# RainischLecture3Notes by lanyuehua

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```									Lecture #3,                          Date: 1/14/2013
Thermo. Exercises.
Example 1:
A vessel with the volume of 100 ft3 contains 1 ft3 of
saturated liquid and 99 ft3 of saturated water vapor at
14.7 psia. Heat is transferred to the vessel until it is
filled with saturated vapor.
Determine the heat transfer for this process.
1Q2   =U2– U1+m(V22 -V22)/2gc+ mg(Z2 -Z1)/gc+ 1W2
mg(Z2 - Z1)/gc = Change in potential energy
m(V22 – V22)/2gc = Change in Kinetic Energy
Since this is a closed system, we can use the
following equation, namely:
1Q2   = U2 – U1 + 1W2
No ���������������� ������������ this process, therefore:
1Q2   = U 2 – U1
From the saturation pressure steam tables we find ����f
= 0.0167 ft3/lbm, and ����g =26.80 ft3/lbm
U1 = m1 liquid u1 liq + m1 vapor u1 vapor
m1liquid = Vliq/����f = 1/0.0167 = 59.81 lbm
m1liquid = Vvap/����g = 99/26.8 = 3.69 lbm
uf = 180.0 Btu/lbm            ug = 1077.5 Btu/lbm
U1 = (59.8)( 180.0) + (3.69)( 1077.5) = 14,740 Btu
mtotal = m1liq + m1vap = 59.81 + 3.69 = 63.5 lbm
v2 = Volume/Mass = V/m = 100/63.5 =1.575 ft3/lbm
We also know that the final state is when the vessel
is fully filled with saturated vapor, thus ����2 = ����g =
1.575 ft3/lbm, and find from steam tables (saturation
pressure corresponding to ����g = 1.575 ft3/lbm) that
the final pressure is 294 psia.
Therefore: u2= 1117.0 Btu/lbm
U1 = m u2 = (63.5 lbm)(1117.0 Btu/lbm) = 70,930
Btu
1Q2   = U2 – U1 = 70,930 Btu - 14,740 Btu = 56,190
Btu
Process Heat Transfer is 56,190 Btu
Example #2
Steam at 100 psi, 400ºF, enters an insulated nozzle
with a velocity of 200 ft/sec.
It leaves at a pressure of 200 lbf/in2 and a velocity
of 2,000 ft/sec.
Determine the final temperature or quality of the
steam.
(If superheated, temperature is the significant
property. If saturated quality is the significant
property).

 No work crossing the boundary of the system.
inWexit = 0

 Since the nozzle is insulated it is assumed that
there is no heat transfer. Q = 0
 No significant change in elevation between inlet
and exit of the nozzle, the change in potential
energy is negligible.
The energy balance equation becomes:
hi + V2i/2gc = he + V2e/2gc
he = hi + V2i/2gc - V2e/2gc
At inlet steam superheated, since 400ºF is higher
than saturation temperature for 100 psi.
Tsat(100 psi) = 327.8ºF hi=1227.6 Btu/lbm
he = 1227.6 Btu/lbm +
+ (200)2/2 (32.2lbm-ft/lbf-sec2)(778 ft-lbf/Btu)
- (20002/2 (32.2lbm-ft/lbf-sec2)(778 ft-lbf/Btu)
he = 1227.6 + 0.80 – 79.8 = 1148.4 Btu/lbm
he = 1148.4 Btu/lbm,    P = 20 psi
h = hf + x(hfg)
1148.4 = 196.16 + x(960.1)
x= 99.2 % quality

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