EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
9th Edition
ISBN: 8220106796979
Author: CENGEL
Publisher: YUZU
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Chapter 16.6, Problem 48P

a)

To determine

The amount of heat transfer for the dissociation of water.

a)

Expert Solution
Check Mark

Answer to Problem 48P

The amount of heat transfer for the dissociation of water is 2056kJ/min.

Explanation of Solution

Write the expression for the stoichiometric reaction for reaction 1.

H2OH2+12O2 … (I)

Here, the stoichiometric coefficient for O2(vO2) is 0.5, for H2O is 1, and for H2(vH2) is 1.

Write the expression for the stoichiometric reaction for reaction 1.

H2O12H2+OH (II)

Here, the stoichiometric coefficient for H2(vH2) is 0.5, for H2O is 1, and for OH (vOH) is 1.

Write the expression for the actual reaction for the reported process.

H2OxH2O+yH2+zO2+wOH (III)

Here, the equilibrium composition contains x kmol of H2O(NH2O), y kmol of H2(NH2), z kmol of O2(NO2), and w kmol of OH (NOH).

Write the hydrogen balance equation From Equation (III).

2=2x+2y+w (IV)

Write the oxygen balance equation From Equation (III).

1=x+2z+w (V)

Write the expression for the formula for total number of moles (Ntotal).

Ntotal=NH2O+NH2+NO2+NOH (VI)

Here, number of moles of H2O is NH2O, number of moles of H2 is NH2, number of moles of O2 is NO2, and number of moles of OH is NOH.

Write the expression for the equilibrium constant for reaction 1 (Kp1).

Kp1=NH2vH2NO2vO2NH2OvH2O(PNtotal)(vH2+vO2vH2O) (VII)

Here, pressure is P.

Write the expression for the equilibrium constant for reaction 2 (Kp2).

Kp2=NH2vH2NOHvOHNH2OvH2O(PNtotal)(vH2+vOHvH2O) (VIII)

Write the energy balance equation to determine the heat transfer (Qout) for the dissociation process.

Qin=NP(h¯fo+h¯h¯o)PNR(h¯fo+h¯h¯o)R (IX)

Here, number of moles of products is NP, number of moles of reactants is NR, enthalpy of vaporization is h¯fo, and the enthalpy of formation is h¯.

Write the expression to calculate the molar flow rate of water (N˙).

N˙=m˙M (X)

Here, mass flow rate of water is m˙ and the molar mass of water is M.

Write the expression to calculate the rate of heat transfer (Q˙out).

Q˙in=N˙Qin (XI)

Conclusion:

Refer Table A-28, “Natural logarithm of equilibrium constants”, select the value of lnkp as 3.086 for reaction 1 (lnKp1) at 3000 K. Hence, the equilibrium constant for reaction 1 is 0.04568.

Refer Table A-28, “Natural logarithm of equilibrium constants”, select the value of lnkp as 2.937 for reaction 2 (lnKp2) at 3000 K. Hence, the equilibrium constant for reaction 2 is 0.05302.

Substitute x for NH2O, y for NH2, z for NO2, and w for NOH in Equation (VI).

Ntotal=x+y+z+w

Substitute 0.04568 for kp1, x for NH2O, y for NH2, z for NO2, x+y+z+w for Ntotal, 1 atm for P, 1 for vH2, and 0.5 for vO2 and 1 for vH2O in Equation (VII).

0.04568=yz0.5x(1 atmx+y+z+w)(1+0.51)0.04568=yz0.5x(1x+y+z+w)(0.5) (XII)

Substitute 0.05302 for kp2, x for NH2O, y for NH2, w for NOHx+y+z+w for Ntotal, 1 atm for P, 0.5 for vH2, 1 for vO2 and 1 for vH2O in Equation (VIII).

0.05302=wy2x(1atmx+y+z+w)(1+0.51)0.05302=wy2x(1x+y+z+w)0.5 (XIII)

Solve Equations (IV), (V), (XII), and (XIII) simultaneously and find the values of x, y, z, and w, as 0.7835, 0.1622, 0.05396, and 0.1086 respectively.

Substitute 0.7835 for x, 0.1622 for y, 0.05396 for z, and 0.1086 for w in Equation (III).

H2OxH2O+yH2+zO2+wOHH2O0.7835H2O+0.1622H2+0.05396O2+0.1086OH

Refer the table A-26, “Enthalpy of formation table”, obtain the enthalpy of H2O (h¯f)H2O, O2 (h¯f)O2,OH (h¯f)OH, and H2 (h¯f)H2 as 241,820kJ/kmol, 0, 0, and 39,460kJ/kmol.

Refer the table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at 3000 K (h¯)H2O and 298 K (h¯)H2O as 136,264kJ/kmol and 9,904kJ/kmol

Refer the table A-22, “Ideal gas properties of H2 gas”, obtain the enthalpy of H2 gas at 3000 K (h¯)H2 and 298 K as (h¯)H2 as 97,211kJ/kmol and 8,468kJ/kmol.

Refer the table A-19, “Ideal gas properties of O2 gas”, obtain the enthalpy of O2 gas at 3000 K (h¯)O2 and 298 K (h¯)O2 as 106,780kJ/kmol and 8,682kJ/kmol.

Refer the table A-25, “Ideal gas properties of OH gas”, obtain the enthalpy of OH gas at 3000 K (h¯)OH and 298 K (h¯)OH as 98,763kJ/kmol and 9,188kJ/kmol.

Write the heat input equation (Qin) using Equation (X).

Qin={NH2O(h¯f+h¯h¯)H2O+NH2(h¯f+h¯h¯)H2+NO2(h¯f+h¯h¯)O2+NOH(h¯f+h¯h¯)OH(h¯f)H2O} (XIV)

Here, number of moles of product is NP, number of moles of reactants is NR, number of moles of H2O is NH2O, number of moles of H2 is NH2, number of moles of O2 is NO2, number of moles of OH is NOH, enthalpy at reference state is h¯f, sensible enthalpy at specified state is h¯, and sensible enthalpy at reference state is h¯.

Substitute 241,820kJ/kmol for (h¯f)H2O, 136,264kJ/kmol for (h¯3000K)H2O, 9,904kJ/kmol for (h¯)H2O, 0 for (h¯f)H2, 97,211kJ/kmol for (h¯3000K)H2, 8,468kJ/kmol for (h¯298K)H2, 0 for (h¯f)O2, 106,780kJ/kmol for (h¯3000K)O2, 8,682kJ/kmol for (h¯298K)O2, 9,188kJ/kmol for , (h¯298K)OH, 98,763kJ/kmol for (h¯3000K)OH,0.05396 kmol for NO2, and 0.7835 kmol for NH2O, 0.1622 for NH2 , and 0.1086 for NOH in Equation (XIV).

h¯R={0.7835kmol(241,820kJ/kmol+136,264kJ/kmol9,904kJ/kmol)+0.1622kmol(0+97,211kJ/kmol8,468kJ/kmol)+0.05396kmol(0+106,780kJ/kmol8,682kJ/kmol)+0.1086kmol(39,460kJ/kmol+98,763kJ/kmol9,188kJ/kmol)(241,820kJ/kmol)}=185,058kJ/kmol

Substitute 0.2kg/min for m˙ and 18kg/kmol for M in Equation (X).

N˙=0.2kg/min18kg/kmol=0.0111kmol/min

Substitute 0.0111kmol/min for N˙ and 185,058kJ/kmol for Q˙in in Equation (XI)

Q˙in=(0.01111kmol/min)(185,058kJ/kmol)2056kJ/min

Thus, the amount of heat transfer for the dissociation of water is 2056kJ/min.

b)

To determine

The amount of heat transfer for the absence of dissociation of water.

b)

Expert Solution
Check Mark

Answer to Problem 48P

The amount of heat transfer for the absence of dissociation of water is 1553kJ/min.

Explanation of Solution

Write the expression to determine the heat transfer for the absence of dissociation of Water.

Q˙in=N˙(h¯3000 Kh¯298K)H2O (XV)

Conclusion:

Substitute 0.01111kmol/min for N˙, 136,264kJ/kmol for (h¯3000K)H2O, and 9904kJ/kmol for (h¯298K)H2O in Equation (XV).

Q˙in=0.01111kmol/min(136,264kJ/kmol9904kJ/kmol)=1404kJ/min

Thus, the amount of heat transfer for the absence of dissociation of water is 1404kJ/min.

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Chapter 16 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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