A tank has two rooms separated by a membrane. Room A has 1 kg or air and
a volume of 0.5m3 ; room B has 0.75 m3 of air with a density of 0.8 kg/m3 .
The membrane is broken, and the air comes to a uniform state. Find the final
density of the air.
Given: mA = 1 kg, VA = 0.5m3 , VB = 0.75m3 , ρB = 0.8kg/m3.
Find: ρTotal
Solution:
m
V
mA + mB
=
VA + VB
mA + VB ρB
=
VA + VB
ρ=
ρTotal
ρTotal
ρTotal =
ρTotal
1 kg + 0.75 m3 0.8 kg/m3
0.5 m3 + 0.75 m3
1.6 kg
=
1.25 m3
3
ρTotal = 1.28 kg/m
A barometer measures 760 mm Hg at street level and 735 mm Hg on top of
a building. How tall is the building if we assume air density is 1.15 kg/m3 .
Given: Barometer readings at the top and bottom of a building. The density
of air.
Find: The height of the building
Solution:
∆P , Air = ∆P , Hg
∆P , Air = −ρAir g∆zBuilding
∆P Hg = −ρHg g∆zHg
ρAir g∆zBuilding = ρHg g∆zHg
ρHg g∆zHg
∆zBuilding =
ρAir g
ρHg
∆zBuilding = ∆zHg
ρAir
3
∆zBuilding = 0.025 m
13580 kg/m
3
1.15 kg/m
3
∆zBuilding = 0.025 m
∆zBuilding = 295 m
13580 kg/m
1.15 kg/m3
Fillout th ethe
P[kPa]
a.
500
b.
500
c.
1400
d.
Solution:
following table for substance water:
T[◦ C] v[m3 /kg]
x
20
0.2
200
300
0.8
a. Tsat |P =500kPa = 151.86◦C. T < Tt extsat, therefore, compressed liquid. Use
the compressed liquid approximation: v ≈ vf |Tsat =T = 0.001002 m3 /kg. x is
undefined.
b. Need to see where v is with respect to the saturation dome. vf |P =500 kPa =
0.001093 m3 /kg. vg |P =500 kPa = 0.37489 m3 /kg. vf < 0.2 m3 /kg< vg ,
therefore it is a a saturated mixture. T = Tsat |P = 500 kPa = 151.86◦C.
x = (v − ff ) / (vg − vf ) = (0.2 − 0.001093)/(0.37489 − 0.001093) = 0.532.
c. Tsat |P =1400kPa = 195.07◦C. T > Tsat , therefore, superheated vapor. Use
table B.1.3. v = 0.143 m3 /kg. x is undifined.
d. x is given so system is a saturated mixture P = Psat |T =300◦ C = 8581 kPa.
v = vf + xvf g = 0.001404 m3 /kg + (0.8) 0.02027 m3 /kg = 0.01762 m3 /kg.
A pressure cooker has a lid screwed on tight. A small opening with A=5 mm2 is
covered with a petcock that can be lifted to let steam escape. How much mass
should the petcock have to allow boiling at 120◦ Cwith and outside atmosphere
at 101.3 kPa? Given: Boiling temperature, outside pressure, area of petcock.
Find: Mass of petcock
Assumptions: Friction is negligible, contents of pressure cooker can be approximated as a saturated mixture of water/steam.
Solution:
X
F = 0 = PCooker A − PAtm A − mg
m=
(PCooker − PAtm ) A
g
PCooker = PSat |T =120◦ C = 198.5 kPa
Pa
(198.5 kPa − 101.3 kPa) 1000
5 mm2
1 kPa
m=
9.81 N/kg
m = 0.0495 kg
2
1m
1000 m