Explanation Write the actual reaction for the dissociation process. C+1 ( O 2 +3 .76N 2 ) → { 0 .97CO 2 + 0 .03CO + 0 .515O 2 + 3 .76N 2 } (I) From Equation (1), infer that the equilibrium composition contains 0.03 kmol of carbon monoxide ( N CO ) , 0.515 kmol of oxygen ( N O 2 ) , 3.76 kmol of nitrogen ( N N 2 ) , and 0.97 kmol of newly formed carbon dioxide ( N CO 2 ) . Refer table 28, “Natural logarithm of equilibrium constants”, obtain the value of ln k p for carbon dioxide as –103.762 at 298 K. Write the equilibrium constant ( k p ) for carbon dioxide. k p = e − 103.762 = 8.6 × 10 − 46 (II) Refer table 28, “Natural logarithm of equilibrium constants”, obtain the value of ln k p for oxygen as –186.975 at 298 K. Write the equilibrium constant ( k p ) for oxygen. k p = e − 186.975 = 6.2 × 10 − 82 (III) Refer table 28, “Natural logarithm of equilibrium constants”, obtain the value of ln k p for nitrogen as –367.48 at 298 K. Write the equilibrium constant ( k p ) for nitrogen. k p = e − 367.48 = 0 (IV) From Equations (II), (III), and (IV), infer that the equilibrium constants for CO 2 , O 2 , and N 2 are almost zero. Neglect the terms and rewrite the actual reaction for the dissociation process. Rewrite the Equation (I) as, C+1 ( O 2 +3 .76N 2 ) → 0 .97CO 2 + 3 .76N 2 (V) Write the energy balance equation for the combustion process. − Q o u t = ∑ N P ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) P − ∑ N R ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) R = N N 2 ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) N 2 + N CO 2 ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) CO 2 (VI) Here, heat released during combustion is Q o u t , number of moles of product is N P , number of moles of reactants is N R , number of moles of N 2 is N N 2 , number of moles of N 2 is N N 2 , number of moles of CO 2 is N CO 2 , enthalpy at reference state is h ¯ f ∘ , sensible enthalpy at specified state is h ¯ , and sensible enthalpy at reference state is h ¯ ∘ . Conclusion: Refer table A-28, “Enthalpy of formation”, obtain the enthalpy of N 2 ( ( h ¯ f ∘ ) N 2 ) as 0 . Refer table A-18, “Ideal properties of nitrogen gas table”, obtain the enthalpy of nitrogen gas. Enthalpy of nitrogen gas at 1000 K, ( h ¯ ) N 2 = 30 , 129 kJ / kmol

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

See similar textbooks

Videos

Question

Chapter 16.6, Problem 93RP

To determine

The amount of heat released per kg of carbon.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 16.6, Problem 92RP

Chapter 16.6, Problem 94RP

Students have asked these similar questions

motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

First monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³

2/Y Y+1 2Cp Q1/ Show that Cda Az x P1 mactual Cdf Af R/T₁ 2pf(P1-P2-zxgxpf) Q2/ A simple jet carburetor has to supply 5 Kg of air per minute. The air is at a pressure of 1.013 bar and a temperature of 27 °C. Calculate the throat diameter of the choke for air flow velocity of 90 m/sec. Take velocity coefficient to be 0.8. Assume isentropic flow and the flow to be compressible. Quiz/ Determine the air-fuel ratio supplied at 5000 m altitude by a carburetor which is adjusted to give an air-fuel ratio of 14:1 at sea level where air temperature is 27 °C and pressure is 1.013 bar. The temperature of air decreases with altitude as given by the expression The air pressure decreases with altitude as per relation h = 19200 log10 (1.013), where P is in bar. State any assumptions made. t = ts P 0.0065h

Chapter 16 Solutions

Thermodynamics: An Engineering Approach

Ch. 16.6 - Write three different KPrelations for reacting...Ch. 16.6 - A reaction chamber contains a mixture of CO2, CO,...Ch. 16.6 - A reaction chamber contains a mixture of N2and N...Ch. 16.6 - A reaction chamber contains a mixture of CO2, CO,...Ch. 16.6 - Which element is more likely to dissociate into...Ch. 16.6 - Prob. 6P Ch. 16.6 - Prob. 7P Ch. 16.6 - Prob. 8P Ch. 16.6 - Prob. 9P Ch. 16.6 - Prob. 10P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.