Department of Civil & Environmental Engineering

MICHIGAN STATE UNIVERSITY

DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING

HOMEWORK SUBMISSION

SUBMITTED BY:

Monther Dwaikat

SUBMITTED TO: Prof. V. Kodur

HOMEWORK TITLE: HW. (2)

CE 891-605 Problem 1 Given:

a) H c  16 MJ/kg M = 200 kg Qp = 9.0 MW Medium t2-fire. b) 2 minutes after the first fire started M = 15 kg 2 Fast t -fire assume the fuel is wood with H c  16 MJ/kg

Solution: a) Calculate the total energy E  MH c  16* 200  3200 MJ Medium fire k = 300 s/MW0.5

Calculate the time to reach the peak heat release

t1  k Q p  300* 9  900 sec.

Energy released in time t1 t Q 900*9 E  1 p   2700MJ 1 3 3 E1 < E steady state burning occurs Energy release rate in steady state burning E2  E  E1  3200  2700  500 MJ Duration of the steady state burning t2  E2 / Q p  500 / 9  55.56 sec The total time of burning is t  t1  t2  900  55.6  955.6sec t 3002 MW ,t  900sec  Q1(t)    9MW ,900sec  t  955.6sec Where t is the time since the onset of the first fire.

b) Calculate the total energy Assume the calorific value of the ceiling material to be 16 MJ/kg E  MH c  15*16  240 MJ Fast fire k = 150 s/MW0.5 Assume Qp = 9.0 MW for the second fire. Calculate the time to reach the peak heat release

t1  k Q p  150* 9  450 sec.

Energy released in time t1 t Q 450*9 E  1 p   1350 MJ 1 3 3 E1 > E steady state burning does not occur Time for all fuel burn

1 1 1 2 3 2 3 tm  3Ek   3* 240*150   253.03sec Heat release rate at that time is 2 2  tm   253.03  Qm        2.846 MW  k   150  2  t 120  Q2 (t)    MW ,  150  120sec (2min)  t  373.03sec (2min 253.03sec) Where t is the time from the onset of the first fire

c) Q(t)  Q1(t)  Q2 (t) t 3002 ,t  120sec    t 3002  t 120 1502 ,120sec  t  373.03sec Q(t)   MW t 3002 ,373.03sec  t  900sec   9 ,900sec  t  955.6sec  The obtained heat release rates for the two fires and their combination are shown in the following figure

) 8 W 7 M (

t 6 a r

e 5 s a e

l 4 e r

t 3 a e

H 2 1 0 0 200 400 600 800 1000 1200 Time (sec)

The heat release rates for the two fires & their combination

2 Problem 2 Given: a) 4 layers of Gypsum Wall Board each of 15.9 mm thickness Tf = 600ºC Tun = 150ºC k = 0.48W/m.K 2 b) q˙   4 kW/m Tun = 150ºC

Solution: a) Temperature difference = 600-150 = 450ºC = 450 K Wall thickness = 4*15.9 = 63.6 mm Temperature gradient = 450/0.0636 = 7075.47 K/m dT q˙   k  0.48 7075.47  3396.23 W/m2 = 3.396 kW/m 2 dx

b) Temperature difference = (T-150) K Wall thickness = 4*15.9 = 63.6 mm Temperature gradient = (T-150)/0.0636 K/m dT T 150 q˙   k  0.48  4000 W/m2 dx 0.0636 The temperature of the fire exposed side will be T = 680 ºC

Problem 3 Given: ASTM E119 fire for 30 min  = 0.9  = 5.67x10-8W/m2K4 Maximum radiant flux = q˙   12kW / m2

Solution: The fire temperature after 30 min Assume ambient temperature (T0) = 20ºC th = 30 min = 0.5 hour T  T  7501 e3.79553 th  170.41 t 0   h T  20  7501 e3.79553 0.5  170.41 0.5  839.27 ºC   Assume the following 1) One opening in the room where the fire started. 2) The opening to be in the wall opposite to the other building 2 3) The opening area (A1) to be 2*2 = 4m . 4) The temperature in the opening is equal to the fire temperature.

Hence, the temperature in the opening will be T  839.27 ºC = 1112.27 K. Radiant heat flux q˙   T 4    0.9 5.67 108  1112.274  12000 W Hence, the configuration factor will be

3   0.15364 Assume the spacing between the two buildings to be large relative to the emitting surface (opening) A 4   1   0.15364  r 2  r 2 Hence, the spacing between the two buildings must be at least 2.88 m

Problem 4 Given: Room dimensions 6m x 6m x 2.5 m high Two openings; 1.5 m high x 2 m wide and 1.5 m high x 4m wide. Fast t2-fire.

Solution: Using Thomas’ Flashover Criterion 2 Area of internal surfaces At  2 (6 6  6 2.5  6 2.5)  132 m 2 Area of ventilation Av  1.5 2 1.5 4  9 m Height of ventilation H v  1.5 m

The flashover heat release rate Q fo  0.0078At  0.378Av H v

Q fo  0.0078132  0.378 9 1.5  5.2 MW

Fast t2-fire, k = 150 s/MW0.5

t fo  k Q fo  150 * 5.2  342.1sec = 5.7 min

Problem 5 Given: Room dimensions 6m x 6m x 2.5 m high Two wall openings; 1.5 m high x 2 m wide and 1.5 m high x 4m wide, 0.5 m above finished floor. One ceiling opening having 3 m2 area. Fuel is moist wood (mc = 12%) H c for dry wood =19MJ/kg Fuel energy density = 800 MJ/m2 floor area.

Solution: 10012 a) and b) Moisture content m   10.7% c 100 12 H c,n  H c (1 0.01mc )  0.025mc

H c,n  19 (1 0.0110.7)  0.02510.7  16.7 MJ/kg

Kawagoe’s approach 2 Area of ventilation, Av  1.5 2 1.5 4  9 m

4 Height of ventilation, H v  1.5 m Height above window mid-height, h = 1.25 A H  A H  2.3A h Modified opening parameter  v v fict v v h A H  9 1.5  2.3 3 1.25  18.74 2.5  v v fict m

Rate of burning, m˙  0.092Av H v fict  0.09218.74  1.724 kg/sec

Ventilation controlled HRR, Qvent  m˙ H c,n  1.72416.7  28.79 MW

Total energy, E f  e f A f  800 36  28800 MJ

E f 28800 Duration of burning, tb    1000.35sec  16.67 min Qvent 28.79

Law’s approach Assume the horizontal area of ventilation is neglected, because Law’s formula does not include opening in the ceiling. 2 Area of ventilation, Av  1.5 2 1.5 4  9 m Height of ventilation, H v  1.5 m 2 Area of internal surfaces At  2 (6 6  6 2.5  6 2.5)  132 m At  Av 132  9 Opening factor,     11.16 m-0.5 Av H v 9 1.5 Room width = 6 m Room width = 6 m W Rate of burning, m˙  0.18A H 1 e0.036  v v D 6 m˙  0.18 9 1.5 1 e0.03611.16  0.656 kg/sec 6 Ventilation controlled HRR, Qvent  m˙ H c,n  0.65616.7  10.96 MW

Total energy, E f  e f A f  800 36  28800 MJ

E f 28800 Duration of burning, tb    2627.7sec  43.8min Qvent 10.96 c) Law’s approach Assume the horizontal area of ventilation is neglected, because Law’s formula does not include opening in the ceiling. 2 Area of ventilation, Av  1.5 2 1.5 4  9 m Height of ventilation, H v  1.5 m 2 Area of internal surfaces At  2 (6 6  6 2.5  6 2.5)  132 m

At  Av 132  9 Opening factor,     11.16 m-0.5 Av H v 9 1.5 0.1 Maximum temperature, Tmax  60001 e / 

5 0.111.16 Tmax  60001 e / 11.16  1207.7 ºC Check reduction facto for fuel Total energy, E f  e f A f  800 36  28800 MJ 10012 Moisture content, m   10.7% c 100 12 H c,n  H c (1 0.01mc )  0.025mc

H c,n  19 (1 0.0110.7)  0.02510.7  16.7 MJ/kg

E f 28800 Fire load, L    1724.55 kg H c 16.7 L 1724.55 Temperature parameter,     51.83kg/m2 Av At  Av  9 132  9 Reduced maximum temperature, 0.05 0.0551.83 T  Tmax 1 e  1207.71 e  1117.2 ºC

Swedish approach 2 Area of ventilation, Av  1.5 2 1.5 4  9 m Height of ventilation, H v  1.5 m Height above window mid-height, h = 1.25 A H  A H  2.3A h Modified opening parameter  v v fict v v h A H  9 1.5  2.3 3 1.25  18.74 2.5  v v fict m 2 Area of internal surfaces At  2 (6 6  6 2.5  6 2.5)  132 m A H  v v fict 18.74 0.5 Ventilation factor, Fv    0.142 m At 132 Total energy, E f  e f A f  800 36  28800 MJ E f 28800 2 Fuel load, et    218.2 MJ/m At 132 Interpolation and extrapolation:

0.5 At Fv  0.08 m 2 At et  200 MJ/m Tmax  1000 ºC 2 At et  400 MJ/m Tmax  1050 ºC 2 Hence, for et  218.2 MJ/m 1050 1000 T  1000  218.2  200  1004.5ºC max 400  200

0.5 At Fv  0.12 m

6 2 At et  150 MJ/m Tmax  1000 ºC 2 At et  225 MJ/m Tmax  1030 ºC 2 Hence, for et  218.2 MJ/m 1030 1000 T  1000  218.2 150  1027.3 ºC max 225 150 0.5 Therefore, at Fv  0.142 m , and using extrapolation 1027.3 1004.5 T  1004.5  0.142  0.08  1039.8 ºC. max 0.12  0.08 Problem 6 Given: Room dimensions, 6m x 6m x 2.5 m high One opening, 1.5 m high x 2 m wide. 2 Fire load ef = 800 MJ/m The conductivity, density and specific heat of concrete are 1.6 W/m.K, 2200 kg/m3, and 840 J/kg.K, respectively. The conductivity, density and specific heat of gypsum are 0.48 W/m.K, 1410 kg/m3, and 880 J/kg.K, receptively.

Solution: a) Eurocode formula Thermal inertia of concrete 6 2 4 2  kc p  22001.6840  2.957 10 W .s/m .K Thermal inertia of gypsum 6 2 4 2  kc p  1410 0.48880  0.59610 W .s/m .K Area of concrete = 6*6 = 36 m2 Area of gypsum board = 6*6 +2*(6*2.5 + 6*2.5) = 96 m2 Weighted average thermal inertia 36 2.957106  96 0.596106  kc    1.24106 W2.s/m4.K2 p av 36  96 Thermal inertia factor  kc  1.24106  1113.55 W.s0.5/m2.K (medium)  p av

kb = 0.055 Window height, H v  1.5 m Window width, B = 2 m. 2 Window area, Av  21.5  3 m Horizontal ventilation area, Ah  0 2 Floor area, Af  6 6  36 m

Av 3 v    0.0833 Af 36

7 Ah 0  h    0 Af 36 2 2 bv  12.5110v v  12.5110 0.0833  0.0833  22.83 Height of room, H r  2.5 m 0.3  4   6.0  900.4 v  Ventilation factor, w    0.62    0.5  H r   1 bvah  0.3  6.0   900.4  0.08334  w    0.62    1.984 m-0.3  2.5   1 0 

Equivalent fire severity, te  e f kbw  800 0.0551.984  87.28 min. b) CIB formula Thermal inertia of concrete 6 2 4 2  kc p  22001.6840  2.957 10 W .s/m .K Thermal inertia of gypsum 6 2 4 2  kc p  1410 0.48880  0.59610 W .s/m .K Area of concrete = 6*6 = 36 m2 Area of gypsum board = 6*6 +2*(6*2.5 + 6*2.5) = 96 m2 Weighted average thermal inertia 36 2.957106  96 0.596106  kc    1.24106 W2.s/m4.K2 p av 36  96 Thermal inertia factor  kc  1.24106  1113.55 W.s0.5/m2.K (medium)  p av

kc = 0.07 Window height, H v  1.5 m Window width, B = 2 m. 2 Window area, Av  21.5  3 m 2 Floor area, Af  6 6  36 m 2 Area of internal surfaces At  2 (6 6  6 2.5  6 2.5)  132 m

A f 36 Ventilation factor, w    1.635 m-0.25 Av At H v 3132 1.5

Equivalent fire severity, te  e f kbw  800 0.07 1.635  91.54 min. c) Law formula Assume wood with net calorific value, H c  16 MJ/kg. 2 Window area, Av  21.5  3 m 2 Floor area, Af  6 6  36 m

8 2 Area of internal surfaces At  2 (6 6  6 2.5  6 2.5)  132 m e A t  f f Equivalent fire severity, e 0.5 H c Av At  Av  800 36 te   91.5 min. 163132  30.5 d) The obtained results are summarized in the following table: Formula Eurocode CIB Law

te (min) 87.28 91.54 91.5

It is clear that both the CIB method and Law’s method give approximately the same fire severity for this particular problem. On the other hand, Eurocode formula gives slightly lower fire severity. Although the three methods are empirical, they result in similar equivalent fire severity within a relatively small range of time for this particular problem. This is probably because the ventilation conditions in this problem are similar to those in which the equations were obtained.