Modern Engineering Thermodynamics

158 CHAPTER 5: First Law Closed System Applications
EXAMPLE 5.7 (Continued )
The unknowns are, after 5 seconds have passed, (a) T 2 =?and(b)
U 2 − U 1 = ? The system is closed, and the material is the helium gas
in the sphere.
The basic equations here are the closed system energy balance (EB)
T ∞ = 15.0°C
0.100 m diameter
System boundary
1Q 2 − 1 W 2 = U 2 − U 1 + KE 2 − KE 1 + PE 2 − PE 1
Helium
Neglect
and neglecting _ KE and _ PE, the closed system energy rate balance (ERB) is
h = 3.50 W/(m 2 .K)
_Q − _W = _U
The auxiliary equations needed here are
FIGURE 5.7
Example 5.7.
1. The mechanical work mode 1 W 2 = _W = 0 (a rigid hollow sphere).
2. The convective heat transfer mode _Q = −hAðT s − T ∞ Þ (the negative sign is necessary here because the helium loses heat).
3. Assuming the helium to be an ideal gas, the internal energy can be represented as du = c v ðdTÞ and _U = m _u = mc v
_T s :
Putting these results into the energy rate balance equation gives (assuming T ∞ = constant)
or
_T s = dT s
dt
_Q = −hAðT s − T ∞ Þ = _U = mc v
_T s
= dðT s − T ∞ Þ
dt
= − hA
mc v
ðT s − T ∞ Þ
This is a first-order ordinary differential equation with the initial condition T s = T 1 when t = 0. Its solution is
T s = ðT 1 − T ∞ Þ exp − hAt
+ T ∞
mc v
where T 2 = T s at t = 5.00 s. The remaining part of the solution is given by the energy balance equation as
The auxiliary equations and calculations are
U 2 − U 1 = mc v ðT 2 − T 1 Þ
h = 3:50: W/ðm 2 ⋅KÞ
T ∞ = 15:0°F
V = π 6 ðDÞ3 = π 6 ð0:100 mÞ3 = 5:24 × 10 − 4 m 3
A = πD 2 = πð0:100 mÞ 2 = 0:0314 m 2 ðthe surface area of a sphereÞ
c v = 3:123 kJ/ðkg⋅KÞ ðfrom Table C:13bÞ
and from the ideal gas equation of state for helium, m = PV/RT, Table C.13b gives R = 2:077 kJ/ðkg⋅KÞ. Then,
and at t = 5.00 s,
m =
ð140: kN/m 2 Þð5:24 × 10 − 4 m 3 Þ
½2:077 kJ/ðkg⋅KÞŠ½ð200: + 273:15ÞKŠ = 7:46 × 10−5 kg
hA ½3:50 W/ðm 2 ⋅KÞŠð0:0314 m 2 Þ
=
mc v ð7:46 × 10 − 5 = 0:472 s−1
kgÞ½3:123 kJ/ðkg⋅KÞŠ
T 2 = ðT 1 − T ∞ Þ expð− hAt/mc v Þ + T ∞
= ½ð200: − 15:0Þ°CŠ exp½ −ð0:472 s − 1 Þð5:00 sÞŠ + 15:0°C
= 32:5°C
Then,
U 2 − U 1 = mc v ðT 2 − T 1 Þ
= ð7:46 × 10 − 5 kgÞ½3:123 kJ/ðkg⋅KÞŠ½ð32:5 − 200:Þ KŠ
= −0:039 kJ