Modern Engineering Thermodynamics

390 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.14 (Continued )
Since this is a closed fixed volume system,
v 1 = v 2 = v = V/m = 0:100 m 3 /15:6kg = 6:40 × 10 − 3 m 3 /kg
Table C.13b, gives the gas constant for methane as R = 0.518 kJ/kg·K, then
and
v′ R = v/v′ c = vp c /RT c = ð6:40 × 10 − 3 m 3 /kgÞð4640 kPaÞ/½ð0:518 kJ/kg .KÞð191:1KÞŠ = 0:300
T R = T/T c = ð1000: + 273:15Þ/191:1 = 6:66
Using these values of v′ R and T R in Figure 11.7, we find that p R = p 2 /p c ≈ 32.0, and the worst case pressure is (p 2 ) worst case =
p R p c = 32.0(4.64) = 148 MPa.
Exercises
40. Determine the final (worst case) pressure in the CNG tank in Example 11.14 if the tank temperature reached only 500.°C.
Answer: (p 2 ) worst case = 90. MPa.
41. If we decrease the size of the CNG tank in Example 11.14 from 0.100 m 3 to 0.0500 m 3 and decrease its temperature in the
fire from 1000.°C to 200.°C, determine the final (worst case) pressure in the tank. Answer: (p 2 ) worst case = 8.00 MPa.
Compressibility factor data can be used to estimate the specific enthalpy and entropy pressure dependence of
substances as follows. Integrating the specific enthalpy total differential given in Eq. (11.22) from a reference
state at p o , v o , T o , and h o to any other state at p, v, T, and h gives
Z T
h = h o + c p dT +
T o
Then, arbitrarily choosing h o , T o , and p o to be zero gives
h =
Z T
0
= h* +
0
Z p
v − T ∂v
dp
∂T p
p o
Z p
c p dT + v − T ∂v
0 ∂T p
Z p
v − T ∂v
dp
∂T p
where h* is the ideal gas specific enthalpy defined earlier in the discussion of the gas tables. From Eq. (11.39),
we can write
so that
Then,
and
∂v
∂T p
v − T ∂v
∂T p
v = ZRT/p
dp
= ZR/p + ðRT/pÞð∂Z/∂TÞ p
(11.40)
= ZRT/p − ZRT/p − ðRT 2 /p
= −ðRT 2 /pÞð∂Z/∂TÞ p
h = h* − R
Z p
0
ðT 2 /pÞð∂Z/∂TÞ p
dp
Þð∂Z/∂T
Þ p