
Solving for t
1
gives
t
1
¼ t ln
(T
0
þ RQ) T(0)
(
T
0
þ RQ) (T
d
þ D)
¼ 0:24 ln
(50 þ 28:8) 50
(50 þ 28:8) (75 þ3)
¼ 0:86 h (2:155)
From Equation 2.153 with T(0) ¼508F, the temperature response is
T(t) ¼ 50e
t=0:24
þ 78:8(1 e
t=0:24
), 0 t 0:86 (2:156)
(d) The furnace shuts off when the temperature reaches T
d
þD ¼788F and the subsequent cooling
from 788FtoT
d
D ¼728F follows the step response in Equation 2.153 with Q ¼0 and T(0) ¼
T
d
þD ¼788F. Thus,
T(t) ¼ (T
d
þ D)e
(tt
1
)=t
þ T
0
[1 e
(tt
1
)=t
], t
1
t t
2
(2:157)
¼ 78e
(t0:86)=0:24
þ 50[1 e
(t0:86)=0:24
], 0:86 t t
2
(2:158)
where t
2
is the time when the building temperature is T
d
D 728F. Note the (