Question 8.6: A CONTINUATION OF EXAMPLE 5.2, WITH THE ADDED MATERIAL SHOW ...
CONTINUATION OF EXAMPLE 5.2, WITH THE ADDED MATERIAL SHOW IN ITALIC TYPE
An incandescent lightbulb is a simple electrical device. Using the energy rate balance and the entropy rate balance on a lightbulb, determine
a. The heat transfer rate of an illuminated 100. \text{W} incandescent lightbulb in a room.
b. The rate of change of its internal energy if this bulb were put into a small, sealed, insulated box.
c. The value of the entropy production rate for part a if the bulb has an isothermal surface temperature of 110.°\text{C}.
d. An expression for the entropy production rate as a function of time for part b.
First, draw a sketch of the system (Figure 8.6).
Here, the unknowns are (a) \dot{Q}, (b) \dot{U}, (c) \dot{S}_ P, and (d) an expression for \dot{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). \dot{Q}= −100. \text{W}.
b). \dot{U}= 100. \text{W}.
The answer to part c can be obtained by the indirect method of the SRB. From Eq. (8.2), we have
\dot{S}_P=\dot{S} -\frac{d}{dt} \int_{Σ}(\frac{\overline{d} Q}{T_b} )_\text{act}
Since we are given that the surface temperature of the bulb is isothermal,
\frac{d}{dt} \int_{Σ}(\frac{\overline{d} Q}{T_b} ) = \frac{1}{T_b}\frac{d}{dt}\int_{Σ}\overline{d} Q =\frac{\dot{Q}}{T_b}
and
\dot{S}_P=\dot{S} – \frac{\dot{Q}}{T_b}
In part a, we have steady state operation so \dot{S} = 0, then the answer to part c is
\dot{S}_P= – \frac{\dot{Q}}{T_b} = \frac{100. \text{W}}{(110.+273.15 \text{K})} = 0.261 \text{W/K}
The solution to part d is obtained by recognizing that. in part b. The bulb is insulated, so \dot{Q} = 0, then
\dot{S}_P = \dot{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_ \text{ref }= c \text{ln}(T/T_\text{ref }) = c \text{ln}(T) – c \text{ln}(T_\text{ref })
where s_ \text{ref } and T_\text{ref } are values chosen at some arbitrary reference state. Then,
\dot{S} = m\dot{s} = \frac{mc}{T} (\frac{dT}{dt} ) = \frac{mc}{T} (\dot{T})
Similarly, we can write
m(u – u_\text{ref }) = c(T – T_\text{ref })
so that
\dot{U} = m\dot{u} = mc(\frac{dT}{dt}) = mc(\dot{T})
Therefore
\dot{S} = \frac{\dot{U}}{T} = \dot{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 \dot{U} = 100. \text{W}.
Therefore,
\dot{T} = \frac{dT}{dt} = \frac{\dot{U}}{mc} = \text{constant}
and integration of this equation gives
T = \frac{\dot{U}}{mc}t +T_0
where T_0 is the bulb temperature immediately before the insulation is applied. Therefore, the answer to part d is
\dot{S}_P =\frac{\dot{U}}{\frac{\dot{U}t}{mc} + T_0} =\frac{mc}{t + \frac{mc T_0}{\dot{U}} }
Since \dot{U}= 100. \text{W}. and m, c, and T_0 are all constant measurable quantities, it is clear from this result that \dot{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.
