Modern Engineering Thermodynamics

258 CHAPTER 8: Second Law Closed System Applications
EXAMPLE 8.6 (Continued )
Here, the unknowns are (a) _Q , (b) _U , (c) _S P , and (d) an expression for _S P as a function of time. The system is the lightbulb
itself, and apparently, we do not need to know any specific system properties to solve this problem.
The answers for parts a and b can be found in Example 5.2 as
a). _Q = −100: W:
b). _U = 100: W:
The answer to part c can be obtained by the indirect method of the SRB. From Eq. (8.2), we have
_S P = _S − d Z
dQ
dt Σ T b act
Since we are given that the surface temperature of the bulb is isothermal,
Z
d dQ
= 1 Z
d Q
dQ =
_ dt Σ T b T b dt Σ T b
and
Q _S P = _S − _ T b
In part a, we have steady state operation so _S = 0, then the answer to part c is
Q _S P = − _ =
100: W
= 0:261 W/K
T b ð110: + 273:15 KÞ The solution to part d is obtained by recognizing that. in part b. The bulb is insulated, so _Q = 0, then
_S P = _S
The surface and internal temperatures of the bulb are not constant here. If we recognize that most of the mass of the bulb is
made up of incompressible material (glass and tungsten wire), then we can write
s − s ref = c ln ðT/T ref Þ = c ln ðTÞ− c ln ðT ref Þ
where s ref and T ref are values chosen at some arbitrary reference state. Then,
_S = m_s = mc
dT
= mc _T
T dt T
Similarly, we can write
mu− ð u ref
Þ = cT− ð T ref Þ
so that
_U = m _u = mc dT
= mc _T
dt
Therefore,
_S =
U _ T = _S P
The temperature in this equation is the mean temperature of the bulb and can be evaluated from the answer to part b,
where we find that _U = 100: W:
Therefore,
_T = dT
dt
= _ U
mc = constant
and integration of this equation gives
U
T =
_ mc t + T 0
where T 0 is the bulb temperature immediately before the insulation is applied. Therefore, the answer to part d is
_S P =
_U
=
_Ut
mc + T 0
mc
t + mcT 0
_U
Since _U = 100:W: and m, c, and T 0 are all constant measurable quantities, it is clear from this result that _S p slowly decays to
zero as time t goes to infinity. However, the bulb temperature increases linearly with time, so the bulb overheats and burns
out quickly.