need help please thank you A rigid tank contains 10 kg ofa 2-phase mixture of H2

Question

need help please thank you

A rigid tank contains 10 kg ofa 2-phase mixture of H20 at 400 kPa, with a quality of 80%. The top of the tank is connected to a discharge line through a valve. The valve is now opened and water vapor from the top of the tank is allowed to flow out. In order to maintain a constant system pressure of 400 kPa during this process, heat is transferred with the surroundings and electrical work is done. After 100 minutes, it is observed that all the H20 in the tank has become saturated vapor, at which time the valve is closed and the process ends. The voltage and current associated with the electrical work are V 3110 volts, IE 7 amps. Note: The magnitude of the electrical work is found by the equation: VI 4t A) Calculate the mass of H20 that exits the tank during this process B) Calculate the total amount of heat transfer during this process.

Explanation / Answer

part (A)

since it's a rigid tank volume is constant

initial volume=m[(1-x)*vf+x*vg)

where m=mass,vf,vg=specific volume of saturated water,liquid respectively,x= quality;

m=10kg,vf @400Kpa saturated =0.001084m^3/Kg

vg @400Kpa saturated=0.46242 m^3/Kg,x=80%=0.8

so initial volume =10[(0.2*(0.001084)+0.8*(0.46242)]=3.701m^3

in final state x=100% because only vapour is left

and mass equal to mass left let it be m-x

where x is mass exited tank

so final volume=(m-x)[(1-x)*vf+x*vg)

final volume=(10-x)*(0.46242)m^3

equte initial and final volume

3.701=(10-x)*(0.46242)

so,x=1.996Kg

mass left tank = 1.996 kg

part (B)

apply conservation of energy for control volume

W electric+ Q heat transfer= change in enthalphy of system_(1)

W electric = VIdt=110*7*6000=4620Kj(since 100 minutes =6000sec)

enthaphy of system=m[(1-x)*hf+x*hg]

m=initial mass=10kg,x=80%=0.8,hf @ 400Kpa saturated=604.66Kj/Kg

hg@ 400Kpa saturated= 2738.1Kj/Kg

initial enthalphy=10*[(0.2)*(604.66)+(0.8)*(2738.1)]=23114.12Kj

final mass = m-x=8.004

final enthalphy=8.004*(2738.1)=21915.7524Kj

change in enthaphy=(21915.7524-23114.12)Kj=-1198.3676Kj

so substitute in (1)

heta transfer=change in enthaphy - W electric=-1198.3676-4620=5818.3676Kj

so heat transfer= 5818.3676Kj from tank to surroundings

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