Sheffield International college Lectures and Seminars

1. 1
FC025 W5 Seminar (Solutions)
Fluids
Useful conversion factors:
Papsi
mft
76.68941
305.01
=
=
Q1) The volume of air in a typical classroom is on the order of 2 3
10 m . Multiply the mass by
the acceleration of gravity to calculate the weight.
( )( )( )3 2 2 3
1.29 kg/m 10 m 9.81 m/s 10 NW mg Vgρ 3
= = = =
Q2) ( )( )( )
2
5 2 8m
1.01 10 N/m 360 ft 160 ft 5.40 10 N
3.281 ft
F PA
 
= = × = × 
 
Q3) 2 2 2
atm g 14.7 lb/in 9.9 lb/in 24.6 lb/inP P P= + = + = kPa6.169=
Q4 a)
F mg
A
P P
= =
tire tire4
4
mg mg
A A A
P P
= = ⇒ =
( )
( )( )5
2
2
2
tire
1.01 10 Pa2
14.7 lb/in
1320kg 9.81 m/s
0.0135 m
4 35.0 lb/in
A
×
= =
b) Since the area and pressure are inversely proportional, as the pressure increases the
area of contact decreases.
c)
tire4
mg
P
A
=
( )
( )( )
( )
2 2
5 2
2 5
2 1 m
100 cm
1320kg 9.81 m/s 14.7 lb/in
2.79 10 Pa 40.6 lb/in
1.01 10 Pa4 116cm
P
 
= = × = 
× 
Pa5
108.2 ×=

2. 2
Q5)
a)
g
PP
hghPP at
at
ρ
ρ
−
=⇒+=
( )( )
3 5
3 2
116 10 Pa 1.01 10 Pa
1.90 m
806 kg/m 9.81 m/s
h
× − ×
= =
b)
3 3
2 4 2
2.05 10 m
0.314 m
65.2 10 m
h
−
−
×
= =
×
total 2 1.897 m 0.314 m 2.211 mh h h= + = + =
( )( )
at
5 3 2
1.01 10 Pa 806 kg/m 9.81 m/s (2.211 m) 118 kPa
P P ghρ= +
= × + =
Q6)
at w w Hg Hg
at at w Hg2 (1.0 m )
P P gh gh
P P g d gd
ρ ρ
ρ ρ
= + +
= + − +
at
w
Hg w
(1.0 m)
P
g
d
ρ
ρ ρ
−
=
−
( )
5
3
2
4 3 3
1.01 10 Pa
1000 kg/m 1.0 m
9.81 m/s 0.74 m
1.36 10 kg/m 1000 kg/m
d
×
−
= =
× −
Q7) Hot air inside the balloon is less dense than the surrounding air (cooler air). Thus, the
difference in density causes the balloon to be lifted off the ground due to the buoyant
force created by the surrounding air.
hotairocb WWF += arg
The upward buoyant force = the weight of the cool air displaced by the balloon.
( ) ( ) ( )
( ) ( ) ( )
hotairhotairoccoolaircoolair
hotairoccoolair
hotairoccoolair
hotairoccoolair
VmV
mmm
gmgmgm
WWW
×+=×
+=
×+×=×
+=
ρρ arg
arg
arg
arg
Since the volume of the hot air = the volume of the displaced surrounding air.
3arg
arg
12.1
11430
1890
29.1 mkg
V
m
VmV
VVV
oc
coolairhotair
hotairoccoolair
hotaircoolair
=−=−=∴
×+=×
==
ρρ
ρρ

3. 3
Q8)
a)
w
0w B
B w
F T F W
F Vg W Tρ
= + − =
= = −
∑
( )( )
4 3w
3 2
w
20.0 N 17.7 N
2.34 10 m
1000 kg/m 9.81 m/s
W T
V
gρ
−− −
= = = ×
b)
( )( )
3 3
2 4 3
20.0 N
8.70 10 kg m
9.81 m/s 2.345 10 m
m W
V gV
ρ −
= = = = ×
×
Q9) ( ) ( )
1 1 2 2
2 2
1 1 2 2/ 4 / 4
A v A v
d v d vπ π
=
=
( )
2 2
1
2 1
2
3.4 cm
1.1 m/s 39 m s
0.57 cm
d
v v
d
   
= = =   
  
Q10)
2
4
m d
Av v
t
π
ρ ρ
 ∆
= =  
∆  
( ) ( )
2
3 (0.038 m)
1000 kg/m 2.1 m/s 2.4 kg s
4
m
t
π ∆
= = 
∆  
Q11)
a)
2 2
1 1 1
1 1 2 2 2 1 1 1
2 2 2
/ 4
/ 4
A d d
Av A v v v v v
A d d
π
π
   
= ⇒ = = =   
   
( )
2
1
2 1 11
12
4 4 1.4 m/s 5.6 m s
d
v v v
d
 
= = = = 
 
b)
( ) ( ) ( )
2 2
1 1 2 2
2 2
2 1 1 2
2 23 3
2
1 1
2 2
1
( )
2
1
110 10 Pa+ 1000 kg/m 1.4 m/s 5.6m/s
2
95 kPa
P v P v
P P v v
P
ρ ρ
ρ
+ = +
= + −
 = × −
 
=
Q12) ( ) smmsmghv 0.75.2)81.9(22 2
===
Q13)

4. 4
a) P1 = air pressure inside the window, P2 = pressure of the air outside the window.
( ) ( )( )( ) kPavvPPP
vPvP
64.18017029.1
2
1
2
1
2
1
2
1
22
1
2
212
2
22
2
11
=−=−=−=∆∴
+=+
ρ
ρρ
b) ( )( ) ( )( ) NAPF 25.19571042251064.18 43
=×××=∆=∆ −