The product stream from a combustion unit at 700 degree C contains 85 mole% N_2(
ID: 1050164 • Letter: T
Question
The product stream from a combustion unit at 700 degree C contains 85 mole% N_2(g), and 15 mole% CO_2(g) on a dry basis. The dew point of the product stream is 51.3 degree C. The stream enters a condenser and exits at 30 degree C. The heat from the combustion product is transferred to cooling water that runs through the condenser. A partial flowchart is shown below. The cooling water liquid enters the condenser at 2 degree C and leaves as a liquid at 28 degree C. Determine The total heat lost from the combustion product stream and The required flowrate of cooling water. You need to prepare an enthalpy table to get credit for your solution.Explanation / Answer
Dew point is 51.3 deg.c
At dew point, partial pressure of water vapor = vapor pressure of liquid = 97.4 mm Hg
Moles of water vapor/moles of dry air = partial pressure of water vapor/ partial pressure of dry air
Basis : 1 mole of dry combustion product
Hence
Moles of water vapor/ 1= 97.4/(760-97.4)=0.15
Moles of water vapor = 0.15
Water need to be cooled from 700 deg.c to 100 deg.c for it to get condensed ( also known as super heat of vapor) = mass* specific heat* temperature difference =0.15*18* 2 j/g.K *(700-100)=3240 joules
The gas is assumed to be saturated at leaving conditions of 30 deg.c
Vapor pressure = partail pressur =31.8 at 30 deg.c
Moles of water vapor/ moles of dry air = partial pressrue of water vapor/partial pressure of dry air
Moles of water vapor/1= 31.8/(760-31.8) =0.044
Moles of water vapor = 0.044
Moles of water condensed = 0.15-0.044=0.106
Mass of water condensed = 18*0.106=1.91 gm
Latent heat of vaporizatino of water = 2265 J/gm
Heat removed during condensatino = 1.91*2265 =4326 joules
From 100 deg,c to 30 deg,c , sensible heat needs to be removed which is =0.15*18*4.18*(100-30)=790 joules
Total heat that needs to be removed = 4326+790+3240 =8356 joules
Sensibe hea of 85%N2and 15% CO2 need to be removed
Moles N2 =0.85 and mles of CO2= 0.15
Cp data
N2 : Cp/R =3.280+0.503*10-3 T
Enthalpy change is obtanied by integration of CP/R between 700 deg.c =973.15K and 30 deg.c =303.15K
=8.314 [0.85* { 3.280* (973.15-303.15) +0.503*10-3*( 973.152= 303.152)/2} ]=17050 joules
For CO2 Cp/R= 5.457+1.045*10-3T
Enthalpy change = 8.314[0.15*{ 5.457*(973.15-303.15) +1.045*10-3(973.152= 303.152)/2}] =5116.84 joules
Total enthallpy change = heat loss = enthalpy change of water + enthalpy change of N2+ enthalpy change of CO2= 8356+17050+5116.84=30532.84 joules/mole of wet gas
This much heat is removed by cooling water which =mass of water* specific heat* temperature difference = m*4.184*(28-2)= 30532.84 , m= 280.674 gm / mole of wet gas
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.