A stream of humid air enters a condenser in which 95% of the water sapor in the air is condensed. The flow rate of the condensation (the liquid leaving the condenser). 2254/he Dry air may be taken to contain 21 mole fu oxygen with the balance p stream nitrogens Calculate the flow rate of the gar leaving the condenser and the mole frachunt of oxygen, mitrogen, and water in this stream. 16 is the An additional of information entenns air contains 10.09midle to water prece n (1/h) 0.100 noltholio) F0,900 mst dry air/n) 0.79 mot N/moldry duf: 5 vanabler cir air -3 specier -I vol. flow rate -1952 fractant Condensade O dof hog ₁ (moto) •M₂ (mil N₂/h) n's 122521₂0(1)/2 n₂ (mult₂o/h) 22 (95% of water in feed) silve these eens below EESTILOL BUD low mot next page Sulve the See more on the equations Age on the
A stream of humid air enters a condenser in which 95% of the water sapor in the air is condensed. The flow rate of the condensation (the liquid leaving the condenser). 2254/he Dry air may be taken to contain 21 mole fu oxygen with the balance p stream nitrogens Calculate the flow rate of the gar leaving the condenser and the mole frachunt of oxygen, mitrogen, and water in this stream. 16 is the An additional of information entenns air contains 10.09midle to water prece n (1/h) 0.100 noltholio) F0,900 mst dry air/n) 0.79 mot N/moldry duf: 5 vanabler cir air -3 specier -I vol. flow rate -1952 fractant Condensade O dof hog ₁ (moto) •M₂ (mil N₂/h) n's 122521₂0(1)/2 n₂ (mult₂o/h) 22 (95% of water in feed) silve these eens below EESTILOL BUD low mot next page Sulve the See more on the equations Age on the
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
Chemical engineering: solve the equations
![**Degree-of-Freedom Analysis**
A stream of humid air enters a condenser in which 95% of the water vapor in the air is condensed. The flow rate of the condensation (the liquid leaving the condenser) is 225 lb/h. The air may be taken to contain 21 mole% oxygen with the balance nitrogen. Calculate the flow rate of the gas stream leaving the condenser and the mole fraction of oxygen, nitrogen, and water in this stream.
An additional piece of information is that the entering air contains 10.0 mole% water.
---
**Diagram: Process Flow**
The diagram illustrates a system with variables indicating mol/hr for different components:
- \(n_1\) (mol/h entering)
- 0.100 mol H2O/mol
- 0.900 mol dry air/mol
- 0.021 mol O2/mol dry air
- 0.679 mol N2/mol dry air
- \(n_2\) represents the liquid stream with a 95% water fraction in the feed.
- The gas stream leaving the condenser is represented by arrows leading to:
- \(n_3\) (mol of O2)
- \(n_4\) (mol of N2)
- \(n_5\)
- \(n_2\) represents the liquid H2O (lb/h) leaving the system.
---
**Degree of Freedom Analysis:**
- Variables: 5
- Species: 3
- Unknown Flow Rate: 1
- 95% Fractional Condensation: 1
\[
\text{Degrees of freedom} = 0
\]
**Note:** The system is fully specified if all equations are independent.
**Instructions:**
Solve the equations below:
\[
n_2 \left(\frac{\text{mol H}_2\text{O}}{\text{h}}\right) = 225 \left(\frac{\text{lb}}{\text{h}}\right) \quad \text{(95% of water in feed)}
\]
*Equation solving notes are to be continued on the next page.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ba88ecd-654b-4f6b-9783-e93c33ca6a16%2Fb310c4be-a0f9-4c08-a356-6f6833dff77b%2Fucj4qyd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Degree-of-Freedom Analysis**
A stream of humid air enters a condenser in which 95% of the water vapor in the air is condensed. The flow rate of the condensation (the liquid leaving the condenser) is 225 lb/h. The air may be taken to contain 21 mole% oxygen with the balance nitrogen. Calculate the flow rate of the gas stream leaving the condenser and the mole fraction of oxygen, nitrogen, and water in this stream.
An additional piece of information is that the entering air contains 10.0 mole% water.
---
**Diagram: Process Flow**
The diagram illustrates a system with variables indicating mol/hr for different components:
- \(n_1\) (mol/h entering)
- 0.100 mol H2O/mol
- 0.900 mol dry air/mol
- 0.021 mol O2/mol dry air
- 0.679 mol N2/mol dry air
- \(n_2\) represents the liquid stream with a 95% water fraction in the feed.
- The gas stream leaving the condenser is represented by arrows leading to:
- \(n_3\) (mol of O2)
- \(n_4\) (mol of N2)
- \(n_5\)
- \(n_2\) represents the liquid H2O (lb/h) leaving the system.
---
**Degree of Freedom Analysis:**
- Variables: 5
- Species: 3
- Unknown Flow Rate: 1
- 95% Fractional Condensation: 1
\[
\text{Degrees of freedom} = 0
\]
**Note:** The system is fully specified if all equations are independent.
**Instructions:**
Solve the equations below:
\[
n_2 \left(\frac{\text{mol H}_2\text{O}}{\text{h}}\right) = 225 \left(\frac{\text{lb}}{\text{h}}\right) \quad \text{(95% of water in feed)}
\]
*Equation solving notes are to be continued on the next page.*
![### Transcription of Engineering Notes
This document contains calculations related to a chemical process, specifically focusing on condensation, gas balance, and concentrations. Below is a transcription of the handwritten notes:
---
\[
\dot{n}\left(\text{methanol}\right) = \frac{225 \, L \, H_2O}{h} \times \frac{1.00 \, \text{mol MeOH}}{1 \, L} \times \frac{1\, \text{mol H}_2O}{18.0 \times 10^{-3} \, \text{kg}}
\]
**Condensation Calculation:**
- 95% condensation:
\[
\dot{n}_2 = 0.95(0.100)(\dot{n}_1)
\]
**Gas Balance Equations:**
1. **Oxygen Balance:**
\[
\dot{n}_1(0.900)(0.2) = \dot{n}_3
\]
2. **Nitrogen Balance:**
\[
\dot{n}_1(0.900)(0.79) = \dot{n}_4
\]
3. **Water Balance:**
\[
\dot{n}_1(0.100) = \dot{n}_2 + \dot{n}_5
\]
**Total Output Gas Flow Rate:**
- Total: \(\dot{n}_3 + \dot{n}_4 + \dot{n}_5\)
**Output Gas Composition:**
- \(Y_{O_2} = \frac{\dot{n}_3}{\dot{n}_{\text{total}}}\)
- \(Y_{N_2} = \frac{\dot{n}_4}{\dot{n}_{\text{total}}}\)
- \(Y_{H_2O} = \frac{\dot{n}_5}{\dot{n}_{\text{total}}}\)
---
These calculations are used to determine the flow rates and compositions of various gases in the process. The formulas incorporate mass balances for methanol, water, oxygen, and nitrogen, used frequently in chemical engineering to design and optimize industrial processes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ba88ecd-654b-4f6b-9783-e93c33ca6a16%2Fb310c4be-a0f9-4c08-a356-6f6833dff77b%2F2pl6a2q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Transcription of Engineering Notes
This document contains calculations related to a chemical process, specifically focusing on condensation, gas balance, and concentrations. Below is a transcription of the handwritten notes:
---
\[
\dot{n}\left(\text{methanol}\right) = \frac{225 \, L \, H_2O}{h} \times \frac{1.00 \, \text{mol MeOH}}{1 \, L} \times \frac{1\, \text{mol H}_2O}{18.0 \times 10^{-3} \, \text{kg}}
\]
**Condensation Calculation:**
- 95% condensation:
\[
\dot{n}_2 = 0.95(0.100)(\dot{n}_1)
\]
**Gas Balance Equations:**
1. **Oxygen Balance:**
\[
\dot{n}_1(0.900)(0.2) = \dot{n}_3
\]
2. **Nitrogen Balance:**
\[
\dot{n}_1(0.900)(0.79) = \dot{n}_4
\]
3. **Water Balance:**
\[
\dot{n}_1(0.100) = \dot{n}_2 + \dot{n}_5
\]
**Total Output Gas Flow Rate:**
- Total: \(\dot{n}_3 + \dot{n}_4 + \dot{n}_5\)
**Output Gas Composition:**
- \(Y_{O_2} = \frac{\dot{n}_3}{\dot{n}_{\text{total}}}\)
- \(Y_{N_2} = \frac{\dot{n}_4}{\dot{n}_{\text{total}}}\)
- \(Y_{H_2O} = \frac{\dot{n}_5}{\dot{n}_{\text{total}}}\)
---
These calculations are used to determine the flow rates and compositions of various gases in the process. The formulas incorporate mass balances for methanol, water, oxygen, and nitrogen, used frequently in chemical engineering to design and optimize industrial processes.
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