Problem Set 6, PS4 due June 2
PS6-1
3.29 50 kg of water at 150 kpa and 25 C is contained in a piston cylinder device having a
piston cross-sectional area of .1 m2. The water is heated causing part of the water to
vaporize. The piston reaches a linear spring having a spring constant of 100 kN/m when
the volume contained by the piston cylinder reaches .2 m3 . The piston then rises 20 cm
further as more heat is added. Determine using the steam tables and EES a) the final
pressure of the water, b) the work done during the processes. Sketch a schematic of the
apparatus and a pressure volume property diagram of the processes.
PS6-2
3-89 A balloon material has the characteristic that the pressure inside the balloon is
always proportional to the square of the diameter. If a spherical balloon made of this
material contains 10 lbs of air at 30 psia and 800 R determine the work done when the
volume of the balloon doubles as a result of heat transfer.
Problem Set 6, PS4 due June 2
p
PS6-1
3 - 29
3
v1 = v f @25O C = .001003 m 3 /kg
V1 = m × v1 = 50 kg × .001003 m 3 /kg = .05 m 3
2
kx
100 N/m × .2 m
= 150 kPa +
a) p 3 = p 2 +
A
.1 m 2
1
50 kg liquid water
p 3 = 150 kPa + 200 kPa = 350 kPa
150 kPa, 25 C
.2 m 3
= .004 m 3 /kg
v2 =
v
50 kg
V3 = V2 + .1 m 2 × .2 m = .22 m 3
V .22
= .0044 m 3 /kg
v3 = 3 =
b)alternate = area under straight process line
m 50
(p + p 2 ) × (V −V )
T@P = 350 kPa and v = .0044 m 3 /kg PressureTable
W = m p1 (v 2 − v1 ) + 3
3
2
O
= 138.88 C, in two phase region
2
2
3
.2m
(350 + 150) × (.22 − .20)
b)W = m ∫ p dv + ∫ F dx = m p1 (v 2 − v1 ) + ∫ (p 2 A + kx ) dx W = 50 ×150(.004 − .001003) +
2
1
2
0
2
k×x
W = 22.48 + 5 = 27.48 kJ
W = m p1 (v 2 − v1 ) + p 2 × A × x +
2
100 N/m
(.2 m )2
W = 50 ×150(.004 − .001003) + 150 × .1× .2 +
2
W = 22.48 + 3 + 2 = 27.48 kJ
Problem Set 6, PS4 due June 2
PS6-1
Problem Set 6, PS4 due June 2
3-89
PS6-2
p ∝ D2
p1 = C × D12 ,
p = C × D2
π
V= D3 Volume of a sphere
6
1
6 3
D=   V
π
V
2
3
2
3
1
V
W = ∫ pdV = ∫ p1
p 2 =30 × 2 =30×1.587=47.6 psia
p V2 144×30×1.587×197.6
=
=2539.3 R
mR
10×53.35
∆U=10 lb m ×.208 BTU/lb m K (2539.3-800)=36177
T2 =
p C × D2
=
p1 C × D12
p = p1
2
3
V
2
3
dV =
2
V13
p1
2
3
1
V
∫V
2
3
.208 is the c v at about average 1500O F
Q=∆U+W=36177.+717 = 36,894. BTU
mRT
V1 =
p
10 lb m ×53.35 ft lbf / lb m F×800 R
V1 =
144in 2 /ft 2 ×30 psia
V1 = 98.8ft 3
2
V2 = 2 × V1 = 197.6ft 3
5

p 1 3  53
W = 2  V2 − V13 
5

V13 
5
5

3 × 30 × 144 
ft lb f
3
3
−
W=
197.6
98.8
/7
78


2
BTU

98.8 3 × 5 
W = 717.8 BTU
1
10 lb
30 psia
800 K
V2 = 2 × V1