Radiant PTC CE BTA 12.2 Us Mp

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PTC_CE_BTA_12.2_us_mp.mcdx

Mathcad Enabled Content Copyright 2011 Knovel Corp.

Building Thermal AnalysisAndreas Athienitis 2011 Parametric Technology Corp.Chapter 12 Analysis of radiant heating systems

12.2 Thermal Analysis of Ceiling Heating

Disclaimer

While Knovel and PTC have made every effort to ensure that the calculations, engineering solutions,diagrams and other information (collectively Solution) presented in this Mathcad worksheet aresound from the engineering standpoint and accurately represent the content of the book on which the

Solution is based, Knovel and PTC do not give any warranties or representations, express or implied,including with respect to fitness, intended purpose, use or merchantability and/or correctness oraccuracy of this Solution.

Array or ig in:

ORIGIN 0

This section presents a model for a space (one zone) heated with a ceiling heating system (gypsum

ceiling panels with embedded electric heating). It is assummed that the heat delivered by the systemis qaux.

The auxiliary heating qaux is proportional to the error between the setpoint Tsp and the actual roomair temperature T1:

qaux Kp Tsp T 1 Kp ...proportional control constant

The explicit finite difference method is employed to model the system and to determine the heatingload and the room temperature variation.

Mathcad Enabled ContentCopyright 2011 Knovel Corp.All rights reserved.

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1. Weather inputs(a) Outside temperature.(b) Solar radiation.

Note that, since heating equipment should be sized based on extreme weather conditions, solarradiation may be excluded from this analysis. However, if a passive solar analysis is to be performed,the solar gains should be considered.

2. Bui ld ing data

NS =number of surfaces contributing to the zone energy balance.

NSe = " " exterior surfaces (walls and roof).

Ai =area of exterior surface i.

Nw: number of windows (=NSe normally) Awi: area of window i

Window type: U-value or thermal resistance, single or double glazing

and kL value (extinction coefficient x thickness)

Adoor: external door area Rdoor: external door R-value

Wall construction: Wall layer properties. For the interior layer, properties for transient analysis arerequired.

ach: infliltration - air changes per hourhi: inside combined surface heat transfer (film) coefficient for surface i

Internal gains:

Qintr: radiative internal gainsQintc: convective internal gains(optional inputs to be added as heat sources in finite difference model)


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Example:Consider a house which consists of a basement and a ground level floor. Here we consider theground level zone, heated with ceiling heating.

Hh 10 ft

Lh 50 ft

Wh 33 ft

Surfaces contributing to energy balance: NS 6

i , 1 2 NS 1-4 walls, 5-ceiling, 6-floor se 1 5 exterior surfaces

Nw 4 iw , 1 2 Nw (assume four windows -sum the windowareas on each house side)

Window and door areas:

Aw1

86 ft2 Aw2

43 ft2 Aw3

43 ft2 Aw4

43 ft2

Ad1

20 ft2 Ad2

20 ft2 Ad3

20 ft2 Ad4

20 ft2


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Wall net areas: A1

Lh Hh Aw1

Ad1

A2

Wh Hh Aw2

Ad2

A3

Lh Hh Aw3

Ad3

A4

Wh Hh Aw4

Ad4

A5

Wh Lh

A6

A5

Hi 8 ft ...internal height Vol A5

Hi

Door thermal resistance Rd 38.75 ft2

Fwatt

window resistance (double-glazed) Rw 6.588 ft2 F

watt

Film coefficients h1

0.428 wattft2 F

h2

h1

h3

h1

h4

h1

Hot ceiling, hot floor h5

0.361 watt

ft2 Fh

60.48

watt

ft2 F

ach=air changes /hour ach 0.5

Specific heat anddensity of air

cpair 0.239 Btu

lb Fair 0.075

lb

ft3


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Calculation of infiltration conductance:

Uinf ach Vol3600 sec

air cpair =Uinf 34.672 wattF

Thermal Resistance of Walls (includ ing air films):

Vertical Walls

1. gypsum board

L1

0.043 ft 1

50 lb

ft3thickness,density.

k1

0.027 watt

ft Fc

10.179

Btu

lb Fconductivityspec. heat

2. insulationRins 60.063 ft

2 F

watt

3. siding +sheathing

Rsid 7.169 ft2

F

watt

4. exterior film ho 1.135 watt

ft2 F

15% of area is framing ff 0.15 ...fraction of area which is framing


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2 by 4 wood stud with R-value: Rf 14.919 ft

2 F

watt

R1

1

+1 ff

++++

L1

k1

Rins Rsid1

ho

1

h1

ff

++++

L1

k1

Rf Rsid1

ho

1

h1

=R1

57.553 F ft2

watt

Calculation of wall conductance excluding interior layerand film (to be used for admittance calculations):

u1

1

R1

L1

k1

1

h1

Assume that all exterior walls are of the same construction:

ii , 1 2 4

Lii

L1

Rii

R1

uii

u1

kii

k1

ii

1

cii

c1


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Calculation o f roof -ceiling thermal resistance:

Ceiling 1. gypsum boardwith imbedded electricresistance wires

L5

0.059 ft k5

k1

c5

c1

5

1

2. insulation Rinsc 116.25 ft2 F

watt

3. air-film (attic) ha 0.619 watt

ft2 F

Rc 1

+1 ff

+++

L5

k5

Rinsc1

ha

1

h5

ff

+++

L5

k5

Rf1

ha

1

h5

=Rc 71.939

F ft2

watt

Roof 1. exterior air film ho 1.032 watt

ft2 F

2. shingle backer board Rb 3.681 ft2

F

watt


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3. wood shingles Rsh 3.294 ft2

F

watt

Rr 1

+1 ff

+++Rb Rsh1

ho

1

ha

ff

++++Rf Rb Rsh1

ho

1

ha

=Rr 10.521 ft2 F

watt

Assuming a 30 degree slope for the roof, we calculate the ceiling-roof combined resistance per unitceiling area (assuming no ventilation in the attic) as follows:

Ar

A5

cos 30 degR

5

+RcA

5

RrAr

A5

u5

1

R5

L5

k5

1

h5

...for admittancecalculation=R5

81.051 ft2 F

watt

Floor

1. Concrete blocks 5cmthick (over radiantpanels of low R)

L6

0.164 ft k6

0.288 watt

ft F

6137.342

lb

ft3


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2. Insulation & plywood Rins 69.75 ft2

F

watt

c6

0.191 Btu

lb F

3. Air film (horizontalheat flow upward)

ho 0.48 watt

ft2 F

R6

+++Rins

L6

k6

1

ho

1

h6

u6

1

R6

L6

k6

1

h6

=R6

74.486 ft2 F

watt

Calculation of Wall AdmittancesThis is not required for ceiling heating, but admittances give us an idea about the dynamic thermalproperties of the space.

The self-admittance and the transfer admittance will be calculated for each wall, considering thethermal capacity of the room interior layer. Note that the steady-state value of the admittance is equal tothe wall conductance. We will calculate admittances to the interior surface and to the room airpoint.The analysis will be performed for the mean term and three harmonics of the weather inputs and heatsources.

Admittances:

Steady state admittance tointerior surface is equal to wallU-value (excluding interior film);first subscript indicatesfrequency, second subscriptindicates surface number.

Ys,0 i

Ai

Ri

1

hi

Yt,0 i

Ys,0 i


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...admittances fromoutside to room air(steady state)

Y ,0 i

Ai

Ri

Yta ,0 i Y ,0 i

n , 1 2 3 j 1 ,n i

j

2 n

ki

i

ci

86400 sec

Uii

Ai

hi

Uoi

ho Ai...interior and exterior surfaceconductances

Uwiw

+

Awiw

Rw

Adiw

Rd...conductance of double-glazedwindows and doors;

Ys,n i

Ai

+ui

ki

,n i

tanh ,n i

Li

+

ui

ki

,n i

tanh ,n i

Li

1

Yt,n i

Ai

+

cosh ,n i

Li

ui

sinh ,n i

Li

ki

,n i

Y,n i

Ys ,n i Uii

+Ys,n i

Uii

Yta,n i

Yt,n i

Uii

+Ys,n i

Uii

Wall admittances fromoutside to inside air.


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Zone Admittance Yz (from room temperature node):

n , 0 1 3 Yzn

++Uinf iw

Uwiw

i

Y,n i

=||Yz

n||

135.598579.615830.420960.147

watt

FNote that Yz0 is simply the total U-value of the house.

The magnitude |Yz1| for a frequency of one cycle per

day gives us an indication of the daily dynamics of thespace; the higher its value the smaller the room dailytemperature swing.

Outside Temperature

The outside temperature for a day is modeled by a Fourier series based on NTo+1 values that are aninput to the array below. If more detail is required, NTo may be increased. Then, the the Fourier seriesmay be used to generate intermediate values as required by the time step of a finite difference model.

NTo 7 it , 0 1 NTo ...time index tit

it 3 hr ...time

n , 0 1 3 ...harmonics wn

2 n

24 hrj 1

To

10.4

0.43.26.8

12.219.417.615.8

F Tonn

it

To

it

exp j wn

tit

+NTo 1

...Fourier harmoniccoefficients

=Ton0

10.625 F mean daily temperature


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Solar Radiation (approximate model)

Solar radiation transmitted through the windows may be modelled by a half-sinusoid from sunrise tosunset time ts. For initial analysis of radiant heating, the solar radiation may be set to a low value(50 watts per sq.m. of window).

Redefine time array: it , 0 1 23 tit

it hr

Let ts 5 hr ...(time from solar noon to sunset)

Smax 4.645watt

ft2...peak solar radiation (at noon)(assume minimal level forcloudy day)

fit

Smax cos

tit

12 hr

2 ts

S

itif

,,>f

it0.0

watt

ft2f

it0.0

watt

ft2

S(t) may be modelled with a discrete Fourier series as follows:

Snn

it

S

it

exp j wn

tit

24

iw

Awiw

we multiply by total windowarea to determine total solarradiation.

We will assume that the fraction of this radiation absorbed by each interior surface is proportional toits area:

is , 1 2 6 Atot is

Ais

Qr,n i

Snn

Ai

Atot


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Finite Difference Model

The general form of the explicit finite difference formulation corresponding to node i and time interval p,is:

T ,i +p 1 +

t

Ci

+qi

j

T ,j p T ,i p

R ,i j

T ,i p

where C is capacitance, j represents all nodes connected to node i, and q is a heat source such asauxiliary heating or solar radiation.

Critical time step:

tcritical min

Ci

j

1

R,i j

for all nodes i.(the selected time step should belower to ensure numericalstability)

The thermal network is shown below.The floor is discretized into two layers (one thermal capacitanceand two resistances for each layer). The unheated surfaces are represented by node 6. S representssolar radiation transmitted into the room and absorbed by thesurfaces (a low value is assumed). Resistance R1o represents heat loss by infiltration and through thewindows, doors and vertical walls (their thermal capacity is considered negligible as compared to thefloor).


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15/21


R1o 1

+Uinf ii

++

Awii

Rw

Adii

Rd

Aii

Rii

...resistance of unheated surfaces(apart from floor) and infiltration

R12

1

A6 h6 h6r 4

watt

ft2 F...radiation heat transfer coefficientbetween floor and ceiling

R26 1

h6r A6

R56

L5

k5

A5

2...this represents the thermal resistanceof half of the ceiling panels

R161

A5

h5

R5o +R56

2

1

A5

u5

Stability Test to Select Time Step

The time step Dt should

be lower than theminimum of the threevalues in the vector TS

TS

C3

+1

R23

1

R34

C4

+1

Ro

1

R34

C5

+1

R56

1

R5o

tcritical min TS =tcritical 215.390 sec t 200 sec


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16/21


The simulation will be performed for two days (periodic). Thus, weather data have to be generated for

NT times as follows:

NT 86400sec

t2 =NT 864 ...number of time steps for two days

p , 0 1 NT tp

p t ...times at which simulation is to beperformed.

n1 , 1 2 3

Top

+Ton0

2 n1

Re Tonn1

exp j wn1

tp

Ambient temperature

2

46

8

10

12

14

16

18

-2

0

20

10 15 20 25 30 35 40 450 5 50

Top F

tp hr


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Stp

+Sn0

2 n1

Re Snn1

exp j wn1

tp

Solar radiation (total in zone)

100

200

300

400

500

600

700

800

900

110

-100

0

1.110

10 15 20 25 30 35 40 450 5 50

Stp watt

tp hr

Tbp 60.8 F...basement temperature


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18/21


Initial conditions must first be assummed. Also, the size of the heating system qmax has to bedecided together with the proportional control constant Kp. A good estimate for qaux can be

determined by multiplying the zone conductance Yz0 with the maximum indoor - outdoor temperaturedifferential and increasing the result by 50%. Fo example:

qmax ||Yz

0||

30 C 1.5 =qmax 1.098 104 watt

Initial estimates of temperatures:

T,1 0

T,2 0

T,3 0

T,4 0

T,5 0

T,6 0

69.875.278.884.262.664.4

F

qaux0

0 watt Tsp 71.6 F ...setpoint can vary withtime of day if desired.

Kp 2780watt

F...proportional control constant( a good value is qmax/2)


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qaux+p 1

T,1 +p 1

T,2 +p 1

T,3 +p 1

T,4 +p 1

T ,5 +p 1

T,6 +p 1

if ,,>Kp Tsp T ,1 p qmax qmax Kp Tsp T ,1 p

> Tsp T ,1 p

1.0 C

+++T ,2 p

R12

Top

R1o

T ,6 p

R160.7 St

p

++1

R12

1

R1o

1

R16

++

T,3 p

R23

T,6 p

R26

T,1 p

R12

++1

R23

1

R26

1

R12

+t

C3

+

T,4 p

T,3 p

R34

T,2 p

T,3 p

R23

T

,3 p

+

t

C4

+

Tbp

T,4 p

Ro

T,3 p

T,4 p

R34

T ,4 p

+t

C5

++

T,6 p

T,5 p

R56

Top

T,5 p

R5oqaux

p

T

,5 p

+++

T,2 p

R26

T,5 p

R56

T,1 p

R16

0.3 Stp

++1

R56

1

R26

1

R16


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Results for second day (first day is affected by assumed initial conditions):

Heating profile Temperatures

210

310

410

510

610

710

810

910

110

0

110

1.110

2 9 31.5 34 36.5 39 41.5 4 4 4 6.524 26.5 49

qauxp watt

tp hr

8.5

17

25.5

34

42.5

51

59.5

68

76.5

-8.5

0

85

29 31.5 34 36.5 39 41.5 44 46.524 26.5 49

T,1 p F

Top F

T,2 p F

T,5 p F

T,6 p F

tp hr

Note: T1 is room air temperature


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Heating energy consumption by

numerical integration for second day:v ,

NT

2+

NT

21 NT 1

Qh v

+++qauxv

||qaux

v||

qaux+v 1

||qaux

+v 1||

4 t

+v 1t

v

=Qh 5.633 105 Btu

A more accurate result may be obtained if another capacitance is added for the walls and the radiantexchanges are modelled more accurately (based on radiation view factors).

References

1. Athienitis, A.K., 1994, "Numerical model for a floor heating system", ASHRAE Transactions, Vol.100, Pt. 1, pp. 1024-1030 .

2. ASHRAE, 1997, Handbook-Fundamentals, Atlanta, GA.

User Notices

Equations and numeric solutions presented in this Mathcad worksheet are applicable to thespecific example, boundary condition or case presented in the book. Although a reasonable effortwas made to generalize these equations, changing variables such as loads, geometries andspans, materials and other input parameters beyond the intended range may make someequations no longer applicable. Modify the equations as appropriate if your parameters fall

outside of the intended range.For this Mathcad worksheet, the global variable defining the beginning index identifier for vectorsand arrays, ORIGIN, is set as specified in the beginning of the worksheet, to either 1 or 0. IfORIGIN is set to 1 and you copy any of the formulae from this worksheet into your own, you needto ensure that your worksheet is using the same ORIGIN.Engineering and construction code values shown in US Customary units are converted fromoriginal values in Metric units. They are NOT obtained from US codes unless specified.


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