SAMPLE

PROBLEM:
1. Air contained in the tank is under an absolute
pressure
of 60 kPa and has a temperature of 60°c. Determine the mass
of the air in the tank.
R=286.9
J/(kg⋅K)

1.5m
4m

Given: P= 60kPa = 60000

N/m2
T= 60°c + 273 = 333 K

R=286.93/(kg·K)

h3 4m

r = 1.5m

Solution: PV= mRT

m = PV ; V = πr2h
RT

m = 60000 [πT (1-5)2

(4)]
286.9(333)
=17.757 kg

2. The tank contains air at a temperature of

18°C and an
absolute pressure of 3.48m and the temperature rises to
42°C. Determine the mass of air that must be removed
from the tank to maintain the
 160 kPa. If the volume of the
tank is

Same pressure.

鲜
Given: T 18°C + 273 = 291 K

P= 160 kPa

V = 3.48 m2

T2 42°C + 273 = 315 K

PIV

Solution: m
R.T
M2 =
P2V2
R2T2

Am = PIV P2V2
RIT

PV
R2V2

Am3 px (+-
+2)
Am = 160000 (3.48)
(1241-315) = 145.783
R
R

3. Determine the mass of carbon tetrachloride that should

be mixed with 80 kg of glycerin so that the combined mixture
has a density of 1450 kg/m3.
 Pretrachloride =
1592.52 kg/m3
11

Pglycerin 1257.524
kg/m3
Given: mc = 80 kg
Pc = 1450 kg/m3
VG = me
Po
80 1257.524

P1 = 1592.52 kg/m3
PG = 1257.524
kg/m3

Solution: p = m ; P = mc = mr +
me = 1450 kg/m3
Vc =VG +VT
=
Vc
VT + VG

mc=mormy

= 0.064 m3

Pove = POVO
+ P+VT
Pc (VG+ V1) =
PGVG +/+V+
 V1 = 0.086m3

my +80
0.086+0.064
(1450)=137.5
kg

4. An amount of glycerin has a volume of Im when the

pressure is 120 kPa. If the pressure is increased to
400 kPa. Determine the change in volume of this cubic
meter. The bulk modulus For glycerin is Ev=4.52 Gla.
Given: Vi= Im3

P1 = 120 kPa

P = 400 kPa

Ev 4.52 Gla

Solution Ev=-dp--(PE-Pi)
dy

DV= VF - Vi = -
(PF - Pi) V
=-

Ev

(400-120) (10)
(1)
4.52*109
=

-6.19 x 10-5
m3
3

5. The plate rests on top of the thin film of water,

which is at a temperature of 25°C. If a pressure
difference occurs between A and B. and a small
Force F is applied to the plate, the velocity profile
across the thickness of the water can be
described as u = 140y - 800y2 m/s, where
y is in meters. Determine the shear stress
acting on the Fixed surface and on the bottom
of the plate.

A
10mm
U
1

u = 0.897 (103) N·s/m2

17 0.32 m/s

Solution : 2
= 1 24
u =

du

dy
·1404-800y 2
m/s
= 140-1600×

shear stress at Fixed

surface.
Y⋅ O
du
x =
μdy
Given: U = 140y -
800y2 m/s
 =μ [140-
1600(0)]

= 0.897 × 10-3
(140)
=0.12558
N/m2
u= 0.897 (10-3)
N·s/m2

shear stress at the bottom of the

plate.
y= 10mm =
0.01m
T*0.897x10-3 (140-
1600(0.01))
t= 0.111228
N/m2
2

on a

thin Film
6. The 100-kg plate in the figure is resting very OF
SAE 10W-30 oil, which has a viscosity of u= 0.0652
N's /m2. Determine the Force P that must be
applied to the center of the plate to slide it
over the oil with a constant velocity of 0.2 m/s.
Assume the oil thickness is 0.1mm, and the
velocity profile
across this thickness is linear. The bottom
contact area of 0.75m2 with the oil.
0.2m/s
P

30'
of

Given: m = 100
kg
the plate has a

u = 0.0652 N-
s/m2
U = 0.2m/s
t = 0.1mm
W=mg= 106 (9.8)=98/N
U=0.2m/s
↓K
3300
P
A= 0.75m2

Solution:
10.1mm
0.1m/s
du

←
dy
=T

du 0.2 0.1mm dy
0.0001
=2000/s
 du
T =μ

= 0.065
(2000/s)
t = 130.4
N/m2
t =

F = TA = 130.4 (0.75) =
97.8N
ΣFx =0

F= Pcos 30°

P=F
cos 30
97.8 Cos
36
112.930 N

7. The tube rests on a 1.5mm thin film of oil having a

viscosity OF = 0.0586 Ns /m2. If
the tube is rotating at a constant
angular velocity of w = 4.5 rad/s. Determine the
shear stress in the oil at r = 40mm and r = 80mm.
Assume the velocity profile within the oil is linear.
40mm

80mm

Given: t 1.5 mm

u = 0.0586
N.s/m2
w = 4.5
rad/s
Solution:
7= 124
u
90

40
14.5mm

10
U

40

u@40mm::
4.5 rad/s = 4.5 [2π
(0.04)] = 1.131 m/s
duu dy
=

1.131
0.0015

(1.13001
5)
T= 0.0586 0.0015

=44.1844 Nm
@ 80mm:
2

L= 88.369
N/m2

8. Determine the distance h that the column

of mercury
in the tube will be depressed when the tube is inserted
into
mercury
the

a.)
 For
at
a room temperature of 20°C. Given D5.5mm
b.) Plot this relationship of h (vertical
axis) versus (
Imm ≤D ≤ 6mm. Give values for increments of AD = Imm Surface
Tension of mercury σ 0.466 N/m.

40

2 (0.466) (cos(40))
Given: D=5.5mm 5.5x10-3m

Solution: h=13550 (9.81)

(0.00275)
0 = 40°

T=0.466 N/m
P = 13550
kg/m3
r: 0.00275 m

= 1.953x10-3m
or
1.953
mm

SEATWO
RK!
1. The container is filled with water at a temperature
 of 25°C and a depth of 2.5m. If the container has a
mass of 30 kg. Determine the combined weight of
the container and the water.
←lm

2.5m
Given d= 2.5m

T= 25°C

Mc = 30 kg
r = 0.5m

Solution: Wwater =ɣ water

(Vwater)
* = 9810 (π (0-
5)2 (2-5))
= 19261.890 N

mc = 30x 9.81 = 294.3 N

W = 19261.890 +294.3=19556.190N

2. Gasoline is mixed with 3m3 of ethyl alcohol so that

the volume of the mixture in the tank becomes 5.5
m3. Determine the density and the specific
gravity of the mixture at standard temperature
and pressure
PGasoline =726
kg/m3
Palcohol = 789
kg/m3
Given: VA = 3m3
 VG = 2.5m3

matmG
Vc: 5.5m3

Solution: Pc = VA+
VG
P=7; ma = pv = 726 (2-5)=
1815 kg
M1 = Pr = 789
(3)= 2367 kg
Pc=
St
2367 +1815

S = Pw
5.5
=760.364 kg/m3
760.364
1000
=0.760

3. Water in the swimming pool has a

measured depth of 3.03 m when the temperature
is 5°C. Determine its approximate depth when the
temperature becomes 35°C. Neglect losses due to
evaporation.
am

4m
 Solution Ti TF
Given hi= 3.03m

3.03x9x4 hF * 9 x 4
278
14

308
Ti=5°C + 273 278 K

TF = 35°C + 273 = 308

K

A=(9×4)m2

;hF = 3.357m

QUIZ 1:
1. An 8m diameter spherical balloon is filled with helium
that is at a temperature of 28°C and an absolute
pressure of 106 kPa.
Determine the weight of the helium contained in the
balloon. R= 2077 J/(kg⋅K)
Solution: T 28°c + 273 = 301 K
V = 1/3 πr 3 =
256 H
PV=MRT
tt m
 106×103 (256 π) = m (2077)
(301) = 45.454 kg
w = mg
= 45.454 (9.8)=445.903 N

2. Two liquids of different densities (p = 1200

kg/m3, p2 = 600 kg/m3) were poured together
into a Im3 container, filling it. If the resulting density
of the mixture is 800 kg/m3, find the respective
amounts
of liquids used in m3. Also find the weight
of the mixture. Solution: Vi+V2 = 1m3
m, +m2 = m
3

P1V1 + P2
V2 = Pv
V, 21-V2

1200 (1-V2) +600 (V2) =

800 (1)
2 3
V2= V2 = 3
m
0.667m3

HT

V1
=
3

0.333 m3
3
 m3pv = 800(1) =
800 kg
W
= mg
- 800 (9.81)
=

7848 N

3. The tank contains air at a temperature of 18°C and

an absolute

pressure of
160 kPa. If the volume of the tank is 4.5 m3 and the
temperature rises to 52°c. Determine the mass of air
that must
be removed from the tank to maintain the same
pressure.
R=286.9 J/(kg.
k)
Solution: P1 = 160 kPa
V1 = 4.5 m3

T1 = 18+ 273 = 291

K
PV
m=
RT
Pv=MRT ; m
PIV P2-V2
1

Am=m, -m2 = R1T

R2 T2

= 2x (+/+ -
 +/-2)
P2 = 160 kPa

V2 = 4.5m3

T252 273 = 325 K

4. At a point deep in the ocean,

the density of
seawater is

1035 kg/m3. Determine the absolute pressure at this

point if at the surface the density is 1035 kg/m3
and the absolute pressure is P2 = 101.3 kPa. Take
Ev = 2.42 GPa.
Solution:
P = 101.3 kPa

There is no change in
Pressure
Since no change in
change in volume /
Density

5. The Newtonian fluid is confined between the plate

and a Fixed surface. Ip its velocity profile is
 define u = (8y -0.3y2)mm/s.
is in mm. Determine the Force P that
must be applied
where
Y
to the plate to cause this motion. The plate
has a surface area
2

OF 15 000 mm" in contact with the

fluid. Take M = 0.482 Ns /m2.
1
4mm ↓

Solution: u 8y-
0.3y2
du

dy = 8-
0.6yls
@y =
4mm
du
dy
= 5.6/s
+ = dy = 0.482 (5.6) =
2.6992 N/m2
P: TA 2.6992(

6.

water is at a
temperature
OF

30°C. Plot the

height h
OF
 the

Function
OF
the
water as a

For 0.4mm ≤ w ≤ 2. Amm. Use

increments of 0.4mm. Take
gap
w between the two glass
plates

σ = 0.0718 N/m and = 0.

Solution: w
h

0.4
0.037

0.8
0.018

‫به‬
1.2
0.012

1.6
0.009

2.0
0.007

2.4
0.
 SAMPLE
PROBLEM:
1. The mattress of a water bed is 2.00 m long by
2.00m wide and 300 cm deep.
9) Find the weight of the water in the
mattress
b) Find the pressure exerted by the
water bed
on the Floor when
the bed rests in its normal position. Assume the entire
lower surface of the bed makes contact with the
floor.
by a
c.) what if the water bed is
replaced
300 N regular
bed that is supported by Four legs? Each leg
has a circolar cross section
exert
OF
radius 2.00 cm. What pressure
does this bed
on the floor?
Given: V = 2m x 2m x 0.3 m

V = 12m2
 Solution A. 8water = 9810
N/m3
Ws = 8wV = 9810
(12)=117720 N
8. P = = =
WA
117720
2x2

C. AL = 4πT
(0.02)2
P = = = = 300
A
A
=1

T 625
29430 Pal
2

625
m

=59683.1037 Pa

2. A pneumatic jack is used in a service station as shown

in the figure. If the car and lift weight 25
kN, determine the Force
that must be developed by the air compressor at B to
raise the lift at a constant velocity. Air is in the
line from B to A. The
air line at B has an inner diameter of 15mm,
and the post at A has a diameter of 280mm.
 15mm

← 280mm 1000
B

Given: W1 = 25 kN = FA

DA= 280 mm

Do= 15 mm

FA FB
FA
Fo
Solution: AA
=

AB
DA2
푸 DA2 풋 002

FB =
FA DB2
DA2
25 (15)2
=
=

(286)2
0.0717 KN=
71.7474 N

3. The air pressure within the bicycle tire is

determined
a gage to be 70 kPa. If the local atmospheric pressure is
104 kPa. Determine the absolute pressure in the tire.
from
 Solution: Pabs = Patm
+
H

+ Pg
104 +70

=174 kPa
Given: Pg = 70
kPa
Patm 104 kPa

4. The tank and drainpipe shown in the figure are

filled with gasoline and glycerin to the depths
shown. Determine the pressure on the drain plug at
C. Report the answer as a pressure head in
meter. Take Pga = 726 kg/m3 and Pg 12 60 kg/m3.

A
2m

Given: Pga = 726

kg/m3

Gasoline
Im

Pg1 = 1260
kg/m3
Po
 ↓

hga =
Im
Glycerin
1-5 m

hgi =
1.5m

Solution: Pc Pga +
Pgi

= Pgahga +
fg, hgr
726 (9.81)(1) + 1260 (9.81) (1-5)= 2566
2. 96 Pa
Pc
h ==

= 25662.96 = 2.616 m
9810

5. Consider a hydraulic jack being used in a car repair

shop. The piston have an area of A1 = 0.8 cm2 and A2
= 0.04 m2. Hydraulic oil with a specific gravity of 0.870
is pumped in as the small piston
on the left side is
pushed up and down, slowly raising larger piston
on the right side. A car that weights 13,000 N is to
 be jacked up.
the

a.) At the beginning when both pistons are at the same

elevation (h-o), calculate of Force F1 in newtons
required to hold the weight of the car.
b.) Repeat the calculation after the car has been lifted two
meters (h=2). Compare and discuss.
F2P2A2

F=PA

A2
0
P

A1

Solution: F
= F2
A1 A2
F2

F1 =
FRA!
A2

13000 [0·8 (100)] =

26 N
0.04

b. P = P2; P = P2 + Poil;

Soil Poil
Yoil
Given: A1 =
0.8cm2

A2 0.04 m2

F2 = 13000N
 A1
Soil 0.870
F2

A2

F + ɣoil
hail
=

Pwater water

Poil = Soil (water)

Yoil = Poib
9=
Soil (water)
(g)
8oil = SOIL ɣwater

F1 = F2 A1 + A1
(Som) (8 water) (hol)
A2

F1 = 26+0.8 (0.870)
(9810) (2)
10000

27.366 N

6. The top part of a water tank is divided into two

compartments as shown. Now a Fluid with an unknown
density is poured into one side, and the water level rises a
certain amount on the other side to

compensate
for
this effect. Based on the final fluid height
 shown on the Figure. Determine the density of the
fluid added. Assume the liquid does not mix with water.
Given: h = 75 cm

75 Unknown
SK.
liquid

45
Cm

WATER
95cm
hw= 50 cm

Solution Pw Paghw = PL =
P1ghi

Pwghw =
Pight
Phw
PL = hu
1000 (0.5)
0.75

666.667 N/m2
2

7. The maximum blood pressure in the upper arm of a

healthy person is about 120 mm Hg. If a vertical tube open
to the atmosphere is connected to the vein in the arm of the
person, Determine how high the blood will
rise in the tube. Take the density of the blood to
be 1040 kg/m3.
FO ERW
H7Q

Given: hug = 120

mm
Pblood = 1040
 kg/m3 SHg =
13.6

Solution:
P=SHg/waterg Hug
= 13.6(1000) (9.81) (0.12) =
16009.92 Pa
P

h =ɣblood
16009.92
17

10 40 (9.81)
1.569m blood

8. The Funnel in the Figure is filled with oil and

water to the levels shown, while portion CD of
the tube contains mercury. Determine the distance
h the mercury level is from the top of the oil
surface. Take.
Po = 880
kg/m3
Pw = 1000 kg/m3 PHG = 13 550
kg/m3

0.3m

0.4m

C
 Solution: o + PoghAD +Pughoc -
PHaghco =0
0+ (800) (9.81) (0.3)+(1000) (9.81) (0.4)-(13550) (9.81)
(0.3+0.4-1)=0
h=0.653 m

9. Determine the difference in pressure between the

centerline points A and B in the two pipelines the figure,
if the manometer liquid co is at the level shown. The
density of the liquid in AC and DB is p = 800 kg/m3 and
in CD, PCD = 1100 kg/m3.
040mm

6040
65nn
4
,
30

250mm

Solution: PB-PghBD +
Paghoc +pghcA = PA
PB-(800) (9.81)(0.250) + (1100) (9-81)
(0.065)+(800) (9.81 ) (0.03) = PA

Ap= PA-P-1.025 kPa

10. The inclined-tube manometer shown is used

to measure small pressure changes. Determine
the difference in pressure between points A and
E if the manometer liquid, mercury is
 at the level shown. The pipe at A contains water,
and the one at E contains natural gas. For
mercury, PHG = 13 550 kg/m3.

400mm

100mm $

C
20
.E

700mm

Solution: PA tywhAs tyng

hoc -y HgheD
= PE

PA + (1000) (9.81) (0.4) + (13550) (9.81) (0.1) -

(3550) (9.81) (0.7sin 20)
= PE

PA-PE=14607.689 Pa

SEAT
WORK.
 1. The container is partially filled with oil,
water and air.
Pw Determine
the
pressures at A, B and C. Take Pa = 1000
kg/m3,
Po = 830
kg/m3.
{
Air
1.25 Oil

1
2
 →
←
→
1
1.25
1 5
 1.25 m

Given: Pw =
1000 kg/m3
Poil = 830
kg/m3

Water

0.25m1
Pressure @ A
Solution: PA Poil + Pwater
PA = (p ghloil+
(pgh) water
= 830 (9.81 ) (1·25) + 1000 (9.81)
(1.25) = 22 440. 375 Pa
Pressure @ B

P1 = P2

Poil = Po + Pwater

PB = Poil - Pwater
= 830 (9.81)(1-25)-1006 (9.81)
(0.25)=
Pressure @ C
Pc = POIL + P
water
7725.375 Pa

Pc = 830 (9.81) (1.25) + 1000

(9.81) (1)
=19987.875 fa
2. In car lift used in a service station, compressed air
exerts a Force on a small piston that has a
circular cross section 5.00 cm. This pressure is
transmitted by a liquid to a
has a reduce
OF
15.0 cm.
of
radius

piston that

a.) What force must the compressed air

exert to lift a car weighing 13 300 N?
b.) what air pressure produce
this force?
Solution: F1 =

the
13,360 (5)2
152

1477.778 N
13300
b. P = π (0.15)2
ᅲ

188156.511 Pa

3. A deep sea exploration vehicle operates at a

depth where
 of
oil

pressure in
pressure in the surrounding water is equivalent to
30 Ft with a specific gravity (SG) of 0.8.
Calculate this millimeters of mercury
(SG = 13.6)

4. Express Pascal's Law and give a real-world

example of it. Answer: P=E where;
A

P = is the pressure

F= is the applied Force

and
A = is the area over which the Force
is applied.
Real World Example:
Hydraulic Lift
411

A hydraulic car lift uses Pascal's Law to amplify Force,

allowing a small input Force
to lift heavy
 vehicles.
of
5. Someone claims that the absolute pressure in
a liquid constant density doubles
when the depth is doubled. Do
you Agree?
Explain.
Answer: No, because absolute pressure includes
atmospheric
pressure (Po), which stays constant. When the
depth (h) doubles,
only the liquid pressure (pgh) doubles, but the total
absolute pressure does not.

SEAT
WORK #2
1. The U-tube manometer is filled with mercury, having
a density of PHg= 13550 kg/m3. Determine the
differential height h
mercury
when the tank is filled with water.
OF
the
 2m

3m

Solution: P1 =
P2

PHGghHG =
Pwg hw
(13550) (h) = 1000(2th)
= 0.159m
2. The tube is filled with mercury From
A to B, and with water B to C. Determine the
height h
For equilibrium.

1
0.5m

10.1m
130
A
OF
the water column

0.3m
 -0.4m-

Solution: P1 - P2

ɣw hw = 8HG THG +
8water h water

9810 (h) = 13550 (9.81) (0.5 sin 30+

0.1) + 9810 (0.3)
h
=5.0425 m

3. Determine the absolute pressure of the water

in the pipe at 1 if the tank is filled with crude
oil to the depth of
1.5m. PATM = 101
kPa.

1.5m

B
0.6m

10.4
m
8.5m
 Solution: Pass = Palm +
Peage

101000 + Perude oil +

Pwater

101000 + 880 (9.81) (1.1) +

1000 (9.81) (0.9)

=119325.08 Pa

4. Determine the difference in pressure Po-PA

between the Centers
A and B
OF

the pipes, which are filled with water. The

mercury in the inclined-tube manometer has the level
shown. Take SHG 13.55
こ

water

A
100 mm

250mm

Mercury

3. A solvent used for plastic manufacturing

consists of cyclohexanol
in tank A. and ethyl lactate in tank B.
Determine the
on the top of tank A. if the
mercury
pressure

in the manometer is at

the level shown, where h=0.75 m. Take Sc 0.953, SHG = 13.55

and
Sel = 1.03.

1-5m
G

Solution:
mercury
Th
4.5m

QUIZ 2:
 1. A vacuum gage connected to a tank
reads 45 kPa at
a

location

where the barometric reading is 755 mmHg. Determine

the absolute
(in kPa) in the tank. Take Pug
#13,590 kg/m3
pressure
=

Given: Pgage - 45 kia

Parm = 755
mmHg
Solution: Parm 755
mm Hg
P =pgh = 1350 (9.81)
(0.755) = 100655.015 Pa
Pabs = 100. 655 + (-45)
55.655 kPa
а
OF a

open

2. The maximum blood pressure in the upper arm

healty person is about 120mmHg. If a vertical
tube the atmosphere is connected to the vein in the
arm person, determine how high the blood will
rise in the tube.
Take the density of
Given hHG = 120 mm
the blood to be 1040
kg/m3.
 Pblood = 1040
kg/m3
SHO= 13.6

PHG = 13600

Solution: PHG
Pelood
=

phg =
phg
holood PHghing
Polood

13600(0.1
2) 1040

holood 1.569m
13

Of the
to

IF
3. If the
pressure
at a point in the ocean is
GOkPa,what is the
pressure 27 meters below this point? SG of
saltwater =1.03.
Solution: P= 60+ Psw
P = 60000 +1.03 (1000) (9.81) (27)
332.816 kPa
4. Convert 760mm of mercury to an (a) oil with SG of
 0.82 and b water at 4°C.
Solution: Sh, S2h2; SHGING SOIL hair
= Swhw
13.6(0.760)=0.82 (hoir) = 1
(hwater)
holl = 12.605m

hw
1

10.336m

5. Lacking on mercury, an improvised barometer uses a

liquid that weight 0.735 times that of
mercury. Assuming that the unit weight of air is
constant at 12 N/m3, evaluate the
approximate height of the mountain if the
barometer reading are 600 mm at the base and
850mm at the top of the mountain
SHO= 13.6.

Solution: Sliquid -0.735

(13.6)
AP -phg
AP =
Spugh
AP=0.735 (13.6) (1000) (9.81)
(0.25)
AP = 24 515.19 Pa

24515.19
12(h
mountain)

h mountain
 = 204 2.933 m

press

6. A piston of cross-sectional area a is

used in a hydraulic
to exert a small Force of magnitude f on the
enclosed liquid. A connecting pipe leads
to a larger piston of cross-
sectional area A. If the piston diameters are
2.4cm and 50.0cm
(in N) on the small piston will balance a
60.0 KN
what
Force

Force on the large

piston.?
Solution:
F2
=

Fi
2.4
60000 50
F= 138. 24 N

7. Determine the absolute pressure of the water in the

pipe at B if the tank is filled with crude oil to the depth
of
1.5 m. PaTm = 101 kPa.

1-5m
0.6m

0.4m
1

0.5m
 1
3m

131 13
Solution: Parm +
Pgage
= 10 1000 + Po14 +
Pwater

=101000 +880 (9.81) (1.1) + 1000

(9.81) (0.9)
=119325.08 Pa

8. Determine the value of y in the

manometer in the Figure.

Air, 5kPa

oil
(0-80)

0.6

mercury

Solution: Pair + PoIL + PW

- PHG = 0
5000+800 (9.81) (3) +1000 (9.81)
(1-5)-13600 (9.81) (y)=0
Y-0.324m

9. A simple open U-tube contains

mercury. When 13.2 cm of
oil (SG=0.8) is poured into the
right
 right arm
OF
the tube, how
high
above its initial level does the mercury rise in the left
arm?
Solution: P1 = P2

10. The
Soil hoIL =
SHG hHG
0.8 (132) = 13.6 (2h

h= 0.388 cm
gage pressure of the air in the tank shown in
the Figure is measured to be 50 kPa.
Determine the differential height h of the
mercury column.
50 kfa
- oil
75cm
S6=0.72

Air

1-898-10
Water

mercur
y
JG= 13.6
Solution: Pair + Pwater - PHG -
POIL = 0
50000+1000 (9.81) (30) - 13600 (9.81) (h) - 720(9-
81) (75)=0

h= 35.7cm

REVIEW
PROBLEM:
1. A reservoir of glycerin has a mass of 1200
kg and a volume
of 0.952 cu.m. Find its.

a.)
weight
c.) mass
density
b.) unit
weight
d.) specific
gravity
Solution:
11.772 KN
1260
=

3101.97162129779

11.772
=

kg/m3
a.) w = ma = (1200) (9.81) =
11772 N or
b.) P.m c)
Y= = =
 0.952 1260-504kg/m3 or
12.361 KN/m3
0.952
12.36

d.) SG=
9.81

with a
こ

12.366 KN/m3 or
12 60.981
1.260

2. A cylindrical tank 80 cm in diameter and

90cm high is filled liquid. The tank and the liquid
weighed 420 kg. The weight of the empty tank is
40 kg. What is the unit weight of the
liquid in KN/m3 ?
Solution: Wiiquid = 420-40
= 380 kg
E

=380 9.81 37 27.8 N

V=π (0.4)2 (0.9) = 0.452 m3

y =
3727.8 0.452
=8247.345 N/m3 or 8.247
kN/m

OF
3. A vertical cylindrical tank with a diameter of 12m
and a depth
Ameters 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: V1 = # (2)2 (4) = 452.390
m3
V2 = 452.390 (9.769) =
457.058 m3
Av = V2 - Vi
= 457.058-452.390 = 4.668 m3

4. Benzene at 20°C has a viscosity of 0.000651 Pa-s. what

shear stress is required to deform this fluid at a strain
rate
oF
4900s"?

Solution
(0.000651) (4900)
=3.1899 Pa

5. The plate is moving at 0.comm/s when the

Force applied to the plate is 4mN. Ip the
surface area of the plate contact with the liquid is
0.5m2, Determine the approximate viscosity of the
liquid, assuming that the velocity distribution is
linear.
0.6mm/s

↑
4mm +
 solution:
I=N dy
du
H

0.0006
dy 0.004
39

4m N

0.15 s

F/A 4×10-3 11 =
(dv/dy) = (0.5)
(0-15)
0.0533 N's/m3

6. A square block weighing 1.1 KN

and 250 mm on an
edge slides down
on indine on a film of oil 6.0 μm thick.
Assuming a linear velocity profile in the
oil and ignoring the drag (air resistance)
what is the terminal velocity of the block? The
viscosity of oil is 7 mPa-s and the angle of its
inclination is 20°C.
Solution: I =μ
(14)
= 7 × 10 - 3 [
6 -10 × 10 - 0 ];
VT
 = 1166.667 VT
7x10-3
6.0×10-6

6x10-6)

Ff = TA = (1166.667 VT) (0.250)2 =

72. 917 VT
1100 sin 20 = 72.917
VT

=
VT 5.160 m/s

7. The marine water strider, Hylobates, has a mass of 0.36 g. If it

has six slender legs, Determine the minimum contact
length of all OF its legs combined to support itself
in water having a temperature of I-20°C. Take σ =
72.7 mN/m, and assume the legs to be thin
cylinders with a contact angle of 0=0°.
0.36×103(9.81)
Solution: 1 = mg;
20 2(72-7x10)
= 0.024 or [24.289 mm

8. A volume of 8 m3 of oxygen
initially at 80 kPa
of absolute
pressure

and 15°C is subjected to an absolute pressure of 25 kPa

while the temperature remains constant. Determine
 the new density and volume
OF
the
oxygen.
Solution: P1 = PIRT,; 80000 = p,
(259.8) (15+273)
= 1.069
kg/m3
P
P
AA
12 (2)
80
1.069
25
12

P2-10.334
kg/m3
m=p. V1; 1.069(8) =
8.552 kg
m=P2V2; 8.552 = (0.334)
V2
V225.665 m
3

9. At 32°C and 205 kPa gage, the specific

weight of a certain gas
was 13.7 KN/m3.
Determine the
P RT

Solution: p=
R
R=
P
PT
 R = 306.325
(1.40)(305)
KJ
= 0.717 kg-k or
gas
constant
Of this
gas.

J
717.39 кдак

10. A rigid steel container is partially filled

with a liquid at
15 atm. The volume 30 atm the volume
C

OF

OF
the liquid is 1.232 liters. At a
pressure of
the liquid is 1.231 liters. 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
(inverse of Eo)?
Solution: AP = 30-15= 15
atm
AV = 1.232 -1.231 = 1x
10-3
 Es=
-18480
9869.233

B =
1872
=1.872 GPa

= 0.534 GPa

11. An amount of glycerin has a

volume OF
Im3 when the

pressure is 120 kPa. If the pressure is increased to

400 kPa.
Determine the change in volume of this cubic
meter. The bulk
modulus For glycerin is EB
= 4.52 Gfa.
Solution: AP = P2 -
P1
AP(V)
AV= EB J
400-120 = 280 kPa = 280 000 280000(1)
4.52×109

12. A barometer reads 760 mm Hg and a pressure

gage
attached

to a tank reads 850 cm of oil (SG = 0.80). What is the

absolute
pressure in the tank in kPa ?
=

Solution: Pans Parm +

Pgage
= (13.6×9·81)(0.76) + (0.8x9.81)
(8.5)

168.104 kPa abs

 13. If atmospheric pressure is
95.7 kPa and the
gage
attached

to the tank reads 188 mm Hg vacuum, Find the

absolute pressure
within the tank in klaa (kPa abs).
こ

Solution: Pabs Patm +

Pgage
[gage = Ymercury hmercury;
(9.81) (13.6) (0.188)
=25.082 kia vacuum; -25.082 kla

Pabs = 95.7 +(-25.082)=70.618

kPa abs

14. Assuming specific weight of to be constant at 12

N/m3, What is the approximate height of Mt.
Banahaw if a mercury barometer at the base of the
mountain reads 654 mm and at the same time,
the mountain with a value of 480mm.
another reading at the top of the mountain with
Solution: Poot-PTop =
yh

15.

(9.81)(13.6) (0.654) - (9.81) (13.6) (0.48)

= 12h

h = 1.935
 or 1934.532 m

For the tank shown. Find the value of h2. where h, and
h2 is

3m and 4m,
respectively.

OIL (0.84)

water
h3

Solution: 0+0.84 (9.81) (hz) + 9.81 (3) -

9.81(4) = 0
h2 = 1.190 m

16. For the manometer shown, determine the

pressure at the
center of the pipe.

1.5m

ป
oil
(0.80)
Mercury

Solution: y
£1 + 1
(13·55) + 1-5
(0-8) = 33
 0+ 14.75 = P
P3
= 14.75m
y

P2 = 14.75(9.81) = 144.698
kia

17. In the [inclined manometer shown, the pressure at

point m was increased from 70 kPa to 105 kPa. Due to
additional pressure, the top level of mercury moves
20mm in the sloping tube. What is the value of
angle?
Water

'm

Mercury
10

P2

Solution: ++y (186) -

x=m
76

13.6y-x =
9.81
EQUATION I

A: (0.2sin @ty +0.2) (13.6)-

(x +0.2)=
Pm

0+2.72sine
+13.64
+2.72 -x-0.2=
 105
9.81

13.64 -x = 8.183- 2.72 Sine

EQUATION 2

soj
8.183 -2.72 sin = 708
9.81

Sine 6.3852

=22.66°

18. In the Figure below, Find the

difference of pressure
at point A &
B.
Kerosene (0-82)
Air
(0.0012)

250 mm
Benzene (0-88)
310mm
LA.

200mm
90mm

Mercurypa
Solution:
PA
У
PB

y
6
400mm

61
100mm
1
150mm

water

+0.2 (0.88) - 0.09 (13.6) - 0.31

(0.82) +6.25-0.1 (0.0012) =
 = 1.0523m

PA-PB = 9.81 (1.0523)=10. 323

kPa
Po

19.
IF
a 250 N Force is applied to the lower piston (D1: Acm),
what is the magnitude of the Force F2 that can
be resisted by the upper piston (D2 = 12cm).
Neglect the weight of the piston and take h =
2.2m, SG of oil (inside of cylinder) = 0.85.
F2

Solution: PA
4(250)
π (0.04) 2 = 198 943.679 Pa P2 = P1 + (Sywater )
( 21 - 22)
= 198943.679 + (0.85) (9810) (-2.2) = 180 598.979 F2
= P2A2; (180598.979) ((0.12)) =
2042.526 N
20. The steel pipe and steel chamber shown in the figure
together weight 2760 N. What force will have to be
exerted on the chamber by all the bolts to hold it
in place? The dimension
h is equal to 76 cm. Note: There is no bottom on
the
chamber-only a
a plange bolted to the
floor.
pd=0.25h
 steel pipe
4h

(1-2)
steel chamber

D=h
FBD:
Wpc-Weigh of pipe & chamber

WL

TEN
1

Weight of liquid

-pressure of liquid

TFBT - total exerted

by bolt

Solution: Py=0
FN +FB = Wpc
+WL

82h A2 +
Fo = 27
1.2 (9.81) (5
× 0.76) (+
2

kN + G2 [+
(#7)2 (44) +
[= (#)2 (44) +
=(h)" (b)]
GL

(0.7
6)2
4
 T

[+
10:174174(0.76
) + + (0.76) 2 ]
) + 76 •
27+
[10.7619]

= 20.293 +FB = 2.7

+5.073
2.7. ·

FB-12.52 kN or 12.52 kN ↓
[1.2
(9.81)]
4