mechanical lock.

$3\mathrm{.}2.$ Many very large liquefied natural gas $($LNG$)$ storage tanks have been built or are under

construction. The LNG is predominantly liquid methane with a boiling point near $111K$

at $1$ bar. To avoid excessive loss, the tanks are very well insulated. In the construction of

such tanks, to encourage the contractor to do the best job, there is normally a clause written

into the contract which awards the builder a bonus if the heat leak into the tank is below

some agreed value $-$ but there is also a penalty clause if this specified heat leak is exceeded.

Since such penalty $($or bonus$)$ values are large, it is crucial to specify a detailed testing

procedure to "prove" the heat leak after the tank has been built and filled with LNG$.$

Normally in such a proof test, the filled tank is allowed to attain an equilibrium state with

the internal pressure held constant. Ambient conditions should not vary greatly during

this period. At the end of this pretest period, the tank and liquid are assumed to be in

thermal equilibrium at the existing tank pressure. The actual proof test then consists of

measuring the boil$-$off vapor over a period of several days while keeping the tank pressure

equal to the pretest value.

Consider a real test. The LNG tank contains $40,000m^3$ of pure liquid methane. The tank

pressure is $\frac{1\mathrm{.}044}{b}ar.$ Over the test period, the measured boil$-$off rate was $4\mathrm{.}267{10}^4$

$mo\frac{l}{h}.$ Calculate as accurately as you can the heat leak into the tank $(\frac{J}{h}).$ In this particular

case, the contract specified a $$500,000$ penalty clause if the heat leak exceeded $0\mathrm{.}35G\frac{J}{h}$

$(0\mathrm{.}35{10}^9\frac{J}{h}).$ A bonus of the same amount was to be awarded to the contractor if the

heat leak was less than $0\mathrm{.}35G\frac{J}{h}.$

Data $($from the National Bureau of Standards Report NBSIR $73-342)$; For methane at

$112K,1\mathrm{.}044$ bar

$V^V=8\mathrm{.}6036{10}^{-3}\frac{m^3}{m}ol,V^L=0\mathrm{.}0380{10}^{-3}\frac{m^3}{m}ol$