UNIVERSITY OF TORONTO
FACULTY OF APPLIED SCIENCE AND ENGINEERING
MID-TERM EXAMINATION CHE324F – Process Design
Calculator Type: 2
Exam Type: A
Dr. G.W. Norval
October 6, 2014; 12:15 – 13:45
Duration: 1 hr and 30 minutes
Instructions:
1.
There are three questions – attempt all three
2.
Non-programmable calculator only
3.
Do not separate pages from the answer book
Some Useful Information
Water (16°C), μ = 0.0011 Pa s, 1 inch = 2.54 cm.
 v 2  vb2 

  W p  W f  0
  g Z a  Z b    a
2



2
( P2  P1 ) v2  v12
dH  ( Z 2  Z1 ) 

g
2g
2
DP 2 fLV
fLv 2
Wf = f =
(Fanning) =
(Darcy)
r
D
2D
 Pa  Pb

 


P  dH *  * g ;  Power  Work * mass flow
 P  Pb  Pv  ve
 
NPSH A   i
 Z e  Fe
g

 2g
Q  U A TLM
Q = 2 π k L (ΔT)/ln (ro/ri)
Q = k * A * (T/x)
t=
(P * D)
2 * (S * E + P * Y )
v2
h f = (K pipe + K1 + K 2 + K3 + ... + K n ) * ( )
2g
1)
(10 marks)
a)
What is meant by the term Sanction? (1 marks)
b)
What is meant by the term Duty? (1 marks)
c)
Does a spiral heat exchanger operate in counter-current mode? (1 mark)
d)
Explain how centrifugal and positive displacement pumps work (2 marks)
e)
The air in a room contains 2 ppm of styrene (MW = 104 g/mol). The room has dimensions of
10 m x 8 m x 3.5 m high. How many grams of styrene are in the room. (2 marks)
f)
A gaseous stream of 5% N2O4 in air (5C and 200 kPa) enters a heat exchanger. The gas is
heated to a temperature of 75C. At that temperature, the N2O4 decomposes completely to NO2.
N2O4 ® 2 NO2
Is the pressure at the exit of the heat exchanger greater than 200 kPa, equal to 200 kPa or less than 200
kPa? Explain how you know that your answer is correct. (3 marks)
2)
(10 marks)
A continuous steady-state process involves two reactions:
C6 H12 + 6 H 2O ® 6 CO +12 H 2
C6 H12 + H 2 ® C6 H14
250 kmoles of hexane and 800 kmoles of water are fed into the reactor – every hour. The yield of
hydrogen is 40%, and the selectivity ratio of the first reaction to the second reaction is 12.
Heats of formation: C6H12: -45.3 kJ/mol; H2O: -241.8 kJ/mol; CO: -110.5 kJ/mol; H2: 0 kJ/mol; C6H14:
-167.2 kJ/mol
a)
Calculate the molar flowrates of the 5 components in the product stream. (5 Marks)
b)
Is the overall process endothermic or exothermic? (3 marks)
c)
The process output is at 225 kPa and 450 K; what is the volumetric flowrate, and what is a
reasonable pipe diameter for the output pipe? (2 marks)
3)
(6 marks)
Liquid water (500 L/min) is flowing in a 3 SCH40 pipe (ID = 3.042”). The pipe is steel.
a)
Is the friction factor chart a Fanning or a Darcy chart? (1 mark)
b)
How do you know the answer to part a) (1 mark)
c)
What is the pressure drop in the pipe? (Pa per 100 m length) (2 marks)
d)
The piping network has an elevation change of +7 m. The total length of pipe is 2200 m. There
are 14 elbows (L/D = 30), plus 8 gate valves (L/D = 8); the fT for 3” pipe is 0.018. What is the total
pressure drop in the system? (2 marks)
Solutions
1)
(10 marks)
a)
What is meant by the term Sanction? (1 marks)
b)
What is meant by the term Duty? (1 marks)
c)
Does a spiral heat exchanger operate in counter-current mode? (1 mark)
d)
Explain how centrifugal and positive displacement pumps work. (2 marks)
e)
The air in a room contains 2 ppm of styrene (MW = 104 g/mol). The room has dimensions of
10 m x 8 m x 3.5 m high. How many grams of styrene are in the room. (2 marks)
f)
A gaseous stream of 5% N2O4 in air (5C and 200 kPa) enters a heat exchanger. The gas is
heated to a temperature of 75C. At that temperature, the N2O4 decomposes completely to NO2. Is the
pressure at the exit of the heat exchanger greater than 200 kPa, equal to 200 kPa or less than 200 kPa?
Explain how you know that your answer is correct. (3 marks)
a)
Sanction is the term used to define that a project is approved; the project team is authorize to
start to spend the capital dollars to purchase the equipment and the labour to build the project.
b)
Duty is what a piece of equipment is required to do. For pumps and compressors – it is the
mechanical work that the device applies to the moving fluid. For heat exchangers – it is the amount of
heat transferred.
c)
Spiral heat exchangers work in countercurrent flow – one fluid starts at the centre and spirals
outwards, the other starts at the outside and spirals inwards.
d)
Centrifugal pumps have a rotating impellor. The fluid is drawn into the eye of the impellor, and
the mechanical action (rotation) thrusts the fluid out tangentially; it is a continuous action. Positive
displacement pumps have a discontinuous action – either a piston that draws fluid in and then pushes it
out or a diaphragm pump that has an expanding and collapsing diaphragm, or a gear pump that has a
positive gear driven action.
e)
The room volume is (10 m x 8 m x 3.5 m = 280 m3). The ideal gas law gives the number of
moles of gas in the room (n = PV/RT = 101.3 kPa * 280 m3/(8.314 * 298 K) = 11.44 kmol.
The concentration is 2 ppm. This gives 2.29 mmol of styrene. The mass is 2.29 mmol * 104 g/mol =
2.38 gram
f)
The pressure at the exit of the heat exchanger is less than 200 kPa. Wherever there is a flow
system, there is pressure drop – due to the resistance to flow. The pressure at the exit of a unit must
always be lower than at the inlet. Otherwise, the gas would flow in the reverse direction.
2)
(10 marks)
A continuous steady-state process involves two reactions:
C6 H12 + 6 H 2O ® 6 CO +12 H 2
C6 H12 + H 2 ® C6 H14
250 kmoles of hexene and 800 kmoles of water are fed into the reactor – every hour. The yield of
hydrogen is 40%, and the selectivity ratio of the first reaction to the second reaction is 12.
Heats of formation: C6H12: -45.3 kJ/mol; H2O: -241.8 kJ/mol; CO: -110.5 kJ/mol; H2: 0 kJ/mol; C6H14:
-167.2 kJ/mol
a)
Calculate the molar flowrates of the 5 components in the product stream. (5 Marks)
b)
Is the overall process endothermic or exothermic? (3 marks)
c)
The process output is at 225 kPa and 450 K; what is the volumetric flowrate, and what is a
reasonable pipe diameter for the output pipe? (2 marks)
This solution can be done through creating a matrix, or by generating the linear equations.
species
Hexene
Water
CO
Hydrogen
Hexane
In (kmol/hr)
250
800
0
0
0
Change (kmol/hr)
-x-y
- 6x
+6x
+12x – y
+y
Out (kmol/hr)
(250 – x – y)
800 – 6x
6x
+12x – y
+y
Let x be the kmoles/hr of hexane converted in Rx 1, and y be the kmoles/hr of hexane converted in Rx
2.
40% of the inlet hydrogen leaves as H2 gas. But, water is the limiting reagent.
Hence
12x – y = 0.4 * ((800/6) * 6 + 800 * 1) = 640
12 x – y = 640
x:y = 12:1; hence x = 12 y
hence – 12 * (12 y) – y = 640; hence 143 y = 640; y = 4.48 kmol/hr
and x = 53.71 kmol/hr
hexene: 250 – 53.71 – 4.48 = 191.81
Water: 800 – 6 * (53.71) = 477.7
CO: 6*(53.71) = 322.3
H2: 12*(53.71)-4.48 = 640.0
Hexane: 4.48
b)
Reaction 1:
dHRx = 6 * (-110.5) + 12 * (0) – (-45.3) – 6 * (-241.8) = 833.1
Reaction 2
dHRx = -167.2 – (-45.3) – (0) = -121.9
Rx 1 is endothermic, and Rx 2 is exothermic. There are more moles reacting through Rx 1 than 2 – the
process is endothermic.
c)
Total molar flowrate out = 1636.3 kmol/hr.
Vol flowrate = n(flowrate) * (RT/P) = 1636.3 kmol/hr * (8.314 kPa m3/kmol K * 450K/225 kPa)
Vol flowrate = 27.2 x 103 cum/hr
Vol flowrate = 7.55 cum/s
Area = vol flowrate/ velocity (use 20 m/s for a gas) = 0.377 sq m
Area = (PI/4) d^2 – hence d = 0.693 m.
3)
Liquid water (500 L/min) is flowing in a 3 SCH40 pipe (ID = 3.042”). The pipe is steel.
a)
Is the friction factor chart a Fanning or a Darcy chart? (1 mark)
b)
How do you know the answer to part a) (1 mark)
c)
What is the pressure drop in the pipe? (Pa per 100 m length) (2 marks)
d)
The piping network has an elevation change of +7 m. The total length of pipe is 2200 m. There
are 14 elbows (L/D = 30), plus 8 gate valves (L/D = 8); the fT for 3” pipe is 0.018. What is the total
pressure drop in the system? (2 marks)
a)
This is a Darcy Chart
b)
rearrange the friction factor equation
dP/L = fpv^2/(2D)
c)
D = 3.042” = 0.0773 m;
ρ = 1000 kg/m3
μ = 0.0011 Pa s
Velocity: volumetric flowrate = 0.50 m3/min = 0.00833 m3/s; area = PI/4 * 0.0773^2 = 4.69 x 10-3 m2
Velocity = 1.78 m/s
Reynolds number (DVρ/μ) = 125,000
Relative roughness : ε/d = 0.1 mm/77.3 mm = .001
Friction factor = 0.024
dP = L * f * ρ * v^2/(2 * d) = 100 * 0.024 * 1000 * 1.78^2/(2 * 0.0773) = 49 kPa
d)
Pressure drop is the sum of the component issues
elevation = 7 * 1000 * 9.81 = 68.7 kPa
flow = (2200/100) * 49 = 1078 kPa
fittings = 0.018 * (14 * 30 + 8 * 8) = 8.712; factor * v^2/2g = 8.712 * 1.78^2/(2 * 9.81) = 1.41 m2/s2.
And hf * p gives 1.41 m2/s2 * 1000 kg/m3 = 1410 kgm/s2 m2 = 1.41 kPa.
The total losses are 1148 kPa.