Question 8.11: A CONTINUATION OF EXAMPLE 5.7, WITH THE ADDITION SHOWN IN IT...
A CONTINUATION OF EXAMPLE 5.7, WITH THE ADDITION SHOWN IN ITALIC TYPE
A microwave antenna for a space station consists of a 0.100 \text{m} diameter rigid, hollow, steel sphere of negligible wall thickness. During its fabrication, the sphere undergoes a heat treating operation in which it is initially filled with helium at 0.140 \text{MPa} and 200.°\text{C}, then it is plunged into cold water at 15.0°\text{C} for exactly 5.00 \text{s} . The convective heat transfer coefficient of the sphere in the water is 3.50 \text{ W/(m² .K)}. Neglecting any changes in kinetic or potential energy and assuming the helium behaves as an ideal gas, determine
a. The final temperature of the helium.
b. The change in total internal energy of the helium.
c. The total entropy production in the helium.
First, draw a sketch of the system (Figure 8.11).
The unknowns here are, after 5 s have passed, (a) T_2, (b) U_2 – U_1, and (c) _1(SP)_2; the material is helium gas.
The solutions to the first two parts of this problem can be found in Example 5.7 as
a. T_2 = 32.5°\text{C}.
b. U_2 – U_1 = –0.039 \text{kJ}.
The solution to part c can be found again by the indirect method(entropy balance) as follows. From Eq. (8.1), we have
_1(SP)_2= m(s_2 − s_ι) – \int_{Σ} (\frac{\overline{d} Q}{T} )_\text{act}
We can assume that helium behaves as an ideal gas, then since v_2 = v_1 = constant, we can use Eq. (7.36) to produce
s_2 −s_1 = c_v \text{ln} \frac{T_2}{T_1} + R \text{ln} \frac{v_2}{v_1}
= [3.123 \text{ kJ/(kg.K)} ] \text{ln} \frac{32.5+273.15}{200.+273.15} + 0
= −1.37 \text{ kJ/(kg.K)}
and, from Example 5.7, we have
\dot{Q} = −hA (T_s −T_∞) =\frac{\overline{d} Q}{dt}
so
\overline{d} Q = −hA (T_s −T_∞) dt
and
\frac{\overline{d} Q}{T_b} = \frac{\overline{d} Q}{T_∞} = −hA [(T_s −T_∞)/T_∞]dt
where we have set T_b = T_∞ (i.e., we put the system boundary slightly outside the sphere itself). Also, in Example 5.7, we discover that
T_s = T_∞ + (T_1 −T_∞) \text{ exp} ( – \frac{hAt}{mc_v} )
where T_1 = T_s evaluated at t = 0. Then,
\int_{Σ} (\frac{\overline{d} Q}{T_b} )_\text{act} = −hA \int_{0}^{5s} (\frac{T_s −T_∞}{T_∞} ) dt
=mc_v (\frac{T_1 −T_∞}{T_∞} ) [\text{ exp} ( – \frac{hAt}{mc_v} )\mid^{5s}_{0} ]
Now, from Example 5.7, we have the following numerical values:
m = 7.46 × 10^{-5} \text{ kg}
c_v = 3.123 \text{ kJ/kg.RK}
T_1 = 200.°\text{ C }= 473.15 \text{ K}
T_∞ = 15.0°\text{ C } = 288.15 \text{ K}
and hA/(mc_v) = 0.472 s^{-1}. Substituting these values into the preceding integration result gives
\int_{Σ} (\frac{\overline{d} Q}{T_b} )_\text{act} = −1.35 × 10^{-4} \text{ kJ/K}
then
_1(SP)_2 = (7.46 × 10^{-5} \text{ kg})[−1.365 \text{ kJ/(kg⋅K)}] – (−1.35 × 10^{-4} \text{ kJ/K} )
= 3.32 × 10^{-5} \text{ kJ/K} = 0.0332 \text{ J/K}
