Problems

1. Two spherical raindrops of equal size are falling through air at a velocity of 0.08 m/s. If the drops join
together, forming a large spherical drop, what will be the new terminal velocity?

Given

Vt = 0.08 m/s

Required

VT =?

Solution
2r2ρg
Vt = 9η

2R2 ρg
VT = 9η

VT R2
= r2
Vt

VT R 2
= (r)
Vt

When two similar spheres combine together to form a single large sphere, then

V1 + V2 = V
4 4 4
πr3 + 3 πr3 = 3 πR3
3

4 4
2(3 πr3 ) = 3 πR3

2r3 = R3
R3
=2
r3

R 3
( ) =2
r

1
R
= 23
r

VT R 2
= (r)
Vt

1 2
VT
= (23 )
0.08

2
VT = 0.08 (23 )

VT = 0.08 (1.587)

VT = 0.126 = 0.13 m/s

2. Calculate the viscous drag on a drop of oil of 0.1 mm radius falling through air at its terminal velocity.
(Viscosity of air = 1.8 x 10-5 Pa.S; Density of oil = 850 kg/m3)

Given

r = 0.1 mm = 0.1 × 10-3 m

η = 1.8 × 10-5 Pa s

𝜌 = 850 Kg/m3

Required

Fd =?

Vt =?

Solution
2r2ρg
Vt =
9η

2(0.01)2(850)(9.8)
Vt = 9(1.8×10−4)

2(1×10−8 )(850)(9.8)
Vt = 1.62×10−4

1.66×10−4
Vt = 1.62×10−4

Vt = 1.024 m/s

Fd = 6πηrVt

Fd = 6(3.142)(1.8 × 10−5 )(0.1 × 10−5 )(1.024)

Fd = 3.48 × 10-8 N

3. What area must a heating duct have if air moving 3.0 m/s along it can replenish the air every 15 minutes
in a room of volume 300 m3? Assume air density remains constant.

Given

v = 3 m/s

t = 15 mins = 15 × 60 = 900 sec

V = 300 m3

Required

A =?

Solution
V
= Av
t

V
A = vt
 300
A = 3×900

A = 0.11 m2

4. Water circulates throughout a house in a hot water heating system. If the water is pumped at speed of
0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3,0 atm, what will be the
flow speed and pressure in a 2.6 cm diameter pipe on the second floor 5.0 m above? Assume the pipes do
not divide into branches.

Given

V1 = 0.5 m/s

d1 = 4.0 cm = 0.04 m
d1 0.04
r1 = 2
= 2
= 0.02 m

P1 = 3.0 atm = 3.0 (1.01 × 105 ) = 3.03 × 105 Pa

d2 = 2.6 cm = 0.026 m
d2 0.026
r2 = = = 0.013 m
2 2

h = 5.0 m

Required

V2 =?

Solution

A1V1 = A2V2
A1V1
V2 = A2

πr12 V1
V2 =
πr2 2

r1 2V1
V2 = r2 2

(0.02)2 (0.5)
V2 = (0.013)2

(4×10−4 )(0.5)
V2 = 1.69×10−4

2×10−4
V2 = 1.69×10−4

V2 = 1.18 = 1.2 m/s

From Bernoulli’s Equation

1 1
(P1 – P2) = 2 𝜌V2 2 – 2 𝜌V1 2 + 𝜌gh2 – 𝜌gh1
 1 1
(P1 – P2) = 2 𝜌V2 2 – 2 𝜌V1 2 + 𝜌gh

1 1
(P1 – P2) = 𝜌 (2 V2 2 – 2 V1 2 + gh)

1 1
3.03 × 105 – P2 = 1000 [2 (0.12)2 − 2 (0.5)2 + (9.8)(5)]

1 1
3.03 × 105 – P2 = 1000 [2 (0.0144) − 2 (0.25) + 49]

3.03 × 105 – P2 = 1000 [7.2 × 10−3 − 0.125 + 49]

3.03 × 105 – P2 = 1000 (48.9)

P2 = 3.03 × 105 – 4.89 × 104

P2 = 2.541 × 105 Pa

In atm,
2.541 × 105
P2 = 1.01×105

P2 = 2.51 atm

5. What is the volume rate of flow of water from a 1.85 cm diameter faucet if the pressure head is 12 m 2?

Given

d = 1.85 cm = 0.0185 m
d 0.0185
r=2= = 9.25 × 10−3 m
2

h = 12 m

Required

V/t =?

Solution
V
= Av
t

A = 𝜋r2
2
A = 3.14 (9.25 × 10−3 )

A = 2.69 × 10−4 m2

KE = PE
1
mV2 = mgh
2

V2 = 2gh
V = √2gh

V = √2 × 9.8 × 12

V = √235.2

V = 15.33 m/s
V
= (2.69 × 10−4 )(15.33 )
t

V
= 4.13 × 10-3 m3/s
t

6. The stream water emerging from a faucet ‘neck down’ as it falls. The cross-sectional area is 1.2 cm2 and
0.35 cm2. The two levels are separated by a vertical distance of 45 mm as shown in figure. At what rate
does water flow from the tap?

Given

A1 = 1.2 cm2

A2 = 0.35 cm2

h = 45 mm = 4.5 cm

Required

V/t =?

Solution

By the volume rate of flow of liquid

V
= Av
t

V
= A2v
t

2ghA22
V = √A 2 −A 2
1 2

2×980×4.5×0.352
V=√ (1.2)2 −(0.35)2

2×980×4.5×0.1225
V=√ 1.44−0.1225

1080.45
V = √ 1.3175

V = √820.07

V = 28.63 cm/s
V
= A2v
t

V
= (0.35) (28.63)
t
V
= 34.4 cm3/s
t

7. Water leaves the jet of horizontal hose at 10 m/s. If the velocity of water within the hose is 0.40 m/s,
calculate the pressure within the hose. Density of water is 1000 Kg/m 3 and atmospheric pressure is 100000
Pa.

Given

V1 = 10 m/s

V2 = 0.40 m/s

𝜌 = 1000 Kg/m3

P1 = 100000 Pa

Required

P2 =?

Solution

From Bernoulli’s Equation

1 1
(P1 – P2) = 2 𝜌V2 2 – 2 𝜌V1 2 + 𝜌gh2 – 𝜌gh1
1 1
(P1 – P2) = 2 𝜌𝜌2 – 2 𝜌V1 2
1
(P1 – P2) = 2 𝜌 (V2 2 − V1 2 )
1
100000 – P2 = 2 × 1000 [(0.4)2 − (10)2 ]

100000 – P2 = 500 [0.16 − 100]

100000 – P2 = 500 (-99.84)

100000 – P2 = - 49920

P2 = 100000 + 49920

P2 = 1.49 × 105 Pa

8. What is the maximum weight of an aircraft with a wing area of 50 m 2 flying horizontally, if the velocity
of the air over the upper surface of the wing is 150 m/s and that of the lower surface is 140 m/s? Density
of air is 1.29 Kg/m3

Given

A = 50 m2

V1 = 140 m/s

V2 = 150 m/s
𝜌 = 1.29 Kg/m3

Required

W =?

Solution
1 1
(P1 – P2) = 2 𝜌V2 2 – 2 𝜌V1 2
1
(P1 – P2) = 2 𝜌 (V2 2 − V1 2 )
F
∆P =
A

F 1
= 2 𝜌 (V2 2 − V1 2 )
A

1
F = 2 𝜌A (V2 2 − V1 2 )

F=W
1
W = 2 𝜌A (V2 2 − V1 2 )
1
W = 2 (1.29)(50)((150)2 − (140)2 )
1
W = 2 (64.5) (22500 – 19600)
1
W = 2 (187050)

W = 93525 N

9. A liquid flow through a pipe with a diameter of 0.50 m at a speed of 4.20 m/s. What is the rate of flow
in L/min?

Given

d = 0.50 m
d 0.50
r=2= 2
= 0.25 m

Required
V
=?
t

Solution
V
= Av
t

A = 𝜋𝑟 2
V
= 𝜋𝑟 2 V
t

V
= (3.142)(0.25)2 (4.20)
t
V
= (3.142) (0.625) (4.20)
t

V
= 0.8247 m/s2
t

1 m3 = 1000 litre

1 s = 1/60 mins
V
= 0.8247 × 1000 × 60
t

V
= 49486.5 litre/min
t

10. Calculate the average speed of blood flow in the major arteries of the body, which have a total cross-
sectional area of about 2.1 cm2. Use the data of example

Given

Aartries = A2 = 2.1 cm2

raorta = r1 = 1.2 cm

vaorta = 40 cm/s

Required

Vartries =?

Solution

A1V1 = A2V2
A1V1
V2 = A2

πr12 V1
V2 = A2

3.142×1.22 ×40
V2 = 2.1

181
V2 = 2.1

V2 = 86.2 cm/s