Exercise 5.3 - Spray Dryer I (2020)

Spray dryer I
Laboratory of
Food Process Engineering
Start
Spray dryer I
The graph on the right illustrates a spray drying
system which is used for the drying of a product. The
liquid feed is atomised in small droplets, which then
are allowed to fall through hot air. The moisture
evaporates, and when moisture has evaporated, the
dissolved components become solid, and form powder
particles. The conditions of all flows can be found by
clicking on the following link: Conditions spray dryer I
Question: Make a Grassmann diagram of this system. As the solids in the product do not
change, you can leave out the standard exergy of the solids. As you need to know the exergy of
all flows to make a Grassmann diagram, the following sub questions can be defined:
a) What is the total exergy (W) of the cold and hot air? Start with question a
b) What is the total exergy (W) of the natural gas? Start with question b
c) What is the total exergy (W) of the feed and product? Start with question c
d) What is the wma (wma=kg moisture/kg dry air of moist air)? Start with question d
e) What is the total exergy of the moist air? Start with question e
Laboratory of
f) Make the Grassmann diagram. Which
Food Process operator is the least efficient?
Engineering Start with question f
Conditions spray dryer I
In this exercise the given standard conditions are: P0=1 atm, T0=293K, w0=0.008 kg/kg. However, in our Mathcad-
calculations we worked with T0 of 293.15. We are not fully consistent and accurate as for the actual temperatures
of air and product we calculated with +273 instead of +273.15
Be aware that the standard pressure is 1 atm, whereas the actual pressures in this exercise are given in bar (and
calculations accordingly). However, the differences you might find will be minimal.
Laboratory of
Back to full question
Question a: What is the total exergy (W) of the cold air General data:
(ca) and hot air (ha)? All data required to calculate the cpp 1.83 kJ/(kgK)
exergy are provided on this slide. cpw 4.18 kJ/(kgK)
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Cold air Hot air
ϕ (kg dry air/h) 122212 122212 26.142 MJ/kg
T (°C) 58.2 175 49.94 kJ/kg
P (bar) 1 1 527.2 kJ/kg
w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Ma 0.02897 kg/mol
1000 kg/m3
Bca 0W T0 293.15 K
A)
Bha 4.73 W P0 101325 Pa
Bca 7.91 W w0 0.008 kg/kg

B)
Bha 1.06 W
Bca 2.85 W
C)
Bha 3.82 W
Bca 5.80 W
D)
Bha 3.87 W
Laboratory of
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Cold air Hot air
ϕ (kg dry air/h) 122212 122212 26.142 MJ/kg
T (°C) 58.2 175 49.94 kJ/kg
P (bar) 1 1 527.2 kJ/kg
w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Ma 0.02897 kg/mol
1000 kg/m3
Bca 0W T0 293.15 K
A)
Bha 4.73 W P0 101325 Pa
Bca 7.91 W This is not correct. Bca cannot

w0 be zero,
0.008 since the air
kg/kg
B) does not have environmental temperature.
Bha 1.06 W To calculate the exergy, it is recommended to start
Bca 2.85 W with determining which forms of exergy a flow
C) consists. Do the air flows contain:
Bha 3.82 W • Chemical exergy?  No. The composition of
both air flows is the same as the composition of
Bca 5.80 W the environmental air (w=w0).
D)
Bha 3.87 W • Pressure exergy  No: pressure of air flows= P0
Laboratory of
• Thermal exergy?  Yes! The temperature of
both flows is higher than T0. As a result, the
exergy of the cold air can never be equal to 0W.
Correct! Both air flows do not contain pressure and chemical exergy, as P=P 0 and
w=w0. The air flows do contain thermal exergy, as T≠T 0. The thermal exergy (J/kg
dry air) can be calculated by combining equations for dry air and the vapour in air:
• For cold air:

• For hot air:
The total exergy (W=J/s) can be calculated with: . Thus:
• Cold air: W
• Hot air: W
Bca 0W
A)
Bha 4.73 W
Bca 7.91 W
B)
Bha 1.06 W
Bca 2.85 W
C)
Bha 3.82 W Cold air Hot air
ϕ (kg dry air/h) 122212 122212
Bca 5.80 W
D) T (°C) 58.2 175
Bha 3.87 W P (bar) 1 1
Laboratory of w (kg/kg) 0.008 0.008
Food Process Engineering Next
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Cold air Hot air
ϕ (kg dry air/h) 122212 122212 26.142 MJ/kg
T (°C) 58.2 175 49.94 kJ/kg
P (bar) 1 1 527.2 kJ/kg
w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Ma 0.02897 kg/mol
1000 kg/m3
Bca 0W T0 293.15 K
A)
Bha 4.73 W P0 101325 Pa
Bca 7.91 W w0 0.008 kg/kg

B)
Bha 1.06 W
Bca 2.85 W This is not the correct answer. In this question, we
C) want to know the exergy in Watt (=J/sec). Thus,
Bha 3.82 W to solve this question, first you should calculate
the exergy in J/kg dry air and this should be
Bca 5.80 W multiplied with the flow rate (ϕ) in kg dry air/sec
D)
Bha 3.87 W (instead of kg dry air/hr
Laboratory of
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Cold air Hot air
ϕ (kg dry air/h) 122212 122212 26.142 MJ/kg
T (°C) 58.2 175 49.94 kJ/kg
P (bar) 1 1 527.2 kJ/kg
w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Ma 0.02897 kg/mol
1000 kg/m3
Bca 0W T0 293.15 K
A)
Bha 4.73 W P0 101325 Pa
Bca 7.91 W w0 0.008 kg/kg

B)
Bha 1.06 W
This is not the correct answer. It seems you
Bca 2.85 W applied the temperatures in Celsius instead of
C) Kelvin.
Bha 3.82 W Some hints:
• To solve this question, first determine which
Bca 5.80 W form(s) of exergy both air flows consist.
D)
Bha 3.87 W • For calculating the exergy, temperatures in
Laboratory of
Kelvin (instead of Celsius) should be used.
Question b: What is the total exergy (W) of the natural General data:
gas? All data required to calculate the exergy are cpp 1.83 kJ/(kgK)
provided on this slide. cpw 4.18 kJ/(kgK)
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Nat. gas
ϕ (kg/h) 412.44 26.142 MJ/kg
Net calorific value (MJ/kg) 54.00 49.94 kJ/kg
Exergetic content (J/J) 1.04 527.2 kJ/kg

Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
A) Bgas 2.32 W T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) Bgas 6.19 W
C) Bgas 6.43 W
D) Bgas 5.62 W
Laboratory of
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Nat. gas
ϕ (kg/h) 412.44 26.142 MJ/kg

Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
A) Bgas 2.32 W T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) Bgas 6.19 W This is not the correct answer. You calculated the
exergy in J/hr.
The net calorific value is the total amount of
Bgas 6.43 W energy that is released by combustion of a fuel or
C)
gas (at environmental T and P). Thus, for natural
gas, 54 MJ of energy is produced per kg.
D) Bgas 5.62 W In this question, we want to know the exergy

(Watt) of the natural gas flow. Thus, the flow rate
Laboratory of should be expressed in kg/sec (instead of kg/hr).
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Nat. gas
ϕ (kg/h) 412.44 26.142 MJ/kg

Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
A) Bgas 2.32 W T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
This is not the correct answer. You did not convert
B) Bgas 6.19 W
from MJ to J and did not include the exergetic
content of natural gas. The net calorific value is
the total amount of energy that is released by
C) Bgas 6.43 W combustion of a fuel or gas (at environmental T
and P). Thus, for natural gas, 54 MJ (=54 J) of
energy is produced per kg. With this given, you
D) Bgas 5.62 W should be able to calculate the exergy of the
natural gas flow. Hint: do not forget to include the
Laboratory of exergetic content (Joule exergy per Joule energy)
of natural gas into your calculations.
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Nat. gas
ϕ (kg/h) 412.44 26.142 MJ/kg

Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
A) Bgas 2.32 W T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) Bgas 6.19 W Correct! The net calorific value is the total amount
of energy that is released by combustion of a fuel
or gas (at environmental T and P). Thus, for
C) Bgas 6.43 W natural gas, 54 MJ of energy is produced per kg.
Calculations for natural gas:
• flow rate = kg/sec.
• Energy: W (=J/sec)
D) Bgas 5.62 W • Exergy:
Laboratory of
cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
Nat. gas
ϕ (kg/h) 412.44 26.142 MJ/kg

Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
A) Bgas 2.32 W T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) Bgas 6.19 W
Bgas 6.43 W This is not the correct answer. You have forgotten
C)
to include the gas flow rate. The net calorific value
is the total amount of energy that is released by
combustion of a fuel or gas (at environmental T
D) Bgas 5.62 W and P). Thus, for natural gas, 54 MJ of energy is
produced per kg of natural gas. Now you can also
Laboratory of calculate the exergy of the natural gas flow.
Question c: What is the total exergy (W) of the feed and General data:
product flows? All data required to calculate the exergy cpp 1.83 kJ/(kgK)
are provided on this slide. cpw 4.18 kJ/(kgK)
cpa 1.01 kJ/(kgK)
Feed Product cps 1.84 kJ/(kgK)

ϕ (kg/h) 8534 4501 26.142 MJ/kg
T (°C) 80 90 49.94 kJ/kg
P (bar) 200 1 527.2 kJ/kg
%solids 52 97.5 Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
Bfeed 1.42 W T0 293.15 K
A)
Bproduct 1.86 W P0 101325 Pa
Bfeed 9.59 W w0 0.008 kg/kg

B)
Bproduct 1.11 W
Bfeed 8.50 W
C)
Bproduct 1.70 W
Bfeed 1.15 W
D)
Bproduct 1.52 W
Laboratory of
Correct! The exergy of the flows is calculated similarly to the exergy of the feed- and
product flows in the evaporator and conveyor dryer questions. Thus, equation 11,
given in the reader, can be used. The pressure exergy part of this equation is changed
to calculate the pressure exergy in J/kg dry product. The full equation is:
(in J/kg dry product)
Example calculation for the feed: kg water/kg dry matter.
J/kg dry product.
The total exergy: W
Bfeed 1.42 W
A)
Bproduct 1.86 W Feed Product
ϕ (kg/h) 8534 4501
Bfeed 9.59 W T (°C) 80 90
B)
Bproduct 1.11 W P (bar) 200 1
%solids 52 97.5
Bfeed 8.50 W
C)
Bproduct 1.70 W
Bfeed 1.15 W
D)
Bproduct 1.52 W
Laboratory of
cpa 1.01 kJ/(kgK)

ϕ (kg/h) 8534 4501 26.142 MJ/kg
T (°C) 80 90 49.94 kJ/kg
P (bar) 200 1 527.2 kJ/kg
%solids 52 97.5 Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
Bfeed 1.42 W T0 293.15 K
A)

B) This is not the correct answer. By giving this
Bproduct 1.11 W answer, you most likely made a mistake in
calculating w (you need the w for calculating the
Bfeed 8.50 W exergy). The w is the weight of water per kg
C)
Bproduct 1.70 W of dry product. Thus, the w is not equal to the
solids content (%).
Bfeed 1.15 W
D) It can also be useful to take a look at the previous
Bproduct 1.52 W
exercises about the evaporator and conveyor
Laboratory of dryer, because the exergy calculations for the feed
and product flows are similar.

cpa 1.01 kJ/(kgK)

ϕ (kg/h) 8534 4501 26.142 MJ/kg
T (°C) 80 90 49.94 kJ/kg
P (bar) 200 1 527.2 kJ/kg
%solids 52 97.5 Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
Bfeed 1.42 W T0 293.15 K
A)

B)
Bproduct 1.11 W
Bfeed 8.50 W
C) This is not the correct answer. It seems you made
Bproduct 1.70 W a mistake in
A hint: 49940 J/kg. So this value, instead of
Bfeed 1.15 W 49.94, has to be used in calculations.
D)
Bproduct 1.52 W
Laboratory of
cpa 1.01 kJ/(kgK)

ϕ (kg/h) 8534 4501 26.142 MJ/kg
T (°C) 80 90 49.94 kJ/kg
P (bar) 200 1 527.2 kJ/kg
%solids 52 97.5 Mw 0.01802 kg/mol
Ma 0.02897 kg/mol
1000 kg/m3
Bfeed 1.42 W T0 293.15 K
A)

B)
Bproduct 1.11 W
Bfeed 8.50 W
C)
Bproduct 1.70 W
This is not the correct answer. These exergies are
Bfeed 1.15 W the exergies of the feed and product in J/kg dry
D) product. However, in this question we want to
Bproduct 1.52 W know the exergy in Watt. Thus, try to combine
Laboratory of your answer with the flow rate and the solids
content of both flows.
Question d: What is the moisture content wma of moist General data:
air (wma=kg moisture/kg dry air)? All data required to cpp 1.83 kJ/(kgK)
calculate the wma are provided on this slide. cpw 4.18 kJ/(kgK)
Flow rate:8534 kg/h cpa 1.01 kJ/(kgK)

cps 1.84 kJ/(kgK)
Moist air
ϕ (kg dry air/h) 122212 26.142 MJ/kg
T (°C) 90 49.94 kJ/kg
P (bar) 1 527.2 kJ/kg
w (kg/kg) ? Mw 0.01802 kg/mol

Ma 0.02897 kg/mol
Flow rate:4501 kg/h
1000 kg/m3
A) wma 0.033 kg/kg T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) wma 0.041 kg/kg
Hint: be aware that the moisture
content is expressed per kg dry air
C) wma 118.8 kg/kg (and not per kg air)
D) wma 0.025 kg/kg
Laboratory of

cps 1.84 kJ/(kgK)
Moist air
ϕ (kg dry air/h) 122212 26.142 MJ/kg
T (°C) 90 49.94 kJ/kg
P (bar) 1 527.2 kJ/kg

Ma 0.02897 kg/mol
Flow rate:4501 kg/h
1000 kg/m3
A) wma 0.033 kg/kg T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) wma 0.041 kg/kg This is not the correct answer. This answer, 0.033,
is the amount of water (kg) that has been taken
up per kg of dry air. However, apart from the
C) wma 118.8 kg/kg water that has been absorbed, air always contains
some water (w0). Thus, the total amount of
moisture in the moist air is the sum of the amount
of water that has been taken up and the w 0.
D) wma 0.025 kg/kg
Laboratory of

cps 1.84 kJ/(kgK)
Moist air
ϕ (kg dry air/h) 122212 26.142 MJ/kg
T (°C) 90 49.94 kJ/kg
P (bar) 1 527.2 kJ/kg

Ma 0.02897 kg/mol
Flow rate:4501 kg/h
1000 kg/m3
A) wma 0.033 kg/kg T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) wma 0.041 kg/kg Correct! The calculations:
• The total amount of water that is taken up by
the air in 1 hour is: 8534-4501=4033 kg.
• So the total amount of water (kg) taken up per
C) wma 118.8 kg/kg
kg dry air (in 1 hour):
• Drying air always contains a bit of water (w0).
• wma=0.033+0.008=0.041 kg moisture/kg dry
D) wma 0.025 kg/kg
air.
Laboratory of

cps 1.84 kJ/(kgK)
Moist air
ϕ (kg dry air/h) 122212 26.142 MJ/kg
T (°C) 90 49.94 kJ/kg
P (bar) 1 527.2 kJ/kg

Ma 0.02897 kg/mol
Flow rate:4501 kg/h
1000 kg/m3
A) wma 0.033 kg/kg T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) wma 0.041 kg/kg
This is not the correct answer. Apparently you

C) wma 118.8 kg/kg were not consistent in the conversions kg/h to
kg/s.
To calculate the wma, you should determine how
D) wma 0.025 kg/kg much moisture is taken up by the air.
Remember that you should express all flow rates
Laboratory of in kg/h or in kg/sec. If you do not do this, it will
result in wrong answers.


cps 1.84 kJ/(kgK)
Moist air
ϕ (kg dry air/h) 122212 26.142 MJ/kg
T (°C) 90 49.94 kJ/kg
P (bar) 1 527.2 kJ/kg

Ma 0.02897 kg/mol
Flow rate:4501 kg/h
1000 kg/m3
A) wma 0.033 kg/kg T0 293.15 K

P0 101325 Pa
w0 0.008 kg/kg
B) wma 0.041 kg/kg
C) wma 118.8 kg/kg
This is not the correct answer. A hint: The feed

D) wma 0.025 kg/kg releases water and all this water is taken up by
the air. Thus, to calculate the wma, you should
Laboratory of
calculate how much water is taken up by the air.
Question e: What is the total exergy of the moist air? All General data:
data required to calculate the exergy are provided on this cpw 4.18 kJ/(kgK)
slide. cpa 1.01 kJ/(kgK)
cps 1.84 kJ/(kgK)
26.142 MJ/kg
Moist air
49.94 kJ/kg
ϕ (kg dry air/h) 122212
527.2 kJ/kg
T (°C) 90
Mw 0.01802 kg/mol
P (bar) 1
w (kg/kg) 0.041 Ma 0.02897 kg/mol
1000 kg/m3
T0 293.15 K
A) Bma 8.6 W P0 101325 Pa

w0 0.008 kg/kg
R 8.314 J/(Kmol)
B) Bma 2.7 W
C) Bma -1.9 W
D) Bma 3.4 W
Laboratory of
cps 1.84 kJ/(kgK)
26.142 MJ/kg
Moist air
49.94 kJ/kg
527.2 kJ/kg
T (°C) 90
Mw 0.01802 kg/mol
P (bar) 1
w (kg/kg) 0.041 Ma 0.02897 kg/mol
1000 kg/m3
T0 293.15 K
A) Bma 8.6 W P0 101325 Pa

w0 0.008 kg/kg
R 8.314 J/(Kmol)
B) Bma 2.7 W This is not the correct answer. The moist air contains
thermal and chemical exergy. The thermal exergy of
the moist air can be calculated similarly to the
C) Bma -1.9 W thermal exergy of the cold and hot air flows
(question a). For calculating the chemical exergy of
the moist air, you can not neglect the mixing
D) Bma 3.4 W exergy. The reason for this is that the molecular
weight of air and steam are comparable. You can
only neglect the mixing exergy if the molecular
Laboratory of
weight of one of the components is much larger.
cps 1.84 kJ/(kgK)
26.142 MJ/kg
Moist air
49.94 kJ/kg
527.2 kJ/kg
T (°C) 90
Mw 0.01802 kg/mol
P (bar) 1
w (kg/kg) 0.041 Ma 0.02897 kg/mol
1000 kg/m3
T0 293.15 K
A) Bma 8.6 W P0 101325 Pa

w0 0.008 kg/kg
R 8.314 J/(Kmol)
B) Bma 2.7 W This is not the correct answer. You did not include
the chemical exergy. The moist air contains thermal
and chemical exergy, because both the temperate
and the composition of the air flow are different
C) Bma -1.9 W from the environment (T≠T0 and w≠w0). The
thermal exergy of the moist air can be calculated
similarly to the thermal exergy of the cold and hot
D) Bma 3.4 W air flows (question a). For calculating the chemical
exergy of the moist air, you can not neglect the
mixing exergy. The reason for this is that the
Laboratory of
molecular
Food Process Engineering weights of air and vapour are comparable.
cps 1.84 kJ/(kgK)
26.142 MJ/kg
Moist air
49.94 kJ/kg
527.2 kJ/kg
T (°C) 90
Mw 0.01802 kg/mol
P (bar) 1
w (kg/kg) 0.041 Ma 0.02897 kg/mol
1000 kg/m3
T0 293.15 K
A) Bma 8.6 W P0 101325 Pa

w0 0.008 kg/kg
R 8.314 J/(Kmol)
B) Bma 2.7 W
This is not the correct answer, because the total
exergy of a flow can never be negative. Thus, you
probably have made a mistake in your calculations.
C) Bma -1.9 W It seems you used b0w instead of b0s.
Some other hints:
• The air contains thermal and chemical exergy.
D) Bma 3.4 W • The thermal exergy can be calculated similarly to
the thermal exergy of the cold and hot air flows.
Laboratory of• The mixing exergy can not be neglected.
Correct! The moist air contains both thermal and chemical exergy (T≠T 0 and w≠w0).
The thermal exergy can be calculated similarly to the thermal exergy of the cold and
hot air flows (question a). For calculating the chemical exergy, the mixing exergy
should be included. Furthermore, you can not neglect the mixing exergy, as the air
and steam have comparable molecular weights. The calculations:
Thermal:
J/kg dry air
Chemical:
J/kg dry air
Total: W
A) Bma 8.6 W Example calculation mixing exergy (J/kg dry air):
B) Bma 2.7 W
= J/kg dry air
C) Bma -1.9 W
D) Bma 3.4 W
Laboratory of
As you finished all questions, you know the exergy of all flows and are now able
to make the Grassmann diagram. You can open ESankey to make the Grassmann
diagram.
After you finished the Grassmann diagram, you can check it by clicking on the
link: I finished the Grassmann diagram. Determine which operator is least
efficient.
I finished the Grassmann

diagram
Laboratory of
The figure below shows a Grassmann diagram that you can make about the spray
dryer system considered in this question. Your own Grassmann diagram should look
similar. Keep in mind that in this Grassmann diagram, the exergy of the dry matter of
the product is excluded, as the exergy of the dry matter is really big and does not
change.
You can see that the burner is the least efficient part of the spray dryer, as a lot of
exergy is destroyed in the burner. This exergy is irreversibly lost to the environment!
If the moist air out of the spray dryer is not reused and is discarded into the
environment, it will also be lost.
Full answers
Laboratory of
Food Process Engineering End of exercise
Check full answer question a
Check full answer question b
Check full answer question c
Check full answer question d
Check full answer question e
Laboratory of
End of exercise
Full answer question a:
Check full answer question b

Laboratory of
Full answer question b:

Laboratory of
Full answer question c:

Laboratory of
Full answer question d:
Laboratory of
Full answer question e:
End of exercise
Laboratory of

Exercise 5.3 - Spray Dryer I (2020)

Uploaded by

Copyright:

Available Formats

Exercise 5.3 - Spray Dryer I (2020)

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Exercise 5.3 - Spray Dryer I (2020)

Uploaded by

Copyright:

Available Formats

Spray dryer I

T (°C) 58.2 175 49.94 kJ/kg

P (bar) 1 1 527.2 kJ/kg

w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Bca 7.91 W w0 0.008 kg/kg

T (°C) 58.2 175 49.94 kJ/kg

P (bar) 1 1 527.2 kJ/kg

w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Bca 7.91 W This is not correct. Bca cannot

• For cold air:

T (°C) 58.2 175 49.94 kJ/kg

P (bar) 1 1 527.2 kJ/kg

w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Bca 7.91 W w0 0.008 kg/kg

T (°C) 58.2 175 49.94 kJ/kg

P (bar) 1 1 527.2 kJ/kg

w (kg/kg) 0.008 0.008 Mw 0.01802 kg/mol

Bca 7.91 W w0 0.008 kg/kg

Net calorific value (MJ/kg) 54.00 49.94 kJ/kg

Exergetic content (J/J) 1.04 527.2 kJ/kg

A) Bgas 2.32 W T0 293.15 K

Net calorific value (MJ/kg) 54.00 49.94 kJ/kg

Exergetic content (J/J) 1.04 527.2 kJ/kg

A) Bgas 2.32 W T0 293.15 K

D) Bgas 5.62 W In this question, we want to know the exergy

Net calorific value (MJ/kg) 54.00 49.94 kJ/kg

Exergetic content (J/J) 1.04 527.2 kJ/kg

A) Bgas 2.32 W T0 293.15 K

Net calorific value (MJ/kg) 54.00 49.94 kJ/kg

Exergetic content (J/J) 1.04 527.2 kJ/kg

A) Bgas 2.32 W T0 293.15 K

Net calorific value (MJ/kg) 54.00 49.94 kJ/kg

Exergetic content (J/J) 1.04 527.2 kJ/kg

A) Bgas 2.32 W T0 293.15 K

Feed Product cps 1.84 kJ/(kgK)

Bfeed 9.59 W w0 0.008 kg/kg

Feed Product cps 1.84 kJ/(kgK)

Bfeed 9.59 W w0 0.008 kg/kg

and product flows are similar.

Feed Product cps 1.84 kJ/(kgK)

Bfeed 9.59 W w0 0.008 kg/kg

Feed Product cps 1.84 kJ/(kgK)

Bfeed 9.59 W w0 0.008 kg/kg

Flow rate:8534 kg/h cpa 1.01 kJ/(kgK)

T (°C) 90 49.94 kJ/kg

P (bar) 1 527.2 kJ/kg

w (kg/kg) ? Mw 0.01802 kg/mol

A) wma 0.033 kg/kg T0 293.15 K

D) wma 0.025 kg/kg

Flow rate:8534 kg/h cpa 1.01 kJ/(kgK)

T (°C) 90 49.94 kJ/kg

P (bar) 1 527.2 kJ/kg

w (kg/kg) ? Mw 0.01802 kg/mol

A) wma 0.033 kg/kg T0 293.15 K

Flow rate:8534 kg/h cpa 1.01 kJ/(kgK)

T (°C) 90 49.94 kJ/kg

P (bar) 1 527.2 kJ/kg

w (kg/kg) ? Mw 0.01802 kg/mol