A well-insulated rigid tank contains 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized. Determine the entropy change of the steam during this process. Answer: 11.1 kJ/K

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SaurbHurbkj · Accepted Answer

Step 1&amp;nbsp;Given:&amp;nbsp;Substance: A mixture of water and vapor, in which the water is saturatedm=3kg&amp;nbsp;p1=200kPa&amp;nbsp;mf1=34m=2.25kg&amp;nbsp;Process: constant volume&amp;nbsp;x2=1&amp;nbsp;Required:△S=?&amp;nbsp;Step 2&amp;nbsp;Solution:&amp;nbsp;First we need to calculate the quality in state 1.x1=mgm=m−mf1m=3−2.253=0.25&amp;nbsp;Now from the steam TABLE A-5&amp;nbsp;p1=200kPa→vf1=0.001061m3kg&amp;nbsp;vg1=0.88578m3kg&amp;nbsp;sf1=1,5302kJkgK&amp;nbsp;sfg1=5.5968kJkgK&amp;nbsp;Using these data, we can calculate specific entropy in state 1 from the equation:s1=sf1+x1g=1.5302+0.25⋅5.5968=2.9294kJkgK&amp;nbsp;We will obtain entrophy in the state 2 from the FIGURE A-10, but to do so first we have to calculate specific volume in the state 1 which is the same as the one in state 2 because this is a constant volume process.&amp;nbsp;Equation for specific volume is:&amp;nbsp;v1=vf1+x1vfg=0.001061+0.25(0.88578−0.001061)=0.22m3kg=v2&amp;nbsp;Step 3&amp;nbsp;Now we can use Molier diagram for water (Figure A-10) and obtain specific entrophy in the state 2.&amp;nbsp;p2=1v2=10.22=4.55kgm3;x2=1⇒s2=6.65kJkgK&amp;nbsp;Finally entrophy change is calculated from:&amp;nbsp;△S=m(s2−s1)=3⋅(6.65−2.9294)&amp;nbsp;△S=11.16kJK&amp;nbsp;Result&amp;nbsp;△S=11.16kJK

A well-insulated rigid tank contains 3 kg of saturated liquid-vapor mixture of w

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