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Methane:: Δ ˙H +Δ ˙E Δ ˙E Q− ˙W

1. The document provides information about the heating values of methane and water at different temperatures and pressures. 2. An energy balance equation is presented and used to calculate the heat of reaction (ΔH) for a reaction between methane and steam in a mixer. 3. The heat of reaction is calculated to be 277 KW based on the moles and heating values of methane and water entering and exiting the mixer.

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chemicalgeeks
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58 views3 pages

Methane:: Δ ˙H +Δ ˙E Δ ˙E Q− ˙W

1. The document provides information about the heating values of methane and water at different temperatures and pressures. 2. An energy balance equation is presented and used to calculate the heat of reaction (ΔH) for a reaction between methane and steam in a mixer. 3. The heat of reaction is calculated to be 277 KW based on the moles and heating values of methane and water entering and exiting the mixer.

Uploaded by

chemicalgeeks
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Q6: Solution:

According to the given information


Methane:
T = 25o C
P = 1 atm

Heating value of Methane H^ = 801.363 kJ/mol


Water:
T = 15 o C
P = 400 KPa = 3.96 atm
T
kJ kJ
H^ of + CpidT
mol mol
Heating value of Water @ 15 oC H^ = 25 = -241.83 2.1 =
-243.93 kJ/mol
kJ kJ
Heating value of Water @ 180 oC H^ = -241.83 mol + 5.6 mol = -236.23 kJ/mol

2.5 mol steam


Ratio before reformer=
mol of CH 4

Applying energy balance for continuous process

H + Ek + E p =QW shaft ,out


As there is no kinetic energy, potential work and shaft work taking place so neglecting these
terms reduce the equation to

Neglecting the terms


E k , E p and W shaft,out

So the equation reduced to

H =Q= ni H^ i ni H^ i
outlet inlet Equation 1
So in the mixer two components are mixed so Heat of reaction for reaction 1 can be calculated
using
Finding the number of moles
nCH = 16.814
4 ,in =0.27 mol/ s
2980.08206
n H 2O,in = 3.940.0123
=0.0017 0 mol /s
2980.08206

At the outlet condition, we have:

n H 2O,out 2.5 gmol of steam


0.278 gmol of CH 4 = 0.696 gmol of steam/ s
gmol of CH 4

s

nCH 4, out 2.5 gmol of steam


s = 0.276 gmol of CH4/ s
gmol of CH 4

0.69 gm of steam
Substituting the values in Equation 1:

H = (0.696 gmol of CH 4 * 801.63 kJ +0.276 gmol of H 20 243.93 kJ -


S gmol s gmol
gmol of CH 4 kJ 0.00170 gmol of H 20 kJ
(0.276 801.63 + 236.23
S gmol s gmol

H = Q = 277.36 kJ = 277 KW
s

Q7: Solution:
Steam Reformer Power requirement
The reform data is given as
Moles in Hin Moles out Hout
CH4 1002.7 801.63 43.345 -
H20 2506.88 -236.514 1402.9176 -216.91
CO 0 - 814.85 -89.7
H2 0 - 3022.794 -361.2
CO2 0 - 144.561 19.8
Total 3509.64 5428.4676
kJ kJ
Heating value of Water @ 180 oC H^ = -241.83 mol + 5.6 mol = -236.23 kJ/mol

kJ kJ
Heating value of Water @ 700 oC H^ = -241.83 mol + 5.6 mol = -236.514 kJ/mol
2.5 mol steam
Ratio before reformer=
mol of CH 4

Applying energy balance for continuous process

H + Ek + E p=QW shaft ,out


As there is no kinetic energy, potential work and shaft work taking place so neglecting these
terms reduce the equation to

Neglecting the terms


E k , E p and W shaft,out

So the equation reduced to

H =Q= ni H^ i ni H^ i
outlet inlet Equation 1
From the data we have calculated values of number of moles enthalpy
Substituting the values in equation 1

H = (1402.916*-236.514+814.85*-89.7*3022.794*-361.2*144.561*19.8) -

(1002.756*801.63 + 2506.89*-236.23) = 685.8 KW

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