Question 11.14: Compressed natural gas (CNG) is essentially methane (CH4). C...
Compressed natural gas(CNG) is essentially methane (\text{CH}_4) .CNG is currently being used as a replacement fuel for gasoline in some automobiles. This requires replacing the automobile’s gasoline tank with a high- ressure 0.100 \text{m}^3 cylinder filled with CNG. Under normal conditions, the tank pressure is no more than 20.0 \text{MPa} when the tank is filled with a maximum of 15.6 \text{kg} of CNG. However, the worst case condition would be if the automobile were consumed by fire and the tank temperature reached 1000.°\text{C}. Using the compressibility charts, determine the maximum pressure in the CNG tank at this worst case temperature.
First, draw a sketch of the system (Figure 11.8).
The unknown is the maximum pressure in the CNG tank. From Table C.12b, we find the critical state properties of methane to be
T_c = 191.1 \text{K} and p_c = 4.64 \text{MPa}
Since this is a closed fixed volume system,
v_1 = v_2 = v = \sout{V}/\text{m} = 0.100 \text{m}^3 /15.6 \text{kg} = 6.40 × 10^{−3} \text{m}^3\text{/kg}
Table C.13b, gives the gas constant for methane as R = 0.518 \text{kJ/kg.K} , then
v′_R = v/v′_c = vp_c/RT_c = (6.40 × 10^{−3} \text{m}^3\text{/kg})(4640 \text{kPa})/[(0.518 \text{kJ/kg.K})(191.1 \text{K})] = 0.300
and
T_R = T/T_c = (1000. +273.15)/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)_\text{worst case} = p_Rp_c = 32.0(4.64) = 148 \text{MPa}.
