Four kilograms of a two-phase liquid–vapor mixture of water initially at 300°C a
ID: 2991359 • Letter: F
Question
Four kilograms of a two-phase liquid–vapor mixture ofwater initially at 300°C and x1 = 0.5 undergo the two
different processes described below. In each case, the mixture
is brought from the initial state to a saturated vapor state,
while the volume remains constant. For each process,
determine the change in exergy of the water, the net amounts
of exergy transfer by work and heat, and the amount of
exergy destruction, each in kJ. Let T0 = 300 K, p0 = 1 bar,
and ignore the effects of motion and gravity. Comment on
the difference between the exergy destruction values.
(a) The process is brought about adiabatically by stirring the
mixture with a paddle wheel.
(b) The process is brought about by heat transfer from a
thermal reservoir at 610 K. The temperature of the water at
the location where the heat transfer occurs is 610 K.
Explanation / Answer
both the cases initial and final state are same:
T1=300c x1=0.5 x2=1.0
T0=300c P0=1bar
V1=Vf1+X1(Vg1-Vf1)=1.4036x10-3+0.5(0.02167-1.4036x10-3)=0.01154m3/kg
u1=uf1+X1(ug1-uf1)=1332.0+0.5(2563-1332)=1947.5 Kj/Kg
s1=sf1+X1(sg1-sf1)=3.2534+0.5(5.7045-3.2534)=4.479Kj/Kg.K
by assumption V2=V1=Vg(T2)
T2=336.8C u2=2474.2KJ/Kg S2=5.3673Kj/Kg.K
a)vol is constant:
E=m[u2-u1-T0(s2-s1)]=4(260.21)=1040.84Kj
E,w=W-pV=W as V=0 W=-m(u2-u1)=-2106.8Kj
Eq=exergy tansfered accompanying heat transfer=0
E=E,q-E,w-E,d =>E,d=-E,w-E=-(-2106.8)-1040.84=1065.96Kj=exergy destructed
b)same change of state:
hence; E=m[u2-u1-T0(s2-s1)]=4(260.21)=1040.84Kj
since work is zero=E,w=0
exergy tansfered accompanying heat transfer is evaluated at the boundry temp=Tb=610K
thus E,q=(1-To/Tb)Q
in this case m(u2-u1)=Q-W as W=0 =>Q=2106.8Kj
E,q=(1-300/610)(2106.8)=1070.67Kj
from exergy balance:
E=E,q-E,w-E,d as E,w=0
E,d=E,q-E=29.83Kj exergy destruction
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