Question

Liquefied natural gas (LNG) is transported in very large tankers, stored as liquid in equilibrium with its vapor at approximately atmospheric pressure. If LNG is essentially pure methane, the storage temperature then is about 111.4 K, the normal boiling point of methane. The enormous amount of cold liquid can in principle serve as a heat sink for an onboard heat engine. Energy discarded to the LNG serves for its vaporization. If the heat source is ambient air at 300 K, and if the efficiency of a heat engine is 61% of its Carnot value, estimate the vaporization rate in moles vaporized per kJ of power output. For methane,

Answer 1

0.2 mol/kJ

Step-by-step explanation:

Methane is stored at a temperature of 111.4K,
`T_c`. The heat source to vapourization of methane is ambient air which is at 300 K.
`T_H`

Estimate the vaporization rate at the efficiency of heat engine 60% of its carnot value.

Calculate the vaporization rate from the given data by relation shown below:

`Vaporization rate = (Q_c)/([(\delta H_n^(lv))/(W)]) ..........(1)`

here,

`Q_c` is the heat at temperature
`T_c, \delta H_n^(lv)` is the phase transition enthalpy of methane and W is the work

Calculate
`Q_c` from the equation shown below:

`Q_c = Q_n (1 - \eta_(HE)) .............(2)`

where Q_n is the heat at temperature of
`T_n` and
`\eta_(HE)` is the efficiency of heat engine

calculate
`Q_H` from the relation shown below:

`Q_H = (W)/(\eta_(HE)) ..........(3)`

calculate the heat engine efficiency from the given carnot engine efficiency as shown below

`\eta_(HE) = 0.6 * \eta_(carnot) ............(4)`

here,
`\eta_(carnot)` is the carnot engine efficiency

`\eta{carnot} = 1 - (T_c)/(T_H)`

substituting the values of temperature, we have

`= 1 - (111.4K)/(300K)\\= 0.629\\`

substitute values of
`\eta_(carnot)` in equation 4, we get

`\eta_(HE) =0.6 * \eta_(carnot)\\ = 0.6 * 0.629\\ = 0.3774\\\\`

check the attached file for additional solution