Answered: 2200 kg / hr of propane gas is burned…

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

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Question

2200 kg / hr of propane gas is burned with 5% excess air and the process is to be designed for a 95% overall conversion of propane. If 90% of the reacted gas is burned completely and the balance to form CO gas, calculate the flow rate and composition of the flue gas.

Expert Solution

Step 1

given that,

the mass flow rate of propane gas = 2200 kg/hr

percentage of excess air = 5% = 0.05

overall conversion of propane = 95% = 0.95

90% of gas is burned completely that is converted to CO₂ and the rest 10% is converted to CO.

it is required to calculate the flow rate and composition of the flue gas

Step 2

given that the overall conversion of propane is 95%. so the amount of propane gas burned = $0.95 \times 2200 = 2090 \frac{k g}{h r}$

combustion reactions of propane:

$\begin{array}{rcl} C_{3} H_{8} + 5 O_{2} & \to & 3 C O_{2} + 4 H_{2} O \\ C_{3} H_{8} + 3.5 O_{2} & \to & 3 C O + 4 H_{2} O \end{array}$

44 kg of propane requires 160 kg of O₂ for complete combustion

2090 kg of propane requires = $\frac{2090 \times 160}{44} = 7600 k g o f O_{2}$

theoretical O₂ supplied= 7600 kg

actual O₂ supplied = 7600(1+0.05) = 7980 kg

moles of O₂ = $\frac{7980}{32} = 249.375 k m o l$

moles of N₂ entering along with O₂ = $\frac{249.375 \times 0.79}{0.21} = 938.125 k m o l$

mass of N₂ = $938.125 \times 28 = 26267.5 k g$

Step 3

90% of propane gas is converted to CO₂

amount of propane gas is converted to CO₂= $0.9 \times 2090 = 1881 k g$

44 kg of propane produces 132 kg of CO₂

1881 kg of propane produces $\frac{1881 \times 132}{44} = 5643 k g o f C O_{2}$

The amount of propane gas is converted to CO= $0.1 \times 2090 = 209 k g$

44 kg of propane produces 84 kg of CO

209 kg of propane produces = $\frac{209 \times 84}{44} = 399 k g o f C O$

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