2. 𝑆𝑢𝑚𝑚𝑎𝑟𝑦
𝑉𝑠 = 4.669 m3
1.
2. EB = 1872486000 Pa or 1872486 MPa 18 1.872 GPa β = 0.534 𝐺𝑃𝑎−1
𝜌 = 2.586 kg/m3
3.
𝑚 = 1.150 kg
4.
𝑝2 = 50 kPa abs
5. 𝑝2 = 34.657 kPa abs 𝑇2 = 210.023 K or − 62.977°C
6. 𝑈 = 5.160 m/s
𝜏 = 3.190 Pa
7.
𝐹 = 987.622 N
8.
ℎ = 9.922 × 10−4 m.or 0.992 mm.
9.
θ = 64.321°
10.
P = 0.027 N
11.
3. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 1
A vertical cylindrical tank with a diameter of 12m and a depth of 4 m is filled with water to the top with water at 20°C. If the
water is heated to 50°C, how much water will spill over? Unit weight of water at 20°C and 50°C is 9.79 kN/m3 and 9.69
kN/m3, respectively.
Solution
Volume spilled is
𝑉𝑠 = 𝑉𝑓 − 𝑉𝑖
Where:
Vs - Volume Spilled
Vf - Final Volume(50°C )
Vi - Initial Volume (20°C )
𝑉𝑖 =
(𝜋)122
4
)(4)
=
(𝜋)122
4
)(4)
𝑉𝑖 = 452.389 m3
Note:
Volume of water changes when
heated but its weight doesn’t change.
Solving for weight
𝑊 = 𝛾𝑉
𝑊 = (9.79)(452.389)
𝑊 = 4428.892 kN
Using the weight in
solving Vf
𝑉𝑓 =
𝑊
𝛾@50°C
=
4428.892
9.69
𝑉𝑓 = 457.058 m3
𝑉𝑠 = 457.058 − 452.389
𝑉𝑠 = 4.669 m3
4. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 2
A rigid steel container is partially filled with a liquid at 15 atm. The volume of the liquid is 1.23200 L. At a pressure of 30 atm,
the volume of the liquid is 1.23100 L. Find the average bulk modulus of elasticity of the liquid over the given range of pressure
if the temperature after compression is allowed to return to its initial value. What is the coefficient of compressibility?
Solution
Given :
𝑝1 = 15 atm
𝑉1 = 1.232 L
𝑝1 = 30 atm
𝑉1 = 1.231 L
Converting atm to Pa
𝑝1 = 15 atm x
101325 Pa
1 atm
𝑝1 = 1519875 Pa
𝑝2 = 30 atm x
101325 Pa
1 atm
𝑝2 = 3039750 Pa
Solving for, EB
EB = −
1519875 Pa − 303975 Pa
1.232 L − 1.231 L
1.232 L
EB =
∆𝑝
∆𝑉
𝑉
EB = 1872486000 Pa or 1.872 Gpa
Solving for, β
β =
1
EB
β = 0.534 𝐺𝑃𝑎−1
5. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 3
Calculate the density of water vapor at 350 kPa abs and 20°C if its gas constant is 0.462 kPa-m3/kg-K
Solution
Solving for density of water vapor
𝜌 =
𝑝
𝑅𝑇
𝜌 =
350 kPa
0.462
kPa − m3
kg − K
20 + 273 k
𝜌 = 2.586 kg/m3
6. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 4
Air is kept at a pressure of 200 kPa and a temperature of 30°C in a 500-L container. What is the mass of the air?
Solution
Solving for the density of air
𝜌 =
200 𝑥 103
(287)(30 + 273)
𝜌 = 2.300 kg/m3
Solving for the mass of air
𝜌 =
𝑚
𝑉
2.300 kg/m3 =
𝑚
500 L x
1 m3
1000 L
𝑚 = 1.150 kg
7. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 5
If 12 m3 of nitrogen at 30°C and 125kPa abs is permitted to expand isothermally to 30m3, what is the resulting pressure? What
would the pressure and temperature have been if the process had been isentropic? Use k=1.4.
Solution
𝑝1𝑉1 = 𝑝2𝑉2
For isothermal condition
125(12) = 𝑝2(30)
𝑝2 = 50 kPa abs
For isentropic condition
𝑝1𝑉1
𝑘 = 𝑝2𝑉2
𝑘
125(12)1.4
= 𝑝2(30)1.4
𝑝2 = 34.657 kPa abs
𝑇2
𝑇1
=
𝑝2
𝑝1
𝑘−1
𝑘
𝑇2
(30 + 273)
=
34.657
125
1.4−1
1.4
𝑇2 = 210.023 K or − 62.977°C
8. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 6
A square block weighing 1.1 kN and 250 mm on an edge slides down an incline on a film of oil 6.0 µm thick. Assuming linear
velocity profile in the oil and neglecting air resistance. What is the terminal velocity of the block? The viscosity of oil is 7 mPa-
s. Angle of inclination is 20°.
Solution
𝑊
𝑊sin(θ)
θ
θ
Force acting parallel to
the film of oil is Wsin(θ)
7 × 10−3 =
1.1 × 103sin(20)/0.252
𝑈/6.0 × 10−6
𝑈 = 5.160 m/s
𝜇 =
𝐹/𝐴
𝑈/𝑦
9. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 7
Benzine at 20°C has a viscosity of 0.000651 Pa-s. What shear stress is required to deform this fluid at a strain rate 4900 s-1?
Solution
𝑈
𝑦
= 4900𝑠−1
𝜇 =
𝜏
𝑈/𝑦
0.000651 Pa − s =
𝜏
4900/𝑠
𝜏 = 3.190 Pa
10. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 8
A shaft 70 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250
mm long. The clearance, assumed uniform, is filled with oil at 20°C with ν = 0.005 m2/s and sp. gr. = 0.9. Find the force exerted
by the oil in the shaft.
Solution
𝑦 =
70.2 − 70
2
× 10−3
= 0.0001 m
ν =
𝜇
𝜌
𝜇 = 4.491
solving for viscosity
0.005 m2/s =
𝜇
998 × 0.9 kg/m3
∗ Density of water at at given temperature at 20C
𝜌 = 998 kg/m3
Solving for the force
𝜇 =
𝐹/𝐴
𝑈/𝑦
4.491 =
𝐹/9.621 × 10−4
0.4/0.0001
𝐴 = 𝜋(.07) 0.25 = 9.621 × 10−4
𝐹 = 987.622 N
11. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 9
Two clean parallel glass plates, separated by a distance d = 15 mm, are dipped in a bath of water. How far does the water rise
due to capillary action, if σ = 0.0730 N/m?
Solution
W
F
F
F
F
F
F
F
F
F
F
F
F
FF
𝐹 = 𝑊
𝐹 = 𝜎𝐴
𝐹 = 0.0730(2𝑡)
l l
d = 0.015 m
𝑊 = 𝛾𝑉
𝑊 = 9810 𝑑𝑙ℎ
h
𝑊 = 9810 0.015𝑙ℎ
0.0730(2𝑙) = 9810 0.015𝑙ℎ
ℎ = 9.922 × 10−4 m.or 0.992 mm.
12. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 10
Find the angle of the surface tension film leaves the glass for a vertical tube immersed in water if the diameter is 0.25 in and
the capillary rise is 0.08 in. Use σ = 0.005 lb/ft?
Solution
Capillary rise in a tube is
ℎ =
4σ cos ϴ
𝛾𝑑
0.08 in x
1 ft
12 in
=
4(0.005 lb/ft) cos ϴ
62.4lb/ft3(0.25 in x
1 ft
12 in
θ = 64.321°
13. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 11
What force is required to lift a thin wire ring 6cm in diameter from a water surface at 20°C? (σ of water at 20°C = 0.0728 N/m)
Neglect the weight of the ring.
Solution
P
If we try to lift the thin ring using force P, surface tension inside and outside
of the ring will contract the force. Therefore, we have
P = Finside + Foutside
P = σinsideA + σoutsideA
Since the wire is thin, we can assume that there is no difference
between the inside and outside diameter.
P = 0.0728 (πD) + 0.0728 (πD)
P = 0.0728 (π(0.06)) + 0.0728 (π(0.06))
P = 0.027 N
