Extra Credit Problems
CHE 222 - Momentum Transfer
BSChE 2-2, Group 3
Acosta, Deanne
Artoza, Jelo
Bacabac, Roncel
Gagelonia, Clara
Pantaleon, Gianna
Schmitt, Fitz
Submitted to:
Peniel Jean A. Gildo
Department of Chemical Engineering
Pamantasan ng Lungsod ng Maynila
June 30, 2021
Group 3
Extra Credit Problems
Problem 1
A 0.5-lbm flat plate that has a dimension of 4’ by 5’ rests on an inclined plane with elevation
of 35◦ with the horizontal. Between the flat plate and the inclined plane, is a 0.67-in layer
of hexane. The system is at 1 atm and 86 ◦ F. Using standard AES units, determine:
(a) The viscosity of the fluid
(b) The shear rate
(c) The shear stress
(d) The momentum flux (use momentum units and specify the direction of momentum
transfer)
(e) The velocity (magnitude and direction) of the speed of the plate
1
Givens
Unknowns
(a) µ, fluid viscosity
(b) dv/dy, shear rate
(c) τ , shear stress
(d) τm , momentum flux, and its direction
(e) |v|, speed of the plate, and its direction
1
Viscosity: Handbook, 2-274
1
Group 3
Extra Credit Problems
Plan
1. Calculate the shear force F by finding the force exerted on the plate parallel to the
layer of fluid,
2. Calculate the shear stress τ using the formula
τ=
F
A
3. Calculate shear rate du/dy, and the speed of the plate v using the formula
du
τ
= .
dy
µ
Use the diagram to find the direction of the velocity.
4. Calculate τm using the fact that its magnitude is equal to that of the shear stress.
5. FInd the direction of τm using the fact that x-direction momentum is transferred in a
direction perpendicular to the fluid layer all the way to the part of the inclined plane
bounding the fluid.
Calculations
Calculation of F
force exerted on plate
F =
parallel to layer of fluid
mg sin 35◦
=
gc
lbm − ft
= (0.5)(sin 35◦ )
s2
= 0.286788 lbf
Calculation of τ
F
A
0.286788 lbf
=
(4 ft)(5 ft)
= 0.0143394 lbf /ft2
τ=
2
Group 3
Extra Credit Problems
Calculation of du/dy, v, and θ
du
τ
=
dy
µ
du
0.0143394 lbf /ft2
=
dy
1.89 × 10−4 lbftf2-s
du
= 75.869899 s−1
dy
du = (75.869899 s−1 )dy
Z 0
Z v
−1
du = (75.869899 s )
dy
0
−0.67
0.67
−1
v = (75.869899 s )
ft
12
v = 4.236069 ft/s
(Shear rate)
(Speed)
The plate will slide down the inclined plane in a direction parallel to the inclined surface.
So, the direction of the velocity is 35◦ below the horizontal.
Calculation of τm and its direction
|τm | = |τ |
= 0.0143394 (lbm − ft/s)/ft2 − s
The momentum flux is normal to the direction of the flow of the fluid, and starts at the plate
in contact with the fluid and ends at the inclined surface. Based on the coordinate system,
it is 55◦ below the horizontal.
3
Group 3
Extra Credit Problems
Answers
(a)
µ = 1.89 × 10−4
lbf -s
ft2
(b)
du
= 75.869899 s−1
dy
(c)
τ = 0.0143394 lbf /ft2
(d) Momentum flux is normal to the fluid layer; starts at the plate, and ends at the
inclined surface
τm = 0.0143394 (lbm − ft/s)/ft2 − s
(e) Velocity is parallel to the fluid layer
v = 4.236069 ft/s,
Problem 2
A new planet is being studied as one of the candidates for the new home of humans. It
was found that the temperature at the ground is 50°C and the pressure is 20 kPa. The
temperature drops 10°C every 2.5 km of elevation and at an elevation of 15.0 km, a new
layer of atmosphere exists with a constant temperature. The atmosphere is made of 60%
nitrogen, 20 %oxygen, and 20% carbon dioxide. Starting with the equation for hydrostatic
equilibrium, determine:
(a) an equation describing the pressure at any elevation.
(b) the pressure at an elevation of 10 km.
(c) the elevation at which the pressure is 10 kPa abs.
Givens
T0 = 50 ◦ C
P0 = 20 kPa
z = 15.0 km
α = 10 ◦ C / 2.5 km
4
Group 3
Extra Credit Problems
Universal Gas Constant:
R = 8314
m2 -Pa
kmol-K
Composition of Atmosohere:
60% Nitrogen
20% Oxygen
20% Carbon Dioxide
Molar Mass:
M(N2 ) = 28.02 kg/kmol
M(O2 ) = 32.00 kg/kmol
M(CO2 ) = 44.01 kg/kmol
Unknowns
(a) Equation describing the pressure at any elevation
(b) The pressure, P , at an elevation of 10 km (Pa)
(c) The elevation, z, when P = 10 kPa abs(m)
Plan
1. Determine the molar mass of the atmosphere
Matmosphere = (xM )Nitrogen + (xM )Oxygen + (xM )Carbon Dioxide
2. Determine the equation describing the pressure at any elevation by using the equation
of hydrostatic equilibrium.
dP
g
=− ρ
dz
gc
According to the ideal gas equation,
PM
ρ=
RT
When temperature varies linearly
T = T0 − α(z − z0 )
3. Using the derived equation, solve for the pressure at an elevation of 10 km. Use the
given values of T0 , P0 , α, and R, and the calculated molar mass of the atmosphere.
Assumption:
Pressure decreases with increasing altitude, therefore
P10
km
< P0
4. Use the derived equation in order to determine the value of elevation, z, when pressure
is 10 kPa abs. Use the given values of T0 , P0 , α, and R.
Assumption:
z must be within 15 kilometers
0 < z < 15 km
5
Group 3
Extra Credit Problems
Calculations
Solving for the molar mass of the atmosphere:
Matmosphere = (xM )Nitrogen + (xM )Oxygen + (xM )Carbon Dioxide
kg
kg
kg
Matmosphere = (0.60) 28.02
+ (0.20) 32.00
+ (0.20) 44.01
kmol
kmol
kmol
Matmosphere = 32.014 kg/kmol
Determining the Equation Describing the Pressure at any Elevation
Z
P
P0
dP
dz
dP
dz
dP
dz
dP
P
dP
P
g
ρ
gc
gP M
=−
RT
=−
gP M
R[T0 − α(z − z0 )]
gM
dz
=−
R[T0 − α(z − z0 )]
Z
gM z
dz
=−
R z0 T0 − α(z − z0 )
=−
Let u = T0 − α(z − z0 ). Then du = −α dz, and
gM
ln P − ln P0 =
[ln(T0 − α(z − z0 )) − ln(T0 − a(z0 − z0 ))]
αR P
gM
T0 − α(z − z0 )
ln
=
ln
P0
αR
T0
gM
P
T0 − α(z − z0 ) αR
ln
= ln
P0
T0
gM
P
T0 − α(z − z0 ) αR
=
P0
T0
gM
T0 − α(z − z0 ) αR
P = P0
T0
6
Group 3
Extra Credit Problems
Solving for the pressure at an elevation 10 km
gM
T0 − α(z − z0 ) αR
P = P0
T
0
1000 kPa
= (20 kPa)
···
1 kPa
···
 ◦
 50 C −
10 ◦ C
m [(10
[2.5 km][ 1000
1 m ]
km)
1000 m
1 m
kg
9.81 m
( 32.014
)
kmol

s2
◦C
10
1000
m
m3 -Pa
− 0 m]  ( 2.5 km )( 1 m )(8314 kmol-K
)
50 ◦ C


= 5.0144 × 10−3 Pa
Solving for the elevation at which the pressure is 10 kPa abs
gM
T0 − α(z − z0 ) αR
P = P0
T
0
1000 Pa
1000 Pa
(10 kPa)
= (20 kPa)
···
1 kPa
1 kPa
···
 ◦
 50 C −
10 ◦ C
m
[2.5 km][ 1000
1 km ]
kg
9.81 m
( 32.014
)
kmol

s2
◦C
10
m3 -Pa
1000
m
[z − 0 m]  ( 2.5 km )( 1 km )(8314 kmol-K
)
50 ◦ C


z = 884.6180 m, or 0.8846 km
Answers
(a) The equation describing the pressure at any elevation
P = P0
T0 − α(z − z0 )
T0
gM
αR
(b) The pressure at an elevation of 10 km.
P = 5.0144 × 10−3 Pa
(c) The elevation at which the pressure is 10 kPa abs
z = 884.6180 m, or 0.8846 km
7
Group 3
Extra Credit Problems
Problem 3
The heat transfer coefficient of a flowing fluid determines the rate at which heat is transferred
in a flowing pipe. Develop the functional relationship of the heat transfer coefficient h with
units W/(m2 -K) with thermal conductiviy k, units W/m-k, diameter D, density ρ, average
velocity v, viscosity µ, and the specific heat Cp , units J/(kg-K).
Givens
h - Heat transfer coefficient
k - Thermal conductivity
D - Diameter
ρ - Density
v - Average velocity
µ - Viscosity
Cp - Specific heat
Unknowns
(a) π1 , π2 , π3
(b) Functional relationship of h with k, D, ρ, v, µ, and Cp
8
Group 3
Extra Credit Problems
Plan
1. Determine the units of each variable and convert it to basic dimensions
h=
=
=
=
=
W
m2 -K
J
h-m2 -K
kg-m2
s2 -h-m2 -K
M L2
t3 L2 T
M
t3 T
m
s
L
=
t
kg
m3
M
= 3
L
v=
W
m-K
M L2
= 3
t LT
ML
= 3
tT
µ=
ρ=
k=
D=L
kg
m-s
M
=
Lt
J
kg-K
M L2
= 2
t MT
L2
= 2
tT
Cp =
2. Determine the total number of variables n, the number of basic dimensions m, and
number of dimensionless groups π-groups.
N = 7 groups
M = M, L, t, T = 4 groups
π = 7 − 4 = 3 π groups
3. Select the repeating variables. Note that the number of repeating variables is the same
with the number of basic dimensions m.
Repeating variables:
ρ−
v−
D−
Cp −
fluid property
flow property
geometric property
heat property
4. In solving every π-term, assign an exponent variable to each property and multiply all
of them with a non-repeating variable.
π = ρa v b Dc Cpd
9
Group 3
Extra Credit Problems
The variables not chosen as repeating variables are the non-repeating variables. In this
problem, we have h, k, and µ. Multiply π to any non-repeating variables, one by one.
Therefore, we will have 3π-terms.
5. Solve for π1 , we can multiply h to π.
π1 = ρa v b Dc Cpd h
6. Substitute the corresponding basic units to each variable in the π-term
π1 = ρa v b Dc Cpd h
a b
2 d M
L
L
M
c
=
(L)
L3
t
t2 T
t3 T
7. Solve the values of each exponent by determining the position of each basic variables.
If the variable is on the numerator, we will use (+) sign. Likewise, if the variable is on
the denominator, we will use (-) sign. In solving π1 .

M : a+1=0
a = −1



L : −3a + b + c + 2d = 0
c=0
⇒
t:
−b − 2d − 3 = 0
b = −1



T : −d − 1 = 0
d = −1
8. Replace the exponent variable from the π-term with the values solved in Step 7. Then,
arrange the variables properly.
π1 = ρ−1 v −1 D0 Cp−1 h
h
π1 =
ρvCp
9. Apply the same process to the next π–terms.
Calculations
Solving π1
π1 = ρa v b Dc Cpd h
a b
2 d M
L
L
M
c
=
(L)
L3
t
t2 T
t3 T
M:
L:
t:
T :
a+1=0
−3a + b + c + 2d = 0
−b − 2d − 3 = 0
−d − 1 = 0




a = −1
c=0
⇒
b = −110



d = −1
Group 3
Extra Credit Problems
π1 = ρ−1 v −1 D0 Cp−1 h
h
π1 =
ρvCp
Solving π2
π2 = ρa v b Dc Cpd k
2 d a b
M
L
L
ML
c
=
(L)
L3
t
t2 T
t3 T
M:
L:
t:
T :
a+1=0
−3a + b + c + 2d + 1 = 0
−b − 2d − 3 = 0
−d − 1 = 0
π2 = ρ−1 v −1 D−1 Cp−1 k
k
π2 =
ρvDCp




a = −1
c = −1
⇒
b = −1



d = −1
Solving π3
π3 = ρa v b Dc Cpd µ
a b
2 d M
L
M
L
c
=
(L)
3
2
L
t
tT
Lt
M:
L:
t:
T :
a+1=0
−3a + b + c + 2d − 1 = 0
−b − 2d − 1 = 0
−d = 0
π3 = ρ−1 v −1 D−1 Cp0 µ
µ
π3 =
ρvD




a = −1
c = −1
⇒
b = −1



d=0
11
Group 3
Extra Credit Problems
Solving the functional relationship of h with k, D, ρ, v, µ, and Cp
π1 = f (π2 , π3 )
h
k
µ
=f
,
ρvCp
ρvDCp ρvD
µ
k
,
h = ρvCp f
ρvDCp ρvD
Answers
the functional relationship of the heat transfer coefficient h with thermal conductiviy
k, diameter D, density ρ, average velocity v, viscosity µ, and the specific heat Cp is
k
µ
h = ρvCp f
,
ρvDCp ρvD
Problem 4
The open-end manometer shown below indicates that the gauge pressure of Gas A is 15.82
ft of oil. The density of the oil is the same as the one used on the multiple-fluid manometer
shown below. The conditions at which the manometer is used is 28 ◦ C.
1. The unknown h
2. The absolute pressure at points A, B, C, and D
12
Group 3
2
Extra Credit Problems
Givens
The manometer is at 28°C
Pressure at Gas A is 15 ft of oil.
Pressure exerted by Gas A is gauge.
Unknowns
(a) Unknown h
(b) Absolute pressures at A, B, C, D
(c) Density of water at 60°F and 28°C Height of each the oil and the mercury summing to
65 cm.
(d) Density of mercury at 28°C
Plan
1. Determine the pressure at point A using the given pressure of Gas A which is 15 ft of
oil.
2. Formulate the equation of the pressure at point B with its height equal to h + 0.2 m.
2
Densities: Handbook, 2-32
13
Group 3
Extra Credit Problems
3. Pressure at point B is equal to the pressure across the other side of the manometer at
the same elevation. Formulate the other equation of PB using the pressure exerted by
mercury and oil. Height of the mercury is h. Let x equal the height of oil. Therefore,
x + h = 0.65. Height of oil will be x = 0.65 − h.
4. Equate the two equations of PB to find h.
5. Calculate the respective pressures at points A, B, C and D. All of the calculated
pressures are gauge.
6. Add 1 atmosphere to each gauge pressure to find the absolute pressures at points
A, B, C and D.
Calculations
3
Densitites
kg
m3
kg
ρ of water 28 ◦ C = 996.2609 3
m
kg
◦
ρ of mercury 28 C = 13526.26 3
m
ρ of water 60 ◦ F = 999.0354
Pressure at Point A
P1 = (pgh)oil
kg
m
1m
= 999.0354 3 (0.890) 9.81 2
15 ft ·
m
s
3.2808 ft
= 39879.6552 Pa gauge
Equation of PB using the pressure exerted by the water
Height of Water = h + 0.2
PB = PA + (ρgh)water
m
kg PB = 39879.6552 Pa + 996.2609 3
9.81 2 (h + 0.2 m)
m
s
3
Water Densities: Perry’s HB, 9th ed, table 2-32; Mercury Density: Perry’s HB, 9th ed, table 2-31
14
Group 3
Extra Credit Problems
Equation of PB using the pressure exerted by mercury and oil
Height of mercury = h
Height of oil = 0.65 − h
PB = (ρgh)mercury + (ρgh)oil
m
kg m
kg 9.81 2 (h) + (0.890) 999.0354 3
9.81 2 (0.65 m − h)
PB = 13526.26 3
m
s
m
s
Equate the two equations of PB
kg m
PB = 39879.6552 Pa + 996.2609 3
9.81 2 (h + 0.2 m)
m
s
kg
m
kg m
=⇒ PB = 13526.26 3
9.81 2 (h) + (0.890) 999.0354 3
9.81 2 (0.65 m − h)
m
s
m
s
m
kg 9.81 2 (h + 0.2 m)
=⇒ 9879.6552 Pa + 996.2609 3
m
s
kg
m
kg m
= 13526.26 3
9.81 2 (h) + (0.890) 999.0354 3
9.81 2 (0.65 m − h)
m
s
m
s
kg
=⇒ 39879.6552 Pa + 9773.319429 2 2 (h + 0.2 m)
m -s
kg
kg
= 132692.6106 2 2 (h) + (8722.4781749 2 2 )(0.65 m − h)
m -s
m -s
kg
kg
=⇒ 39879.6552 Pa + 9773.319429 2 2 (0.2 m) + 9773.319429 2 2 (h)
m -s
m -s
kg
kg
kg
= 132692.6106 2 2 (h) + 8722.4781749 2 2 (0.65 m) − 8722.4781749 2 2 (h)
m -s
m -s
m -s
kg
kg
=⇒ 39879.6552 Pa + 9773.319429 2 2 (0.2 m) − 8722.4781749 2 2 (0.65 m)
m -s
m -s
kg
kg
kg
= 132692.6106 2 2 (h) − 8722.4781749 2 2 (h) − 9773.319429 2 2 (h)
m -s
m -s
m -s
kg
kg
39879.6552 Pa + 9773.319429 m2 -s2 (0.2 m) − 8722.4781749 m2 -s2 (0.65 m)
=⇒ h =
132692.6106 mkg
− 8722.4781749 mkg
− 9773.319429 mkg
2 -s2
2 -s2
2 -s2
=⇒ h = 0.3167 m or 31.67 cm
Calculate the pressures at points A, B, C, and D
15
Group 3
Extra Credit Problems
@ Point A
PA = 39879.6552 Pa + 101325 Pa
@ Point B
PB = PA + (ρgh)water
PB
PB
PB
PB
kg = 39879.6552 Pa + 996.2609 3
9.81
m
kg 9.81
= 39879.6552 Pa + 996.2609 3
m
= 44929.5294 Pa + 101325 Pa
= 146254.53 Pa
m
(h + 0.2 m)
s2
m
(0.3167 m + 0.2 m)
s2
PB = (ρgh)mercury + (ρgh)oil
kg kg m
m
PB = 13526.26 3
9.81 2 (h) + (0.890) 999.0354 3
9.81 2 (0.65 m − h)
m
s
m
s
m
kg
m
kg
9.81 2 (0.3167 m) + (0.890) 999.0354 3
9.81 2 (0.65 m − 0.3167 m)
PB = 13526.26 3
m
s
m
s
PB = 44930.9518 Pa + 101325 Pa
PB = 146255.95 Pa
Slight discrepancy can be observed, this is due to the decimal places maybe. However, the
percentage difference is insignificant. For this problem, we will use the PB exerted by the
water because it is the one indicated in the diagram.
@ Point C
PC = (ρgh)oil
kg m
PC = (0.890) 999.0354 3
9.81 2 (0.65 m − 0.3167 m)
m
s
PC = 2907.2020 Pa + 101325 Pa
PC = 104232.20 Pa.
@ Point D
Since it is stated that kind of apparatus used is an open-ended manometer. Point D is
atmospheric pressure.
PD = 101325 Pa
16
Group 3
Extra Credit Problems
Answers
(a)
h = 0.3167 m
(b)
Point A = 141164 Pa
B = 145989.05 Pa
C = 106988.83 Pa
D = 101325 Pa
Problem 5
Water at 25◦ C is to be transported from a lake via a pump to the top of a tank. A 6 in. Sch
40 steel pipe is used as a suction line while the discharge line is a 4 in Sch 40 pipe. In the
5 in pipe, there are 3 ells, 1-20◦ butterfly valve and a swing check valve. In the 4 in. pipe,
there are 3 ells and 2 gate valves. The pipe is then reduced to a 2 in. Sch. 40 pipe prior
to the tank at which the flow rate is 8.7 kg/s and the height of the discharge relative to the
water level in the lake is 60 m. The pipe lengths are 12 m, 25 m, and 10 m for the 6 in., 4
in., and 2 in. pipes, respectively. The pump has an efficiency of 57%. Calculate:
1. the velocities of the fluid at the different pipe sections
2. the pressure developed by the pump
3. the work imparted by the pump per kg of water flowing
4. the power required by the pump in hp.
4
Givens
T = 25 ◦ C
ρ = 997.08 kg/m3
µ = 0.8937 cP
4
= 0.0457 mm
∆z = 60 m
Geankoplis 5th ed: T A.2-4-Viscosities; T A.2-3-Densities
17
ṁ = 8.7 kg/s
η = 0.57
Group 3
5
Extra Credit Problems
6
Internal Diameters:
6 in. Sch 40 - 6.065 in.
4 in. Sch 40 - 4.026 in.
2 in. Sch 40 - 2.067 in.
Fittings:
Ell - 0.75
20◦ Butterfly valve - 1.54
Swing check valve - 2.0
Gate valve - 0.17
Lengths:
6 in. Sch 40 - 12 m.
4 in. Sch 40 - 25 m.
2 in. Sch 40 - 10 m.
Unknowns
(a) v2 , v4 , v6 the velocities of the fluid at the different pipe sections
(b) ∆p, the pressure developed by the pump
(c) Wp , the work imparted by the pump
(d) Ẇp , power required by the pump in hp.
5
6
Geankoplis 5th ed. T A.5-1 pg. 1688-1689
Handbook 9th ed. T 6-5 pg.6-17
18
Group 3
Extra Credit Problems
Plan
1. Use the mass flow rate to solve for the velocities at different pipe sections.
ṁ = ρvA
2. Solve for the Reynolds Number in each pipe.
Re =
ρvD
µ
3. Look for the value of surface roughness then solve for the Fanning Friction factor in
each pipe using the Churchill Equation.
1
7 0.9
√ = −4 log 0.27 + ( )
D
Re
f
4. Calculate the total friction losses in each pipe
X
F = Ff + hexp + hcon + hfit
5. Use the Mechanical Energy Balance to solve for the Shaft work Ws
∆p
∆v 2
g∆z X
Ws =
+
+
+
F
ρ
2αgc
gc
Since both point 1 and point 2 is exposed to the atmosphere, the difference of the two
pressure is 0.
Calculations
Velocities
kg
8.7
=
s
kg
997.08 3
m
π
4
π
4
π
4
0.0254 m
6.065 in
1 in
2 !
0.0254 m
4.026 in
1 in
2 !
0.0254 m
2.067 in
1 in
2 !
v6
v6 = 0.4681 m/s
kg
=
8.7
s
kg
997.08 3
m
v4
v4 = 1.0624 m/s
kg
8.7
=
s
kg
997.08 3
m
v2 = 4.0304 m/s
19
v2
Group 3
Extra Credit Problems
Reynolds Number
Re =
ρvD
µ
997.08
kg
m3
m
0.4681 ms 6.065 in 0.0254
1 in
Re6 =
= 80452.84568
0.8937 × 10−3 Pa-s
m
997.08 mkg3 1.0624 ms 4.026 in 0.0254
1 in
Re4 =
= 121208.7008
0.8937 × 10−3 Pa-s
m
997.08 mkg3 4.0304 ms 2.067 in 0.0254
1 in
= 236080.7578
Re2 =
0.8937 × 10−3 Pa-s
Fanning Friction Factor
"
"
0.9 #
0.9 #
1
7
0.0457 × 10−3 m
7
√ = −4 log 0.27 +
= −4 log 0.27
m +
D
Re6
80452.84568
6.065 in 0.0254
f6
1 in
⇒ f6 = 5.04307 × 10−3
"
"
0.9 #
0.9 #
7
7
0.0457 × 10−3 m
1
√ = −4 log 0.27 +
= −4 log 0.27
m +
D
Re4
121208.7008
4.026 in 0.0254
f4
1 in
⇒ f4 = 4.9249 × 10−3
"
"
0.9 #
0.9 #
1
7
0.0457 × 10−3 m
7
√ = −4 log 0.27 +
= −4 log 0.27
m +
D
Re2
236080.7578
2.067 in 0.0254
f2
1 in
⇒ f2 = 5.112619 × 10−3
20
Group 3
Extra Credit Problems
Friction Losses
X
F = Ff + hexp + hcon + hfit
X
F = (Ff,6 + hexp,6 + hcon,6 + hfit,6 ) + (Ff,4 + hexp,4 + hcon,4 + hfit,4 ) + (Ff,4 + hexp,2 + hcon,2 )
X
v62 4f6 l6
A20
F6 =
+ 0.5 1 − 2 + Kf,6
2
D6
Ai
2
(0.4681 m/s) 4(5.04307 × 10−3 )(12 m)
A20
=
+ 0.5 1 − 2 + 3(0.75) + 1.54 + 2.0
m
2
Ai
6.065 in 0.0254
1 in
= 0.8612792875 J/kg
X
v42 4f4 l4
A20
F4 =
+ 0.5 1 − 2 + Kf,4
2
D4
Ai
2
4.0262
(1.0624 m/s) 4(4.9249 × 10−3 )(25 m)
+ 0.5 1 −
+ 3(0.75) + 2(0.17)
=
m
2
2
6.065
4.026 in 0.0254
1 in
= 4.337410551 J/kg
X
A0
v22 4f2 l2
+ 0.5 1 −
F2 =
2
D2
Ai
2
(4.0304 m/s) 4(5.112619 × 10−3 )(10 m)
2.0672
=
+ 0.5 1 −
m
2
4.0262
2.067 in 0.0254
1 in
= 34.6337817 J/kg
X
F =
X
F6 +
X
F4 +
X
F2
= 39.83246801 J/kg
Shaft Work
∆p
∆v 2
g∆z X
+
+
+
F
ρ
2αgc
gc
(4.3042 )
+ (9.81)(60) + 39.83246801 J/kg
=
2
= 636.5545301 J/kg
Ws =
21
Group 3
Extra Credit Problems
Work Imparted by the Pump
Ws
η
636.5545301 J/kg
=
0.57
= 1116.762334 J/kg
Wp =
Pressure Developed by the Pump
∆p
∆v 2
g∆z X
+
+
+
F = Ws
ρ
2αgc
gc
1.06242 − 0.46812
∆p
+
= 636.5545301 J/kg
2(1)(1)
997.08 mkg3
∆p = 634.2423 kPa
Power Required by Pump
Ẇp = Wp × ṁ
= (1116.762334)(8.7)
1 kW
1000 W
= 13.02914 hP
22
1 hP
0.74570 kW
Group 3
Extra Credit Problems
Answers
(a) Fluid Velocities
v6 = 0.4681 m/s
v4 = 1.0624 m/s
v2 = 4.0304 m/s
(b) Pressure developed by pump
∆p = 634.2423 kPa
(c) Work imparted by pump
Wp = 1116.762334 J/kg
(d) Power required by the pump
Ẇp = 13.02914 hP
Problem 6
Cyclohexane at 30°C drains from a large reservoir where the liquid level is essentially constant
at 15 ft. At the bottom of the tank, a 3.25-inch drawn tubing drains the fluid until a depth
of 6 ft. An elbow is then encountered followed by the pipe for 50 more feet. In the horizontal
pipe two gate valves are present. Another elbow is encountered followed by a vertical pipe
for 2.5 ft. Finally, another elbow is countered followed by 40 ft of pipe until discharged to
the atmosphere. Calculate:
(a) the linear velocity of the pipe
(b) total frictional losses
(c) the constant mass flow rate of the cyclohexane
Givens
Cyclohexane at 30 ◦ C
Cyclohexane is drained until 6 ft
Size of the tube all throughout is 3.25-inch
2 gate valves on the 50 ft tube
3 elbows (9 ft and 50 ft, 50 ft and 2.5 ft, 2.5 ft and 40 ft)
23
Group 3
Extra Credit Problems
Unknowns
(a) v, linear velocity
P
(b)
F , summation of friction
(c) ṁ, constant mass flow rate
(d) Density and viscosity of hexane at 30 ◦ C
Plan
1. Determine the density and viscosity of cyclohexane at 30 ◦ C
2. There were no givens from which we can obtain the velocity, and without the velocity
we cannot solve the other unknowns, therefore we will do a trial-and-error method.
3. Let us start by simplifying all the working equations that is needed to determine
velocity for easier solving by using what is given to us. First working equation is the
Reynolds number, followed by the fanning friction factor, and lastly the mechanical
energy balance.
4. After simplifying all of the equations, we would do an iteration starting with an initial
guess. The initial guess will be used to determine the Reynolds number, then the
fanning friction factor, and then with the mechanical energy balance we would get an
approximate linear velocity that would be the used on the second iteration.
5. Iteration will continue until; the value of the approximate linear velocity is constant.
24
Group 3
Extra Credit Problems
6. When linear velocity is obtained, we can then determine the Summation of friction
losses, and the constant mass flow rate.
Calculations
7
Density and Viscosity of Hexane:
ρ @ 30 ◦ C = 47.9854 lbm /ft3
µ @ 30 ◦ C = 5.5164 × 10−4 lbm /ft-s
Reynolds Number
3.25 in · 121 ftin (v) 47.4329
ρvD
Re =
=
m
µ
5.5164 × 10−4 lb
ft-s
lbm
ft3
Re = 23558.9258v
Fanning Friction Factor
1
Churchill Equation: √ = −4 log
f
"
0.27
D
+
0.0457 × 10−3
= 5.536 × 10−4
=
m
D
3.25 in 0.0254
1 in
0.27
= 1.4947 × 10−4
D
"
0.9 #
1
7
−4
√ = −4 log 1.4947 × 10 +
Re
f
(
"
0.9 #)−2
7
f = −4 log 1.4947 × 10−4 +
Re
7
Handbook, 9e, table 2-32
25
7
Re
0.9 #
Group 3
Extra Credit Problems
Linear Velocity
∆L = (6 + 50 + 2.5 + 40) ft
∆L = 98.5 ft
X
∆P
v2
g
+ 2 − ∆z +
F = Ws
ρ
2αgc gc
2
X
vave
g
+
F = ∆z
2αgc
gc
(Mechanical Energy Balance)
v2
∆L
A0
F =
4f
+ 0.5 1 −
+ 2(0.17) + 3(0.75)
2αgc
D
Ai

"
!
#
2
2
X
ft
v
98.5
ft
 4f
+ 3.09 2
F =
1 ft
s
m −ft
3.25
in
·
2(1) 32.174 lb
12 in
lbf −s2
X

2
vave
2αgc
+
v2 =
4f
r
v=
v
"
2
2(1) 32.174
lbm −ft
lbf −s2
 4f
98.5 ft
3.25 in · 121 ftin
1512.178
98.5 ft
1
3.25 in ( 12
ft
in
)
+ 4.09
1512.178
1454.7692f + 4.09
Iterations
26
!
ft2
+ 3.09 2
s
(Friction Losses)
#
= 23.5
lbf − ft
lbm
Group 3
Extra Credit Problems
v = 11.78 ft/s

X
X
ft 2
s
11.78
F =
2(1) 32.174
F = 21.3446
 4(4.683 × 10−3 )
98.5 ft
3.25 in · 121 ftin
!
ft2
+ 3.09 2
s
#
lbf − ft
lbm
ṁ = ρvA =
lbm −ft
lbf −s2
"
lbm 3
47.9854
ft
ft
11.7770
s
2 !
π
1 ft
· 3.25 in ·
4
12 in
ṁ = 32.5565 lbm /s
Answers
(a)
v = 11.78 ft/s
(b)
X
F = 21.3446
lbf − ft
lbm
(c)
ṁ = 32.5565 lbm /s
Problem 7
A laboratory scale glass tube set-up is used to measure the pressure drop of flow using pure
ethanol. The fluid flows at a constant rate and temperature of 50 kg/hr and 35 ◦ C. The tube
is 8.9 mm ID and a gauge pressure measuring device is connected to two points in the tube
which are 1.87 m apart. Calculate:
(a) The Reynolds number and nature of the flow
(b) The frictional losses between the two points of measurement.
(c) The pressure drop to be expected in ft of H2 O.
27
Group 3
Extra Credit Problems
Givens
Material:
Glass Tube
= 0.00152 mm ID = 8.9 mm
8
Fluid
ṁ = 50 kg/hr
T = 35 ◦ C
ρ(Pure Ethanol @35 ◦ C) = 776.41 kg/m3
C2
C5
µ = exp C1 +
+ C3 ln(T ) + C4 T
T
C1 = 7.875
C2 = 781.98
C3 = −3.0418
781.98
µ = exp 7.875 +
+ −3.0418 ln(308.15)
308.15
= 8.9504 × 10−4 Pa-s
Unknowns
(a) Re and Nature of Flow
P
(b)
F , friction losses
(c) ∆P , pressure drop in ft of H2 O
8
Density: Handbook, Table 2-112; Viscosity: Handbook, Table 2-313
28
Group 3
Extra Credit Problems
Calculations
Velocity of the Fluid
1 hr
50 kg
× 3600
ṁ
hr
s
=
kg
ρ
776.41 m3
i
hπ
· (8.9 mm × 10−3 )2
q = Av = v
4
v = 0.2876 m/s
q=
The Reynolds Number and Nature of Flow
776.41
ρvD
Re =
=
µ
kg
m3
0.2876 ms 8.9 mm ×
8.9504 × 10−4 Pa-s
1 m
1000 mm
= 2220.38, Transition Flow
The frictional losses between the two points of measurement
X
F = 4f
∆L v 2
v2
v2
v2
+ Kexp
+ Kcon
+ Kfittings
D 2gc
2αgc
2αgc
2αgc
Laminar
16
16
=
= 7.2060 × 10−3
Re
2, 220.3813
X
1.87 m
−3
F = 4(7.2060 × 10 )
1 m
8.9 mm × 1000
mm
f=
(0.2876 m/s)2
)
2(1.0)(1.0 kg-m
N-s2
!
= 0.2505 J/kg
Turbulent
0.00152 m
1
1.256
√ = −4 log
√
+
(3.7)(8.9 mm) 2220.3813 f
f
f = 0.01199
!
X
1.87 m
(0.2876 m/s)2
F = 4(01199)
= 0.4168 J/kg
1 m
8.9 mm × 1000
2(1.0)(1.0 kg-m
)
2
mm
N-s
The pressure drops to be expected in ft H2 O
Laminar
∆p = ρ × (−
X
kg
J
F) =
776.41 3
−0.2505
+ 101325
m
kg
= 33.96 ft H2 O
29
1 atm
33.90 ft H2 O
×
N
1 atm
101325 m2
!
Group 3
Extra Credit Problems
Turbulent
∆p = ρ × (−
X
kg
J
F) =
776.41 3
−0.4168
+ 101325
m
kg
1 atm
33.90 ft H2 O
×
N
1 atm
101325 m2
= 34.01 ft H2 O
Answers
(a)
Re = 2220.38, Transition Flow
(b)
X
F = 0.2505 J/kg
(Laminar)
X
F = 0.4168 J/kg
(Turbulent)
(c)
∆p = 33.96 ft H2 O
∆p = 34.01 ft H2 O
(Laminar)
(Turbulent)
Problem 8
Water 25◦ C enters a reducing bend horizontally at a rate of 10 kg/s. It then comes out at an
angle 60◦ from the horizontal and is ejected to the atmosphere. The pressure at the entrance
of the bend is 28 ft of oil (SG = 0.89) gauge. The diameter of the fitting to which water
enters corresponds to a 3 in. Sch. 80 and the outlet corresponds to a 1 in. Sch 40 pipe.
Calculate:
a. the velocities of the fluid entering and exiting the bend;
b. the resultant force and direction that the bend resists at steady operaton.
Givens
30
!
Group 3
Extra Credit Problems
Inlet
Outlet
Fluid
pi = 28 ft of oil
3 in. Sch. 80
9
Ai = 42.61 × 10−4 m2
αi = 0◦
po = 0 gauge
1 in. Sch. 40
10
Ao = 5.574 × 10−4 m2
α = 60◦
T = 25◦ C
11
ρ = 997.08 kg/m3
ṁ = 10 kg/s
Assumptions
Fxs , Fys = 0
Fyg = 0
Since water enters at a rate of 10 kg/s, it follows by the law of conservation of mass that for
any cross section of the reducing bend, particularly the outlet, water will also pass through
the section at a rate of 10 kg/s. So, similar to the density, the mass flow rate of the water
is constant at any section. Because the water is ejected to the atmosphere, the outlet gauge
pressure po is zero.
Unknowns
(a) vi , vo , fluid velocities at the inlet and the outlet, respectively.
(b) RCV , The resultant force and the direction that the bend resists.
Plan
Below is the sequence of calculations that will be followed in obtaining the unknowns:
9
Geankoplis, Transport Processes and Unit Operations (3e), 892
Geankoplis, 892
11
Geankoplis, 855
10
31
Group 3
Extra Credit Problems
1. Convert pi to Pascals using the specific gravity of the oil SG = 0.89, and the equation
p=
ρgh
gc
2. Using the equation ṁ = Avρ, calculate the fluid velocities vi , vo . Note that since the
mass flow rate at the inlet is 10 kg/s, the mass flow rate at the outlet is also 10 kg/s
3. Obtain Rx , Ry the x and y components of the resultant force exerted by the solid surface
on the fluid R using the equations, which were derived from the overall momentum
balance for a control volume at steady-state with the shear forces and gravitational
forces neglected,
ṁ
∆(v cos α) + ∆(pA cos α)
β
ṁ
Ry = ∆(v sin α) + ∆(pA sin α)
β
Rx =
4. Write R in Cartesian vector form
5. Calculate the resultant force Rf that the bend resists using the fact that Rf is the
resultant force of the fluid on the solid, and that the resultant force of the fluid on the
solid is equal to −R (Newton’s third law).
6. Calculate the magnitude of Rf :
q
2
2
+ Ry,f
;
|Rf | = Rx,f
and calculate the angle θ it makes with the positive x-axis:
Ry,f
θ = arctan
Rx,f
Calculations
Conversion of p1 to Pascals
pi =
=
ρgh
gc
890 mkg3 9.81 sm2
1 kg-m
N-s2
= 74513.89905 Pa
32
28
3.2808
m oil
Group 3
Extra Credit Problems
Calculation of vi , vo
vi =
=
ṁ
Ai ρ
10 kg/s
(42.61 × 10−4 m2 ) 997.08 mkg3
= 2.353740 m/s
vo =
=
ṁ
Ao ρ
10 kg/s
(5.574 × 10−4 m2 ) 997.08 mkg3
= 17.992977 m/s
Calculation of Rx , Ry
ṁ
∆(v cos α) + ∆(pA cos α)
β
= ṁ (vo cos αo − vi cos αi ) + (po Ao cos αo − pi Ai cos αi )
= ṁ (vo cos 60◦ − vi cos 0◦ ) − pi Ai cos 0◦
v
0
− vi − pi Ai
= ṁ
2 kg
m
17.992977 m/s
= 10
− 2.353740
− (74513.89905 Pa)(42.61 × 10−4 m2 )
s
2
s
= −251.076239 N
Rx =
ṁ
∆(v sin α) + ∆(pA sin α)
β
= ṁ (vo sin αo − vi sin αi ) + (po Ao sin αo − pi Ai sin αi )
= ṁ (vo sin 60◦ − vi sin 0◦ ) − pi Ai sin 0◦
√
3
=
ṁv0
√2
3
=
(10 kg/s)(17.992977 m/s)
2
= 155.823752 N
Ry =
33
Group 3
Extra Credit Problems
Component form of R, and Calculation of Rf
R = Rx î + Ry ĵ
= −251.076239î + 155.823752ĵ
Rf = −R
= 251.076239î − 155.823752ĵ
Magnitude, and Direction of Rf
q
2
2
Rx,f
+ Ry,f
p
= (251.076239)2 + (−155.823752)2
= 295.500118 N
|Rf | =
Ry,f
θ = arctan
Rx,f
−155.823752
= arctan
251.076239
= −31.824700◦
= 31.824700◦ , below the horizontal
Answers
(a) Fluid Velocities
vi = 2.353740 m/s
vo = 17.992977 m/s
(b) Resultant force Rf
Rf = {251.076239î − 155.823752ĵ} N
|Rf | = 295.500118 N
θ = 31.824700◦ , below the horizontal
34
Group 3
Extra Credit Problems
Problem 9
Cyclohexane at 30◦ C is heated to 60◦ C using a double-pipe heat exchanger at a rate of 9.2
kg/s. The inner tube contains steam and the annulus contains the hexane. The inner pipe
is a 2.5 in. Sch. 40 pipe and the outer pipe is a 4 in. Sch 40 pipe. The pipe wall in contact
with hexane is always at 80◦ C. Calculate the pressure drop in psia per ft length of heat
exchanger.
Givens
Fluid
Ti = 30 ◦ C
Tf = 60 ◦ C
ṁ = 9.2 kg/s
Contained in the annulus
Pipes
Inner: 2.5 in. Sch. 40, 12 Dinner = 62.71 × 10−3 m
Outer: 4 in. Sch. 40, 13 Douter = 0.1023 m
Temperature of pipe wall in contact with cyclohexane: 80◦ C, constant
Assumptions
ρ will be assumed as the density of cyclohexane at its average bulk temperature in the
annulus.
Unknowns
∆p, the pressure drop of the heat exchanger (psia per ft. length of heat exchanger).
Plan
1. Calculate the equivalent diameter, Deq for the annulus in which the fluid flows using
Deq = Douter − Dinner ,
and then calculate the cross-sectional area Aeq .
12
13
Geankoplis, 892
Geankoplis, 892
35
Group 3
Extra Credit Problems
2. Follow the method of Sieder and Tate to calculate the fanning friction factor f .
3. Calculate the pressure drop ∆p using the equation
∆p = 4f ρ
∆L v 2
.
D 2gc
Note that ρ is the density of cyclohexane at the average bulk temperature Tb .
Calculations
Calculation of Deq , Aeq
Deq = Douter − Dinner
= (0.1023 − 62.71 × 10−3 ) m
= 0.03959 m
π 2
Deq
4 π
=
· 0.039592
4
= 1.231008 × 10−3 m2
Aeq =
14
Calculation of f
Ti + Tf
2
30 + 60 ◦
=
C
2
= 45 ◦ C
Tb =
Density ρb , and viscosity µb at Tb : ρb = 754.868223 kg/m3 ; µb = 6.50478 × 10−4 Pa-s
ρb Deq
v
µb
ρb Deq ṁ
=
µb ρb Aeq
ṁDeq
=
µb A
(9.2 kg
)(0.03959 m)
s
=
N-s
−4
(6.50478 × 10 m2 )(1.231008 × 10−3 m2 )
Reb =
= 454862.1932, Turbulent
14
Handbook, Densities: 2-92; Viscosities: 2-274
36
Group 3
Extra Credit Problems
The friction factor fb corresponding to Tb will be estimated by the relation involving f and
Re from the Blasius solution
0.079
Re0.25
b
0.079
=
454862.19320.25
= 3.041985 × 10−3
fb =
Viscosity µw at Tw = 80 ◦ C: µ = 5.29966 × 10−4 Pa-s.
0.17
µb
ψ=
µw
0.17
6.50478 × 10−4
=
5.29966 × 10−4
0.17
6.50478
=
5.29966
= 1.035446
fb
ψ
3.041985 × 10−3
=
1.035446
= 2.937850 × 10−3
f=
Calculation of ∆p
v=
ṁ
ρb Aeq
9.2
=
(754.868223
kg
s
kg 3
)(1.231008
m
× 10−3 m2 )
= 9.900470 m/s
∆p = 4f ρ
∆L v 2
D 2gc
kg
= (4)(2.937850 × 10−3 ) 754.868223 3
m
= 3347.165014 Pa
= 0.485465 psia
37
1
( 3.2808
m)(9.900470
(2)(0.03959 m)(1
m 2
)
s
kg-m
)
N-s2
!
Group 3
Extra Credit Problems
Answers
The pressure drop ∆p of the heat exchanger is
∆p = 0.485465 psia/ft
38