Example 8.1-1 Equipment Encyclopedia condenser www.wiley.com/college/felder Energy Balance on a Condenser Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. 100 mol/s 0.669 mol Ac(v)/mol 0.331 mol Ny/mol 65°C, 1 atm 8 CHAPTER 8 Balances on Nonreactive Processes (%) CONDENSER 63.55 mol Ac(1)/s 20°C, 5 atm The process operates at steady state. Calculate the required cooling rate. Use H, instead of Ú, for a closed constant-pressure system, since Q=AH for such systems. 36.45 mol/s 0.092 mol Ac(v)/mol 0.908 mol Nymol 20°C. 5 atm Solution We will follow the procedure given preceding this example. Perform required material balance calculations. None-are-required in this example. Write and simplify the energy balance. + W, = AH+AE+AEp. There are no moving parts in the system and no energy is transferred by electricity or radiation, so W, = 0. No significant vertical For this open steady-state system mille

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Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Here are the enthapies for each compound.  For acetone vapor, H(1)in = 35.7 kJ/mol and H(2)out= 32.0 kJ/mol. For nitrogen, H(3)out= -1.26 kJ/mol.  

The written stuff on paper is additional information.  

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Example 8.1-1
Equipment Encyclopedia
condenser
www.wiley.com/college/felder
427 of 695
HA
Te Canice w UC for you to understand une rest of the chapter and to solve the cha-of-chapter
problems.
Energy Balance on a Condenser
Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and
the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown
below.
100 mol/s
0.669 mol Ac(v)/mol
0.331 mol N₂/mol
65°C, 1 atm
408 CHAPTER 8 Balances on Nonreactive Processes
Ô(J/s)
63.55 mol Ac(1)/s
20°C, 5 atm
The process operates at steady state. Calculate the required cooling rate.
Q Search
CONDENSER
2 Use Ĥ, instead of U; for a closed constant-pressure system, since Q=AH for such systems.
Solution We will follow the procedure given preceding this example.
36.45 mol/s
0.092 mol Ac(v)/mol
0.908 mol N₂/mol
20°C, 5 atm
■
Perform required material balance calculations. None are required in this example.
Write and simplify the energy balance.
For this open steady-state system + W, = AH + AÈk +AEp. There are no moving parts in
the system and no energy is transferred by electricity or radiation, so W, = 0. No significant vertical
distance separates the inlet and outlet ports, so AĖp ~0. Phase changes and nonnegligible
temperature changes occur, so AE 0 (relative to AH). The energy balance reduces to
r
P
<
Ơ
J
I
ENG
Sign in
100
7:07 AM
5/6/2023
•
la
+
O
Transcribed Image Text:||! PDF *Elementary Principles of Chemic X + 46°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdf Draw (T) Read aloud Example 8.1-1 Equipment Encyclopedia condenser www.wiley.com/college/felder 427 of 695 HA Te Canice w UC for you to understand une rest of the chapter and to solve the cha-of-chapter problems. Energy Balance on a Condenser Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. 100 mol/s 0.669 mol Ac(v)/mol 0.331 mol N₂/mol 65°C, 1 atm 408 CHAPTER 8 Balances on Nonreactive Processes Ô(J/s) 63.55 mol Ac(1)/s 20°C, 5 atm The process operates at steady state. Calculate the required cooling rate. Q Search CONDENSER 2 Use Ĥ, instead of U; for a closed constant-pressure system, since Q=AH for such systems. Solution We will follow the procedure given preceding this example. 36.45 mol/s 0.092 mol Ac(v)/mol 0.908 mol N₂/mol 20°C, 5 atm ■ Perform required material balance calculations. None are required in this example. Write and simplify the energy balance. For this open steady-state system + W, = AH + AÈk +AEp. There are no moving parts in the system and no energy is transferred by electricity or radiation, so W, = 0. No significant vertical distance separates the inlet and outlet ports, so AĖp ~0. Phase changes and nonnegligible temperature changes occur, so AE 0 (relative to AH). The energy balance reduces to r P < Ơ J I ENG Sign in 100 7:07 AM 5/6/2023 • la + O
Acetone (denoted as Ac) is partially condessed out of a gar
stream containing 66.9% acetone vapor and the balance nitrogen.
Calculate the required cooling rate.
[:liquid v-vapor & (5/5)
Joonol/s
6.669 mal Acco) not)
0.33/nal N₂ /no1
65°C 1 atm
Ас
Acid
& (J/s) 36.45 m³/s
B
Condenser
a
0.908 mol N₂t not
२००८ Sath
63.55 MASCA
20°G Safm
references. Ac (1, 20°C, 5 atr), N₂ lg 65° 4 lata)
2
bout Hout
66.9 H₁ 3.35 8/₂
63.550
33.1 33.1 M₂
O
/ Mol
Here is the process path for H
днь
Ac(1, 20°G 5atm) Stig Ac (1, 20°C, | atm) SHB Ach, 56°
Ac (0.65°C, latim)
Lath
Dita Ae (v. 56°, la tn)
M₁ = 0 Hpath
DH
A₁² Allis + Hibt OHICT OHLA
2
оинст дни
J
756°C
1₁² VAC(D() atn - 5 atm) + for ECRACIA THAHDAG
20°2
Show the calculations for Hi
also, show the paths for H₂ and Hy
and calculate the and My or well.
Also, use fabler Br and B₂
f
65
56
(Ep) dJ
Acl
Transcribed Image Text:Acetone (denoted as Ac) is partially condessed out of a gar stream containing 66.9% acetone vapor and the balance nitrogen. Calculate the required cooling rate. [:liquid v-vapor & (5/5) Joonol/s 6.669 mal Acco) not) 0.33/nal N₂ /no1 65°C 1 atm Ас Acid & (J/s) 36.45 m³/s B Condenser a 0.908 mol N₂t not २००८ Sath 63.55 MASCA 20°G Safm references. Ac (1, 20°C, 5 atr), N₂ lg 65° 4 lata) 2 bout Hout 66.9 H₁ 3.35 8/₂ 63.550 33.1 33.1 M₂ O / Mol Here is the process path for H днь Ac(1, 20°G 5atm) Stig Ac (1, 20°C, | atm) SHB Ach, 56° Ac (0.65°C, latim) Lath Dita Ae (v. 56°, la tn) M₁ = 0 Hpath DH A₁² Allis + Hibt OHICT OHLA 2 оинст дни J 756°C 1₁² VAC(D() atn - 5 atm) + for ECRACIA THAHDAG 20°2 Show the calculations for Hi also, show the paths for H₂ and Hy and calculate the and My or well. Also, use fabler Br and B₂ f 65 56 (Ep) dJ Acl
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