EKC 212
FLUID FLOW FOR CHEMICAL
ENGINEERING
CHAPTER 8 (SOLUTION WIP EXERCISE):
TRANSPORTATION SYSTEM &
FLUID METERING
Dr Mohd Azmier Ahmad
Tel: +60 (4) 5996459
Email: [email protected]
1. Benzene at 37.8oC is pumped through the system of Fig. 1 at the rate of
0.012 m3/min. The reservoir is at atmospheric pressure. The gauge pressure
at the end of the discharge line is 345 kN/m2. The discharge is 4.48m & the
pump suction 2.12m above the level in the reservoir. The discharge line is
0.5-in. Schedule 40 pipe. The friction in the suction line is known to be 5.45
kN/m2 & that in the discharge line is 27.9 kN/m2. The mechanical efficiency
of the pump is 0.75 (75%). The density of benzene is 865 kg/m3, and its
vapour pressure at 37.8oC is 26.2 kN/m2.
Fig. 1
Calculate (a) Developed head of the pump;
(b) Total power input.
(c) If the pump manufacturer specifies a required NPSH of 3.05 m,
will the pump be suitable for this service?
Answer:
a)
The upstream station a is at the datum height.
∴V a ' , Z a ' = 0
The velocity at point a is negligible because of the larger
diameter of the tank in comparison with that of the pipe.
V b is found from Appendix 5.
For a 0.5 in Sch. 40 pipe, a velocity of 1 ft/s (0.3048 m/s) corresponds to
a flowrate of 0.945 gal/min (3.578x10-3m3/min),
0.012m 3 / min
∴V b =
Χ 0.3048m / s = 1.022m / s
−3
3
3.578 x10 m / min
Also α b = 1.0 (turbulent)
Pump work, Wp is found using Bernoulli Equation.
2
gZ
P
α V b'
W pη =
+ b' + b'
+ h f − a'
gc
ρ
2gc
ρ
Pb '
(101.3 + 345)Χ 10
Wη =
3
N / m2
865kg / m 3
p
1.022 2
+ (9.81N / kg )(4.48m) +
2(1)
(5.45 + 27.9) Χ103 N / m 2 101.3Χ103 N / m 2
+
−
= 481.87mN / kg = ΔH
3
3
865kg / m
865kg / m
b). The mass flow rate is
m = ρ v = (865 kg/m3) (0.0122 m3/min) (1min/60s) = 0.173 kg/s
The power input is :
PB =
mΔH
η
=
(0.173)(481.87)
= 111.15W
0.75
c) Use Eq. 8.7 for NPSH
pi
ρ
=
3
2
101.3 x10 N / m
= 117.1 J / kg
3
865kg / m
NPSH =
gc
g
⎞
⎛ pi − pv
⎜⎜
− h fs ⎟⎟ − Z a
⎠
⎝ ρ
Vapor pressure correspond to a head,
26 . 2 x10 3 N / m 2
=
= 30 . 2 J / kg
3
ρ
865 kg / m
pv
The friction in the suction line,
5 .45 x10 3 N / m 2
= 6 .3 J / kg
h fs =
3
865 kg / m
The value of NPSH,
NPSH =
1
(117 .1 − 30 .2 − 6 .3 ) − 2 .12 = 6 .09 m
9 . 81
The available NPSH is considered smaller than the minimum required
value of 6.25m, so the pump is not suitable for the proposed service.
2. Give 4 considerations for pipe selection.
Answer:
(i) Costs of piping (ii) Power (iii) Maintenance
(iv) Pipe stocking
(v) Pipe fittings
3. What type of joining is needed if the pipes have :
(i) thick wall (ii) thin wall and (iii) pipe of brittle materials
Answer:
9 Thick-walled : screwed fittings by flanges or welding
9 Thin-walled : by soldering or compression fittings
9 Pipe of brittle materials (carbon or cast iron): by flanges
4. Sketch the schematic diagram of (a) orifice meter, (b)
venturi meter and (c) pitot tube.
Answer:
(a) orifice meter
(b) venturi meter
1. (c) pitot tube
5. Sketch 3 types of valve plugs with their characteristics.
Answer:
6. Sketch (i) direct acting and (ii) reverse acting of actuator.
Answer:
7. Positive displacement pump force the liquid by
(i)…………………. It can be classified into two general
categories; (ii)……………….and (iii) …………………
Answer :
(i)Changing volume
(ii)Reciprocating pump
(iii)Rotary pump
8. (i)…………, (ii) ………. and (iii)…………are examples
of reciprocating pump.
Answer :
Piston pump, plunger pump and diaphragm pump
9. Centrifugal pump adds momentum to the fluid by means of
………………………….
Answer:
Fast moving blades (centrifugal force)
10. Power lost in the pump are due to the (i)……………, (ii)
…………………, (iii) ……………and (iv)……………………
Answer :
1. Fluid friction which result from the mechanical energy
conversion into the form of heat.
2. Leakage which reduces the volume of the discharge.
3. Disk friction occurs between the outer surface of the
impeller & the liquid within the casing.
4. Bearing losses occur from the mechanical friction in the
bearing & stuffing boxes of the pump.
12. Show that from continuity equation, the fluid pass over
the venturi meter can be expressed as:
C D ρA1 A2
G=
Answer:
From continuity equation:2
Δu 2
+ gΔZ + v ∫ dP + WP + F = 0
2α
1
WP , F , ΔZ = 0 introducing
correction factor, α
u
2
2
2α 2
−
2
1
2
u
= −v ∫ dP ------------ (8.9)
2α 1
1
mass flowrate is constant, G = ρ 1u1 A1 = ρ 2 u 2 A2
G=
u1 A1 u2 A2
=
v1
v2
------------ (8.10)
Where v is specific volume (ρ-1)
A −A
2
1
2
2
2 ghv
For incompressible fluid, v1 = v2
⎛ A2 ⎞ --------- (8.11)
∴ u1 = u 2 ⎜⎜ ⎟⎟
⎝ A1 ⎠
Substitute (8.11) into (8.9)
u 22
⎡ α 2 A22 ⎤
= v (P1 − P2 )
⎢1 −
2 ⎥
2α 2 ⎣ α 1 A1 ⎦
∴ u2 =
2α 2 v (P1 − P2 )
α 2 A22 ---- (8.12)
1−
α 1 A12
Substitute (8.12) into (8.10)
G=
A2
v
2α 2 v (P1 − P2 )
------ (8.13)
α 2 A2 2
1−
α 1 A12
G =
C
D
2 v ( P1 − P 2 )
A 2 A1
2
2
A1 − A 2
v
2
Introducing coefficient of discharge, CD
to eliminate α1 & α2.
C A
∴G = D 2
v
G =
C D A2
v
G =
2v (P1 − P2 )
⎛A ⎞
1 − ⎜⎜ 2 ⎟⎟
⎝ A1 ⎠
C
2
G =
A 2 A1
2
2
A1 − A 2
⎞
⎟⎟
⎠
)
2 ρ ( ρ gh
C D ρ A1 A 2
A
2 v (P1 − P2
⎛ A 12 − A 22
⎜⎜
2
A1
⎝
D
2
1
− A
2
2
v
2 gh v
)
13. A venturi meter of 320 mm diameter at inlet and 85 mm
diameter at the throat is used for measuring the flow of water.
If the pressure drop at convergence is 160 mmH2O and water
mass flowrate is 3.8 kg/s, what is the discharge coefficient at
convergence part of venturi meter.
Answer:
A1
π
= (320x10 )
4
−3 2
= 0.0804 m
Substitute all values
into Eq. 8.17;
3.8 =
G=
A2 =
2
C D ρA1 A2
A −A
2
1
C D (1000)(0.0804)(5.67 x10 −3 )
(
0.082 − 5.67 x10 −3
C D = 0.38
)
2
2
2
π
(
85x10 )
4
2 ghv
2(9.81)(160 x10 −3 )
−3 2
= 5.67x10−3 m2
14. Water is flowing at a velocity of 0.02 m/s in a pipe of 0.25 m in diameter. In
the pipe, there is an orifice with a hole diameter of 0.15 m. What is the measured
pressure drop across the orifice? μwater = 1x10-3 kg/ms; ρwater = 1000 kg/m3.
Answer:
For incompressible fluid, u2 = u1A1 / A2 = (0.02)(0.25)2/(0.15)2 = 0.0556 m/s
Calculating NRe to determine the coefficient of orifice;
Re =
u2 d 2 ρ
μ
=
(0.15)(0.0556)(1000)
= 8340
−3
1x10
From the chart, the corresponding CD at β = 0.5 is 0.65.
Substitute in the Eq.8.21,
ρu 22
(1000)(0.0556) 2
4
2
ΔP =
(
1
−
β
)
=
(
1
−
0
.
6
)
=
3
.
18
kg/ms
2C D2
2(0.65) 2
4
15. Air at a density of 3.5 kg/m3 is flowing through a pitot tube.
The pressure difference gauge indicates a difference of 42 kg/ms2.
What is the air velocity?
Answer
u1 =
2(P2 − P1 )
ρ
2 (42 )
u1 =
= 4.9 m / s
3.5
16. (i)…………………., (ii) ……………… and (iii)
…………………….. are used for transportation of gases.
Answer:
Fan, blower, compressor