
computing time increases greatly and so it is advantageous to use as large a
time step as possible.
Though the explicit finite difference method is easy to use, there is a
restriction in the maximum time step which can be selected because of
instability issues. The maximum time step is given by the stability criterion
which states that   2 for all interior nodes be selected such that
tmax
For example, for a brick wall,
1 x 2

2 
  0.45 106 m2 /s
If a mesh size of 0.01 m is selected, then the simulation time step is 111 s =
1.85 min.
HCB 3- Chap 8: Transient Analysis
19
Transient response of a wall subjected to a step change forcing function
Consider a concrete wall 20 cm thick of a conditioned space which is subject to an
abrupt change in the outdoor air temperature.
Initially, the wall is at 20o C throughout its thickness. The outdoor air temperature is
abruptly increased to 35o C.
Calculate the time evolution of the temperature profiles inside the wall and determine
the heat flow inwards.
Explicit finite difference assuming the wall to be discretized into 5 nodes (nodal
spacing is 5 cm) - nodes are numbered 0, 1... 4 with 0 being the node representing the
exterior wall surface and node 5 representing the interior wall surface.
T , h

T1
T2


Node
0
1
2
3
4
0
1
2
3
4
5
6
7
hrs
T
HCB 3- Chap 8: Transient Analysis
20
Concept of Superposition for Linear Systems
(Used in most Building Energy Simulation Programs)
x1
y1
x2
y2
c1 x1
c2 x2
c1 y1  c2 y2
A system is said to be linear if and only if, it has the following property: if an input x1 (t) produce
output y1(t), and if an input x2(t) produces an output y2 (t), then an input [c1 x1(t) + c2 x2(t)] produce
output [c1 y1(t) + c2 y2(t)] for all pairs of inputs x1(t) and x2(t) and all pairs of real number constants a1
a2. An equivalent concept is the principle of superposition which states that the response of a li
system due to several inputs acting simultaneously is equal to the sum of the responses of each i
acting alone. This is an extremely important concept since it allows the response of a complex system
be determined more simply by decomposing the input driving function into simpler terms, solving
equation for each term separately, and then summing the individual responses to obtain the des
aggregated response.
HCB 3- Chap 8: Transient Analysis
21
Transfer Function Model Formulation
For instance, If the response Ti is determined by two driving terms, heat input Q , and outdoor temperature
To, then one can write the transfer function model in the form:
ai , 0Ti , t  ai , 1Ti , t 1t 
 ai , nTi , t  nt
 ao , 0To , t  ao , 1To , t 1t 
 ao , mTo , t mt
 aQ , 0Qt  aQ , 1Qt 1t  aQ , 2Qt  2 t 
(8.18)
 aQ , r Qt r t
with the three sets of transfer function coefficients ai,0 to ai,n, ao,0 to ao,m, and aQ,0 to aQ,r.
This equation can be considered an algorithm for calculating Ti,t, hour by hour, given the previous values of
Ti and the driving terms To and Q .
Likewise, if Ti and To were given as driving terms, one could calculate Q as response.
HCB 3- Chap 8: Transient Analysis
22
Conductive Heat Gain
Step 1: Instantaneous heat gain using conduction transfer functions
The 1-D conductive heat gain (or loss) Qcond ,t at time t through the roof and walls is calculated acco


Qcond, t   d nQcond, t n t  A   bnTos , t nt  Ti  cn  (8.19)
n 1
n0
 n0

where
A = area of roof or wall, m2 (ft2)
Δt = time step = 1 h
Tos,t = sol-air temperature of outside surface at time t;
bn, cn, dn = coefficients of conduction transfer function.
HCB 3- Chap 8: Transient Analysis
23
The transfer function coefficients are subject to physical constraints
under steady-state limit:


Qcond  d n  A  Tos  bn  Ti  cn  where d 0  1
n0
n0
 n0

Since in that limit we also have
Qcond  AU (Tos  Ti )
the coefficients of Tos and Ti must be equal
 bn   cn
n0
and the U value is given by
(8.20)
(8.21)
(8.22a)
n0
c
U
d
n0
n0
n
(8.22b)
n
HCB 3- Chap 8: Transient Analysis
24
Conduction Transfer Coefficients- Roofs and Walls
TABLE 7.7
Coefficients of Conduction Transfer Function*
(Layer Sequence Left to Right = Inside to
Outside)
n=0
n=1
n=2
n=3
n=6
Σcn
U
0.05362
0.080
1.63
0.97
0.02684
0.055
2.43
0.94
0.00673
0.043
4.85
0.82
0.01652
0.138
5.00
0.56
0.00000 0.00550 0.057
0.283372 0.372389
0.00000
6.32
1.30
0.60
0.98
0.00002
0.01224
0.00009
−0.02561
0.00005
0.00039
0.00058
0.00000 0.00000 0.01477 0.090
0.00029
0.098947 0.314501
0.00000 0.00000
0.00014
0.00000 0.00000 0.00349 0.140
0.074007 0.692064
0.00000 0.00000
7.16
3.21
0.16
0.91
7.54
5.14
0.15
0.41
0.00024
0.00011
10.44
0.30
n=4
n=5
δ
λ
a. Roofs
Layers E0 A3 B25 E3 E2 A0
bn
0.000487
0.03474
0.01365
0.00036
0.00000
0.00000
0.00000
Steel deck with 3.33-in insulation
dn
1.00000
−0.35451
0.02267
−0.00005
0.00000
0.00000
0.00000
Layers E0 A3 B14 E3 E2 A0
bn
0.00056
0.01202
0.01282
0.00143
0.00001
0.00000
0.00000
Steel deck with 5-in insulation
dn
1.00000
−0.60064
0.08602
−0.00135
0.00000
0.00000
0.00000
Layers E0 E1 B15 E4 B7 A0
bn
0.00000
0.00065
0.00339
0.00240
0.00029
0.00000
0.00000
Attic roof with 6-in insulation
dn
1.00000
−1.34658
0.59384
−0.09295
0.00296
−0.00001
0.00000
Layers E0 B22 C12 E3 E2 C12 A0
bn
0.00059
0.00867
0.00688
0.00037
0.00000
0.00000
0.00000
1.67-in insulation with 2-in h.w. concrete RTS
dn
1.00000
b. Walls
Layers E0 E5 E4 B12 C14 E3 E2 A0
bn
0.00000
- Layers E0 A3 B1 B13 A3 A0
3-in insulation w/4-in l.w. conc. deck and susp. clg.
dn
1.00000
Steel siding with 100 mm (4-in) insulation
Layers E0 E5 E4 C5 B6 E3 E2 A0
bn
0.00001
- Layers E0 E1 B14 A1 A0
1-in insul. w/4-in h.w. conc. deck and susp. clg.
dn
1.00000
Frame wall with 13 mm (5-in) insulation
Layers E0 E5 E4 C13 B20 E3 E2 A0
bn
0.00001
- Layers E0 A6 C5 B3 A3 A0
6-in h.w. deck w/0.76-in insul. and susp. clg.
dn
1.00000
100 mm (4-in) h.w. concrete with 50 mm (2-in) insulation
Layers E0 E5 E4 B15 C15 E3 E2 A0
bn
0.00000
−1.11766
0.23731
−0.00008
0.00000
0.00000
0.00000
0.00024
bn 0.04361
−1.40605
dn 1.00000
0.00066
bn 0.00089
−1.24348
dn 1.00000
0.00060
bn 0.00561
−1.39181
dn 1.00000
0.00000
0.00217
0.00251
0.19862 0.04083
0.58814 −0.09034
−0.24072 0.00168
0.00163
0.00049
0.03097 0.05456
0.28742 −0.01274
−0.93389 0.27396
0.00197
0.00086
0.04748 0.02052
0.46337 −0.04714
−0.93970 0.04664
0.00002
0.00014
0.00055
0.00002
0.00032
0.00444 −0.00006
HCB 3- Chap 8: Transient Analysis
0.00002
0.00053
0.034
25
Example 8.2. Conduction transfer function analysis for a wall
Calculate the conductive heat gain per square meter of an exterior vertical wall, of dark color and facing west, for summer design
conditions (July 21) if Ti = 25° C.
Given:
Exterior wall of 0.10 m (4 in) concrete with 0.05 m (2 in) insulation on the outside
Ti = 25° C, Sol-air temperatures are specified
Lookup values: The transfer function coefficients for walls and roofs are listed in Table 8.3. This wall corresponds to the third
entry under the walls section.
b0 = 0.00561
b1 = 0.04748
b2 = 0.02052
b3 = 0.00039
 cn = 0.074007
d0 = 1.0000
d1 = −0.93970
d2 = 0.04664
d3 = 0.0
The d's are dimensionless, and the b's and c's are in W/(m2. K). All other coefficients are zero, and the U value is 0.693 W/(m2. K)
For the first (Qcond ,t / A) entry at t = 1, the value is calculated as:
Q cond ,1
d .Q
d .Q
  1 cond ,11  2 cond ,12  b0 .Tos ,10  b1.Tos ,11  b2 .Tos ,12  b3.Tos ,13  Ti  cn
A
A
A
n 0
=  ( 0.93970)  0.00  0.04664  0.00
 0.00561  24.4  0.04748  25.0  0.02052  27.1  25  0.074007 = 0.020
To initiate the second day calculations, the conductive gains for the last three hours of the previous day are copied as into the first
three rows of the column corresponding to the second day, and so on …
HCB 3- Chap 8: Transient Analysis
26
t
Tos ,t
h
°C
−2
−1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Average
27.2
26.1
25.0
24.4
24.4
23.8
23.3
23.3
25.0
27.7
30.0
32.7
35.0
37.7
40.0
53.3
64.4
72.7
75.5
72.2
58.8
30.5
29.4
28.3
27.2
26.1
25.0
37.95
Table 8.5:
Calculation
results of
Example 8.2
Conduction
transfer model
Q cond , t
A
Q cond , t  24
A
Q cond , t  48
A
Q cond , t  72
A
Q cond , t  96
A
13.438
12.117
10.860
9.661
8.541
7.529
6.598
5.734
4.965
4.378
4.038
3.927
4.027
4.308
4.756
5.404
6.734
8.801
11.325
13.889
16.010
17.022
16.248
14.830
13.440
12.119
10.861
13.440
12.119
10.861
9.662
8.542
7.530
6.599
5.735
4.966
4.379
4.039
3.927
4.028
4.308
4.756
5.404
6.734
8.801
11.325
13.889
16.010
17.022
16.249
14.830
13.440
12.120
10.862
8.965
W/m2
0.00
0.00
0.00
0.020
-0.013
-0.061
-0.136
-0.240
-0.335
-0.325
-0.134
0.225
0.743
1.393
2.170
3.110
4.699
6.995
9.723
12.468
14.749
15.903
15.25 6
13.949
12.659
11.426
10.246
12.659
11.426
10.246
9.116
8.058
7.100
6.218
5.397
4.666
4.113
3.803
3.718
3.842
4.143
4.610
5.274
6.619
8.699
11.234
13.809
15.939
16.959
16.192
14.780
13.396
12.080
10.827
13.396
12.080
10.827
9.631
8.515
7.506
6.577
5.716
4.949
4.364
4.026
3.916
4.017
4.299
4.748
5.397
6.728
8.796
11.320
13.885
16.006
17.019
16.245
14.827
13.438
12.117
10.860
HCB 3- Chap 8: Transient Analysis
27
Sol-air temp.
HCB 3- Chap 8: Transient Analysis
28
Comparison of Transient Responses of Lightweight and Heavy Walls
Sol-air temp.
Figure 8.10 shows this effect for the light-weight steel-siding wall (wall entry #1 in Table 8.3b) and for the
300 mm hw concrete (wall entry #7), both of which have about the same U-value. As expected, the former
has a small time lag of less than 2 hours and the amplitude is very slightly lower (from Table 8.3,
  1.3 h and  = 0.98 ), whereas the concrete wall has almost damped out the entire excitation function (in
this case,   11.4 h and  = 0.10 ).
HCB 3- Chap 8: Transient Analysis
29
Conduction Time Series (CTS) Model Formulation
Used to calculate hourly 1-D conductive heat flows through opaque building
elements such as roof and wall surfaces is part of the Radiant Transfer Series
(RTS) method (described in Chap 9)
CTS model is akin to the conduction transfer function model in the sense that
both of them use a time series model involving lagged values, and also assume
that the indoor space temperature is held constant.
The conduction transfer function model uses an iterative process to calculate
the conductive heat flow across the opaque elements of the building envelope
The novelty of the CTS model is that it replaces this iteration with a simple
summation which does not involve lagged values of the conductive heat gains
HCB 3- Chap 8: Transient Analysis
30
CTS Model
23
Qcond ,t  UA. c j (T os , t -n. t - Ti )
(8.23a)
j 0
23
Subject to
c
j 0
j
 1.0
(
where U is the steady-state overall heat transfer coefficient,
A is the surface area,
cj is the jth conduction time series (CTS) factors which are normalized to unity,
T os , t -n.t is the lagged sol-air temperature from n hours ago, and
Ti is the space temperature assumed to be constant.
HCB 3- Chap 8: Transient Analysis
31
Conduction Time Series Factors (%)
The CTS factors, comprising of 24 coefficients, for various types of walls and
roofs can be found from lookup tables similar to those created for the CTF
coefficients (Table 8.6). These factors can be viewed as hourly adjustments to
the steady-state heat transfer expression which correct for heat absorbed in the
wall or roof during previous hours and released during this hour.
50
40
Wall type 6 (stud wall)
30
Wall type 27
(100 mm concrete wall)
20
10
0
1
3
5
7
9
11 13 15 17 19 21 23
Hour of Day
Figure 8.11 Conduction time series factors for light and heavy walls
HCB 3- Chap 8: Transient Analysis
32
Example 8.3. Conduction time series analysis for a wall
Calculate the conductive heat gain per square meter for a wall type 27 (similar wall construction as that assumed in
Example 8.2) and identical orientation. The solar-air values are the same and the interior temperature is kept constant
at 25° C.
Given: The sol-air temperatures and CTS values are given in Table 8.6. The U value = 0.673 W/(m 2.K), Ti = 25°
C.
Find:
Q cond
SOLUTION
We start by creating a spreadsheet (Table 8.6) with three long columns (one for t, one for CTS and one for Tos,t. The
last column assembles the computed values of Q cond ,t , per unit wall area A, according to Eq. (8.23). Notice that no
iterations are needed. To illustrate the calculations, let us consider hour 22:00. The cell formula is given by:
Q cond ,1
 0.673  [0.01  (27.2  25)  0.1  (28.3  25)  0.2  (29.4  25)  0.18  (30.5  25)
A
 0.14  (58.8  25)  0.1  (72.2  25)  0.07  (75.5  25)  0.05  (72.7  25)
 0.04  (64.4  25)  0.03  (53.3  25)  0.02  (40  25)  0.02  (37.7  25)
0.01  (35  25)  0.01  (32.7  25)  0.01  (30  25)  0.01  (27.7  25)  0  ...] = 14.02
and so on.
HCB 3- Chap 8: Transient Analysis
33
TABLE
8.6
Calculation
results for Example 8.3 meant
to illustrate the use of the
conduction time series model.
The wall is a 100 mm concrete
light weight wall with board
insulation and gypsum board.
- Conductive gain reaches its
maximum at 19:00, which is 3 h after
the peak of the sol-air temperature.
- As a check, verify that the solution
satisfies the steady-state limit.
Multiplying the average U.(Tos−Ti )=
(37.95−25)o C x 0.673 = 8.71 W/m2
(same as shown in table)
t
CTS
Tos ,t
h
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Average
%
1
10
20
18
14
10
7
5
4
3
2
2
1
1
1
1
0
0
0
0
0
0
0
0
°C
24.4
24.4
23.8
23.3
23.3
25.0
27.7
30.0
32.7
35.0
37.7
40.0
53.3
64.4
72.7
75.5
72.2
58.8
30.5
29.4
28.3
27.2
26.1
25
37.95
HCB 3- Chap 8: Transient Analysis
Q cond , t
A
W/m2
6.32
4.74
3.45
2.48
1.66
0.92
0.50
0.43
0.75
1.44
2.54
3.79
5.22
7.44
11.01
15.42
19.83
23.23
24.48
22.38
18.05
14.02
10.75
8.27
8.71
34
.
Thermal Network
Models
T
Wall
section
T0
i
X
models for a wall
provide less
accurate solutions
and are used with
larger time steps
- 15 min or 1 h
h
R
X
Heat
flow
i
x
kA
h
Ti
Lumped model
0
and C   c p Ax
T0
Ti
R1 
(a)
R
1

2 hi A
C
R1
Ts1
C
R2 
1
R2
Ts2
R
1

2 h0 A
Rn
C2
(b) 2R1C lumped model
Tsn
Rn+1
T0
Cn
(c) n th order lumped model
Thermal networks to model heat flow through a homogeneous plane
wall of surface area A and wall thickness x
HCB 3- Chap 8: Transient Analysis
35
Example of
3-node
discretization
100 mm high
density concrete
Finish
Steel
siding
Insulation
Outdoor
air
Indoor
air
Ti
RA3
REO
RA6
RC5
To
RA0
1/2 of the face brick thermal mass is assigned to the central node
and 1/4th to each of the two outside nodes
RC networks are not unique
Discretion of user involved
T0
R1
T1
C1
R2
T2
C2
R3
T3
R4
T4
C1
HCB 3- Chap 8: Transient Analysis
36
1R1C Networks for Simple One-zone Buildings
Heat balance equation
To  Ti
Q
(8.34)
R
This is a first-order differential equation in the variable Ti. Let us define a quantity τ, called the time constant,
(8.35)
  RC
To explain the reason for the name, we rewrite the equation in the form
 Ti  Ti  To  RQ
(8.36)
CTi 
and solve it for the special case when the driving terms To and Q are constant. In that case, the derivative of the
variable
T (t )  Ti (t )  To  RQ
(8.37)
is equal to Ti , and so, Eq. (8.36) simplifies to
T  T  0
(8.38)
Its solution is a simple exponential
 t 
T (t )  T (0) exp  
(8.39)

 
where T(0) is the initial value. The time constant sets the time scale for the cool-down; after one time constant,
T(t) has decayed to 1/e ≈ 0.368 of its initial value T(0). The longer the time constant of a building, the longer37it
takes to cool down or warm up.
The above eqn can be re- expressed as: T(t+1)+a1 T(t) =0
which is the time series formulation of the transfer function approach
HCB 3- Chap 8: Transient Analysis
Time Constant
25
25
Steady-state value
Steady-state value
20
Instrument reading
Instrument reading
20
15
15
63.2% of change
Dynamic
Small time
constant instrument
responseresponse
Dynamic
10
10
Large time constant instrument
of two instruments
55
Small time
constant
00
0
0
5
Large time
constant
10
15
20
25
5
10Time (sec) from
15 step change20
25
Time from step change in input (seconds)
30
30
HCB 3- Chap 8: Transient Analysis
38
1R1C Networks can be used for rough estimation;
for example, cool-down and set-up times in residences
Example 8.5 Time delay in thermostat set-back
Suppose a building can be described by the 1R1C network of Fig. 8.16. The variables Ti, To, and the heat input Q
are assumed constant until t = 0, when the thermostat is set back. How long does it take for Ti to drop from 20° C
(68° F) to 15° C (59° F)?
Given: R = 5.0 °C /kW, implying Ktot = 200 W/°C)], C = 10 MJ/°C, To = 0° C
Q as necessary to maintain Ti = 20° C until t = 0 ( Q  0 for t > 0)
Find:
Time t at which Ti(t) = 15°C
SOLUTION
Let T(t) = Ti(t) − To. For t > 0, it satisfies Eq. (8.38), and the solution according to Eq. (8.39) is
 t 
T (t )  T (0) exp  
 



with T (0)  20 C  0 C  20 C and time constant
  RC  5.0 o C / kW 10 MJ / o C  5.0 104 s  13.889 h
Fig. 8.16
We are looking for the value of t at which T(t) = 15° C – 0° C = 15o C; hence we have to find the value of t that
solves
 t 
15 o C  20 o C  exp  
 
The solution is
 15 
t    ln    13.889 h  0.28768  4.00 h
 20 
HCB 3- Chap 8: Transient Analysis
39
Example 8.6 Thermostat set-back recovery time
Suppose the heating system of the building in Example 8.5 has been built for design temperatures Ti,des = 20°
C (68° F) and To,des = −20° C (−4° F). How long is the setback recovery time for the conditions of Example
8.5?
SOLUTION
First find the heat input Q ,
Ti ,des  To,des 20 C  (20 C)
Q

 8.0 kW (27.3 kBtu / h)
R
5.0 C / kW
Take the start of the warm-up as t = 0. During warm-up, the conditions of constant driving terms are again
satisfied, and so we can again use the exponential solution (Eq. 8.39)
 t 
T (t )  T (0) exp  
 
If we take
T (t )  Ti (t )  To  RQ with RQ  5.0 o C / kW  8.0 kW  40 o C
Then, the initial value is
T (0)  Ti (0)  To  RQ  15 C  0 C  40 C  25 C
We want to know at what time t we reach Ti(t) = 20°C, or
T (t )  20 C  0 C  40 C  20 C
Hence, we have to solve the following for time t:
 t 
20 C  25 C  exp  
 
The solution is
 20 
t    ln 
  (13.889 h)  (0.2231)  3.10 h
 25 
COMMENTS
The warm-up time depends on To and
. The
To Analysis
and the smaller Q ,
HCBon
3-QChap
8: lower
Transient
40
the slower the warm-up. At To = −10° C, we would have found
Outcomes
•
•
•
•
•
•
Understand basic difference between steady state and dynamic models
Understand instances where dynamic models are necessary
Familiarity with different transient solution methods
Knowledge of the finite difference method
Understand concept of superposition of linear systems
Be able to apply the transfer function model to analyze conduction through
simple building envelop elements
• Be able to apply the conduction time series model to analyze conduction
through simple building envelop elements
• Familiarity with how heat transfer through composite envelop elements can be
modeled as RC networks
• Be able to apply the 1R1C network to analyze heat-up and cool-down of onezone buildings
HCB 3- Chap 8: Transient Analysis
41