Nitrogen escapes from a tank having a volume of 0.5 m throug a small hole of 1 m

Question

Nitrogen escapes from a tank having a volume of 0.5 m throug a small hole of 1 mm 1n area. The escape of nitrogen is so slow that the temperature in the tank remains constant at 300 K. The initial pressure of the nitrogen is 400 kPa. The mass flow rate of nitrogen through the hole is given by 0.75pA(RTIA. where p is the pressure in the tank, A is the area of the hole, R is the gas constant and Tis the temperature in the tank. Determine the time required The gas constant for nitrogen is 296.8 J/kgK for the pressure in the tank to reduce to 100 kPa.

Explanation / Answer

" $ is row" dm/dt=d/dt integ( $dV) + integ( $(v.n)dA) the second term in the right hand side is m dot = 0.75pA/sqrt(RT) and for the first term in the right hand side volume is constant so, 0= d$/dt . V + ( 0.75pA/sqrt(RT) ) also from pV=mRT ----> $=p/RT and hence d$= dp/RT 0= dp/dt . V/RT + ( 0.75pA/sqrt(RT) ) arrenge it to the following form: -dp/p=(0.75Asqrt(RT)/V)dt "integrate"-----> ln(p1/p2)=(0.75Asqrt(RT)/V)t t=(V/0.75Asqrt(RT)).ln(p1/p2)=3097.2s

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