Arhitektonska fizika 1

A R H I T E K T O N S K A F I Z I K A 1

DOKAZNICA RAUNA

Ogradna ploha: NEVENTILISANI VANJSKI ZID- PUNA OPEKAKlimatska zona : Ik = 1,20 W/m2K

R = 0, 66 m2K/W

i = 8,00 W/m2K

e = 23 W/m2K

Ri = 0,13 m2K/W

Re = 0, 04 m2K/W

SLOJEVI OGRADNE PLOHE:MATERIJALd (cm)d (m)

Produni malter2,00 0,0217000,85

Puna opeka25 0,2518000,76

Mineralna vuna80,081000,041

Produni malter2 0,0217000,85

1

k = --------------------------- 1/i + d1/1 + ... + 1/ e

1

1,20 = --------------------------------------------------------------- 0,13 + 0,0235 + 0,0329 + x/0,041 + 0,0235 + 0,04

1

1,20 = ---------------------

0,546 + X/0,024 X= 0,0118 USVOJENO: 8,0 cm zadovoljava k= 0,40 W/m2KR = d /

MATERIJALd / R

Produni malter0,02 / 0,850,0235

Puna opeka0,05 / 0,760,329

Mineralna vuna0,008/0,0411,95

Produni malter0,05 / 0,850,0235

Ri = 0,13

Re = 0,04Rk = 2,496ti = 12,821x Ri =12,821x 0,13 = 1,667

t1 = 12,821x R1 = 12,821x 0,0235 = 0,301

t2 = 12,821x R2 = 12,821x 0,329 = 4,218

t3 = 12,821x R3 = 12,821x 1,95 =25,00

t4 = 12,821x R4 = 12,821x 0,0235 = 0,301

t5 = 12,821x R4 = 12,821x 0,04 = 0,513

ZIMSKI REIM :

ti = + 20 C

te = - 12 C

ti te ti te 20 + 12 tj = ---------- x Rj --------------- = ---------------- = 12,821 C Rk Rk 2,496ti = 12,821x Ri =12,821x 0,13 = 1,667t1 = 12,821x R1 = 12,821x 0,0235 = 0,301t2 = 12,821x R2 = 12,821x 0,329 = 4,218t3 = 12,821x R3 = 12,821x 1,95 =25,00t4 = 12,821x R4 = 12,821x 0,0235 = 0,301t5 = 12,821x R4 = 12,821x 0,04 = 0,513ti = +20 C

tn ti = 20 1,667 = 18,333ti t1 = 18,333 0,301 = 18,032t1 t2 = 18,032 4,218 = 13,814t2 t3 = 13,814 25 = -11,186t3 t4 = -11,186 0,301 = -11,487t4 t5 = -11,487 0,513 = -12 CLJETNI REIM :

Vanjska temperatura : 30 C

Nadtemperatura : 50 C

te ti 30 ---------------- = -------------------- = 12,215 C Rk Re 2,496 0,04ti = 12,215 x Ri = 12,215x 0,13 = 1,588t1= 12,215x R1 = 12,215x 0,0235 = 0,287t2 = 12,215x R2 = 12,215x 0,329 = 4,019t3 = 12,215x R3 = 12,215x 1,95= 23,819t4 = 12,215x R4 = 12,215x 0,0235= 0,287ti = +20 C

tn ti = 20 + 1,588 = 21,588ti t1 = 21,588 + 0,287 = 21,875t1 t2 = 21,875 + 4,019 = 25,894t2 t3 = 25,894+ 23,819 = 49,713t3 t4 = 49,713+ 0,287 = 50 C PARODIFUZIJA :

Vanjska temperatura : 5 C

ti = 20 C

ti - tetj = ------------ x Rj Rk ti - te 15 ------------ = ------------- = 6,010 C Rk 2,496ti = 6,010 x Ri = 6,010x 0,13 = 0,781t1 = 6,010x R1 = 6,010x 0,0235 = 0,141t2 = 6,010x R2 = 6,010x 0,329 = 1,977t3 = 6,010x R3 = 6,010x 1,95 = 11,719t4 = 6,010x R4 = 6,010x 0,0235 = 0,141t5 = 6,010x R5 = 6,010x 0,04 = 0,240ti = +20 C

tn - ti = 20 0,781 = 19,219ti - t1 = 19,219 0,141 = 19,078t1 t2 = 19,078 1,977 = 17,101t2 t3 = 17,101 11,719 = 5,382t3 t4 = 5,382 0,141 = 5,241t4 t5 = 5,241 0,240 = 5,001 C FAKTOR OTPORA DIFUZIJE :

Materijalr = x dr

Produni malter1515 x 0,020,3

Puna opeka1212 x 0,253,0

Mineralna vuna11 x 0,080,08

Produni malter1515 x 0,020,3

r = 3, 68i = 60 %

e = 90 %

PRITISCI ZASIENJA :

ti = 20 pi = 2,337 kPa

t1 = 19,219 p1 = 2,223 kPa

t2 = 19,078 p2 = 2,210 kPa

t3 = 17,101 p3 = 1,937 kPa

t4 = 5,382 p4 = 0,897 kPa

t5 = 5,241 p5 = 0,890 kPa

te = 5 pe = 0,872 kPa

PARCIJALNI PRITISCI :

pi = i x pi = 0,60 x 2,337 = 1,402 kPa

pe = e x pe = 0,90 x 0,872 = 0,785 kPa

pi pe 1,402 0,785pj = pj 1 = ------------- = -------------------- = 0,168 r 3,68p1 = pi ( X x r1) = 1,402 (0,168 x 0,30) = 1,352p2 = p1 ( X x r2) = 1,352 (0,168 x 3) = 0,848p3 = p2 ( X x r3) = 0,848 (0,168 x 0,08) = 0,835p4 = p3 ( X x r4) = 0,835 (0,168 x 0,3) = 0,785STABILNOST KONSTRUKCIJE NA LJETNI REIMVanjski neventilisani zid

MaterialbSDU

Produni malter1231,76710,470,2468,902

Puna opeka1121,8569,5363,1379,536

Termoizolacija58,6864,9889,7274,988

Produni malter1231,76710,470,2466,772

Koeficient prolaza toploteb=b1==1231,767b2==1121,856b3==58,686

b4==1231,767Koeficient upijanja toplote od strane materijala S=b; S24=85*10-4b (dnevni period oscilovanja temperature)

S1=85*10-4*1231,767= 10,47S2=85*10-4*1121,856= 9,536S3=85*10-4*58,686= 0,4988S4=85*10-4*1231,767= 10,47

D - karakteristika toplinske inercije materijala [ W / m2K ]

i=8 ; e=23 D = R x S

D1 = 0,0235 x 10,47 = 0,246D2 = 0,329 x 9,536 = 3,137 D2> 1D3 = 1,95 x 0,498 = 0,973D4 = 0,0235 x 10,47 = 0,246

D = 4,602U1 = R1 x S12 + i / 1 + R1 x i = 0,0235 x 10,472 + 8 / 1 + 0,0235 x 8 = 10,576 / 1,188 = 8,902U2 = S2= 9,536U3 = R3x S3 + U2 /1+ R3 x U2 =1,95x0,4988 +9,536/1+1,95x9,536=0,511U4 = R4 x S42 + U3 / 1 + R4 x U3 = 0,0235x 10,472 + 0,511/ 1 + 0,0235 x 0,511U4 = 7,564/ 1,117 = 3,05Faktor priguenja amplitude oscilacijev = 0,9 ex S1 + i / S1+U1 x S2+U1 / S2+U2 x S3+U2 / S3+U3 x S4+U3 / S4+U4 x e+U4 / e

v = 0,9 ex 10,47 + 8/ 10,47+ 8,902 x 9,536 + 8,902 / 9,536+ 9,536 x 0,4988 + 9,536 / 0,4988 + 0,511 x 10,47 + 0,511 / 10,47 + 3,05 x 11,5 +3,05/11,5 = 0,9 ex x0,953x 0,966x 9,94x0,812x 1,265=23,274x9,402= 219,13v- min 15 za vanjske ogradne plohe v= 219,13>15 6

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