4.127 An open cooking pot containing 0.5 liter of water at 20oC, 1 bar sits on a stove burner.
Once the burner is turned on the water is gradually heated at a rate of 0.85 kW while pressure
remains constant. After a period of time, the water starts boiling and continues to do so until all
of the water has evaporated. Determine
(a) the time required for the onset of evaporation, in s.
(b) the time required for all of the water to evaporate, in s, once evaporation starts.
KNOWN: Water is heated in an open pot on a burner.
FIND: (a) The time required for the onset of evaporation, in s, and (b) the time required for all
of the water to evaporate, in s, once evaporation starts.
SCHEMATIC AND GIVEN DATA:
Part (a)
Process 1-2
State 1
Water
T1 = 20oC
p1= 1 bar
V1 = 0.5 liter
State 2
Q  0.85 kW
Water
Saturated Liquid
T2 = Tsat @ p2
p2 = p1
Part (b)
Process 2-3
State 2
Water
Saturated vapor exits at
Te = Tsat @ pe
pe = p1
State 3
e
t = t2
Water
Saturated Liquid
T2 = Tsat @ p2
p2 = p1
Q cv  0.85 kW
t = t2 + Dt
1
t = t3
The temperature-specific volume diagram below indicates the states associated with the overall
process:
p = 1 bar
T
2
99.63oC
20oC
1
e
•
•
•
v
ENGINEERING MODEL:
1. For part (a) the liquid water inside the pot defines the system during process 1-2 as shown by
the dashed line on the accompanying figure. For part (b) the control volume is defined by the
boundary of the liquid water inside the pot during process 2-3 as shown by the dashed line on
the accompanying figure.
2. For the system and control volume, kinetic and potential effects can be ignored.
3. As shown in the figure of part (b), saturated vapor exits the control volume at 1 bar.
4. Water is assumed to remain at 1 bar throughout the entire process.
5. Work associated with system boundary movement (part a) and work associated with control
volume boundary movement (part b) are not ignored.
ANALYSIS:
(a) During process 1-2, heat transfer results in a temperature increase from 20oC to 99.63oC (the
saturation temperature that corresponds to 1 bar). Since evaporation has not yet started, there is
no mass flow of water into or from the system, and the water can be analyzed as a closed system.
Since the mass is constant during process 1-2 and with assumption 2 in the engineering model,
the energy rate balance for process 1-2 reduces to
DU  Q  W
or
m(u2  u1)  Q Dt  W
1
Solving for Dt gives
Dt 
m(u2  u1)  W
Q
The mass of water can be determined from the volume of the liquid water and its corresponding
specific volume at state 1. At p1 = 1 bar and T1 = 20oC, water is compressed liquid. From Table
A-2, v1 ≈ vf1 = 0.0010018 m3/kg
V
m 1 
v1
103 m3
= 0.50 kg
1L
m3
0.0010018
kg
(0.5 L)
2
The specific internal energy at state 1 is determined from Table A-2, u1 ≈ uf1 = 83.95 kJ/kg. The
water is saturated liquid at state 2. From Table A-3 at p2 = 1 bar, u2 = uf2 = 417.36 kJ/kg.
As the liquid water is heated, its volume increases slightly. Thus, work is done by the system at
the top surface during process 1-2
2
W   pdV  mp( v2  v1)
1
From Table A-3, v2 = vf2 = 0.0010432 m3/kg. Solving for work
m3
m3 100 kPa
W  (0.50 kg)(1 bar)(0.0010432
 0.0010018
)
kg
kg 1 bar
1000
N
1 kJ
m2
1 kPa 1000 N  m
W = 0.00207 kJ
Substituting values and solving for Dt give
(0.50 kg) (417.36
Dt 
kJ
kJ
 83.95 )  0.00207 kJ
1 kW
kg
kg
= 196.1 s (3.27 min)
kJ
0.85 kW
1
s
(b) During process 2-3, heat transfer results in phase change from liquid to vapor at 99.63oC (the
saturation temperature that corresponds to 1 bar). Mass flow from the control volume is the
saturated vapor resulting from evaporation of the liquid. The mass rate balance takes the form
dmcv
 m e
dt
The energy rate balance for process 2-3 reduces to
dU cv 
 Qcv  Wcv  mehe
dt
Combining mass and energy rate balances results in
dU cv 
dm
 Qcv  Wcv  cv he
dt
dt
By assumption 3 of the engineering model, the specific enthalpy at the exit is constant.
Accordingly, integration gives
DUcv  Qcv Dt  Wcv Dt  Dmcv he
3
m3u3  m2u2  Qcv Dt  Wcv  (m3  m2 )he
Mass at state 2 is the system mass during process 1-2: m2 = 0.50 kg. Since no mass remains in
the control volume at state 3, m3 = 0. The equation reduces to
 m2u2  Qcv Dt  Wcv  m2he
2
Solving for time associated with evaporation during process 2-3 yields
Dt 
m2 (he  u2 )  Wcv
Q cv
The specific enthalpy for saturated vapor at the exit is obtained from Table A-3 at pe = 1 bar:
he = hge = 2675.5 kJ/kg.
The liquid volume decreases as vapor exits the control volume. Thus, work is done on the
control volume at the moving portion of the boundary during process 2-3
3
Wcv   pdV  p(V3  V2 )   pV2  m2 p2v2
2
Substituting values and solving for work give
Wcv  (0.50 kg)(1 bar)(0.0010432
3
m 100 kPa
)
kg 1 bar
1000
N
1 kJ
m2
1 kPa 1000 N  m
Wcv = −0.0522 kJ
Solving for time associated with evaporation during process 2-3 yields
(0.50 kg) (2675.5
Dt 
kJ
kJ
 417.36 )  (0.0522 kJ)
1 kW
kg
kg
= 1328.3 s (22.1 min)
kJ
0.85 kW
1
s
1 For process 1-2 a combination of the energy balance and the work expression gives an
expression for energy represented in terms of specific enthalpy only.
m(u2  u1)  Q Dt  W  Q Dt  mp(v2  v1)
4
m[(u2  pv2 )  (u1  pv1)]  Q Dt
m(h2  h1)  Q Dt
2 For process 2-3 a combination of the energy balance and the work expression gives an
expression for energy represented in terms of specific enthalpy only.
 m2u2  Qcv Dt  Wcv  m2he  Qcv Dt  (m2 p2v2 )  m2he
 m2 (u2  p2v2 )  Qcv Dt  m2he
 m2h2  Qcv Dt  m2he
 m2 (h2  he )  Q cv Dt
5
PROBLEM 6.8