Assignment #4
9/20/04
PROBLEM 4.1
PROBLEM STATEMENT:
Water (2 kg) is contained in the vertical frictionless piston/
cylinder device shown below. The mass of the piston is such that the H
2
O exists
as a saturated liquid at 10.0 MPa. There is a heat transfer to the cylinder until the
piston reaches some stops, at which point the total volume is 0.02 m
3
. There is
an additional heat is transfer to the water until the H
2
O exists as a saturated
vapor. Determine the total work and the total heat transfer, and show the process
on a Tv diagram.
DIAGRAM DEFINING SYSTEM AND PROCESS:
GIVEN:
Water, m=2 kg, frictionless piston
State 1: Saturated liquid (x
1
= 0), P
1
=10.0 MPa
State 2: V
2
=0.02 m
3
, P
2
=P
1
=10.0 MPa
State 3: Saturated vapor (x
3
= 1), V
3
=V
2
=0.02 m
3
FIND:
Total work and total heat transfer (kJ)
ASSUMPTIONS:
1. The water in the container is a closed system.
3. No net change in kinetic or potential energy.
GOVERNING RELATIONS:
First Law,
IN,ab
OUT,ab
b
a
Q
 W
= U  U
QUANTITATIVE SOLUTION:
Once the piston reaches the stops, there is no change in volume, so
OUT,13
OUT,12
OUT,23
Total work : W
= W
+ W
1
2
1
= mP (v  v )
From Table 11s, v
1
=
v
L
(10 MPa) = 0.001452 m
3
/kg and u
1
=1392.8 kJ/kg.
3
3
2
0.02(m )
v =
0.01m /kg
2(kg)
=
Thus,
3
OUT,12
1000kPa
W
= 2(kg) 10.0(MPa)
(0.01 0.001452)(m /kg)
1MPa
=171.0 kJ
×
×
×


IN,13
IN,12
IN,23
IN,12
2
1
OUT,12
IN,23
OUT,23
Q
= Q
+ Q
Q
= U  U + W
Q
W
∴
3
2
IN,13
3
1
OUT,12
= U  U
Q
= U  U + W
Since state 3 is a saturated vapor, u
3
=u
V
, which from Table 11s, corresponds to
v
3
=v
V
=0.01 m
3
/kg. This value falls between 15 MPa and 16 MPa on Table 11s.
Q
IN,13