Garantie de satisfaction à 100% Disponible immédiatement après paiement En ligne et en PDF Tu n'es attaché à rien
logo-home
Solutions for Combustion Science and Engineering 1st Edition by Annamalai (All Chapters included) €28,84   Ajouter au panier

Examen

Solutions for Combustion Science and Engineering 1st Edition by Annamalai (All Chapters included)

 5 vues  0 fois vendu
  • Cours
  • Combustion Science and Engineering 1e Annamalai
  • Établissement
  • Combustion Science And Engineering 1e Annamalai

Complete Solutions Manual for Combustion Science and Engineering 1st Edition by Kalyan Annamalai; Ishwar K. Puri ; ISBN13: 9780849320712.....(Full Chapters included)...

Aperçu 4 sur 432  pages

  • 26 octobre 2024
  • 432
  • 2007/2008
  • Examen
  • Questions et réponses
  • Combustion Science and Engineering 1e Annamalai
  • Combustion Science and Engineering 1e Annamalai
avatar-seller
SOLUTIONS MANUAL FOR
Combustion Science
and
Engineering




by
Kalyan Annamalai
Ishwar K. Puri

** Immediate Download
** Swift Response
** All Chapters included

, Corrections in Problems, Tables and Charts
PROBLEM 6.7 Operating Pressure : 1000 kpa
PROBLMEM 7.6 CO: AIR MIXTURE
ASSUME T= 1800 K, P=1 BAR ; PLOT [CO] VS T; DO YOU BELIOEVE THE AB FOR BACKWARD
REACT RATE? WHAT IS THE PROBLEM?
Problem 7.7 already workd out; but detaile steps are given

Switch problems 7.8 and 7.9 ( i.e kinetics of problem 7.8 are in 7.9)
Problem 10-14 Solution:

Correct

⎧ ⎢1− ⎥ ⎫
⎡ a⎤
⎧ ⎢1− ⎥ ⎫
⎡ a⎤
⎨(1 + B) − (1 + B ) ⎣ r ⎦ ⎬ ⎨(1 + B) − (1 + B ) ⎣ r ⎦ ⎬
⎭ = 1 − 1 ⎧ 1 + B ⎡⎢⎣1− r ⎤⎥⎦ − 1⎫
a
φ = f* = ⎩ ⎭= ⎩
⎨( ) ⎬
B B B⎩ ⎭

Problem 10.29 RHO D= 2.5e-05



Prob 10-7: R, gas constant in kJ (kg K)

Text Example p 449-450; solution for part (b)

ml /ml0 = =0.80 0.20 = {1 – (3.3x10-6 m2/s t)/ (100x10-6)2 }3/2

t =0.42 = 2 ms !

For all Chapters, uless otherwise stated, ambient T = 25 C and ambient pressure = 100 kpa

Problem 10-14 solution:

Correct

⎧ ⎢1− ⎥ ⎫
⎡ a⎤
⎧ ⎢1− ⎥ ⎫
⎡ a⎤
⎨(1 + B) − (1 + B ) ⎣ r ⎦ ⎬ ⎨(1 + B) − (1 + B ) ⎣ r ⎦ ⎬
⎭ = 1 − 1 ⎧ 1 + B ⎡⎢⎣1− r ⎤⎥⎦ − 1⎫
a
φ = f* = ⎩ ⎭=⎩
⎨( ) ⎬
B B B⎩ ⎭

Problem 10.29 ; correct ρD =2.5e-05




3

,Problem 11.5 Modified



Problem 11.6 : Using .. solution of problem 11.4 5



Problem 11.14 : Use Single Film Model




11.14
Consider a flat carbon surface burning in air with air velocity at about 10 m/s. The air temperature
is 600 K. Assume properties of gas at surface temperature of carbon. Other data are as follows:
cpcarnon 4.39 kJ/kg K, ρcarbon = 1300 kg/m3, Tw = 611 K, hc,I 9203.2 kJ/kg , , hC,III =-14,360
kJ/kg X, Length of plate =0.1 m
Free Stream Gas data : YO2inf 0.23, YCO2inf 0.5, rest is N2 Tinf 600 K, Press
1 bar
Transport Propery data for air at 600 K λgas = 0.0000469 , Pr:=1, cpgas =1.051 kJ/kg K,
μ= 3.06x10-5 N s/m2, νw = 5.27E-05 m2/s ,
Obtain an integral solution for average burn rate. Use the profiles given in Example 5

USE Single film Model

Problem 12.10 and Prob 12.12 are duplicated


Problem 12.10.
Gaseous CH4 fuel is injected with a velocity of 5 cm/s into air. Assume ρD = 6.5x10-5 kg/m s.
HHV =50,000 kJ/kg Assuming incompressible flow, calculate:
a) The YF, YO2, YF and T vs y* profiles for a 2 cm wide slot jet at 50 cm from the wall. Plot the
profiles. Assume ρD = λ /cp for air at 1200 K.
b) The flame length for the jet in part a. (use properties at 300 K), let T ∞ = 300 , P ∞ = 101 kPa.


Problem 12.11.

The diameter of the burner is 0.373 cm. The flow rate of CH4 through the burner is 60 cm3/min.
The inlet temperature is 298 K and pressure is 104 kPa.. Assume ρD = 6.5x10-5 kg/m s
=constant.ν= 1.63E-04 m2/s; cp=1.175 kJ/kg ; . {LHV : 50000 kJ/kg}
a) The flame length for the jet in part a. (use properties at 300 K), let T ∞ = 300 , P ∞ = 100 kPa
b) The φ, YF, YO2, YF and T vs y* profiles at 5 cm from the wall. Plot the profiles.



Problem 12.12

Gaseous CH4 fuel is injected with a velocity of 5 cm/s into air. Assume ρD = 6.5x10-5 kg/m s.
HHV =50,000 kJ/kg Assuming incompressible flow, calculate:
a) The Φ/Φ0 (or = vx,/vx,max,), YF, YO2, YF and T vs y* profiles for a 2 cm wide slot jet at 50 cm
from the wall. Plot the profiles. Assume ρD = λ /cp for air at 1200 K.
b) The flame length for the jet (use properties at 300 K), let T ∞ = 300 , P ∞ = 101 kPa



4

, c) The flame length for the jet in part b is properties ta 300 K are used

( Note dy' = ρ/ρi dy or dy = dy' (ρi /ρ) = dy' (T/Ti); Knowing T or Φ profiles vs y' one can
integrate and obtain T vs y or use Euler's integration formulae in Q-spread sheet)


Problem 13.4
Determine the ignition temperature of cylindrical stream of coal particles of dia. 50 μm. We will
assume that coal particles burn heterogeneously.
coal , O 2 → x CO 2 , y H 2O
The coal composition is given as: ash 8.0%, carbon 76.3%, hydrogen 5.0%; oxygen 8.2% (dry).
Neglect N and S. Use Boie eq. for heating value. Assume Ahet = 450 m/s, E = 66, 000 kJ/kmole,
(a) Let radius of stream, R to be 5 mm. Determine T ∞ , I for A:F in the stream of 1:1, 2:1 and
4:1, (b) let R be 10 mm. Determine T ∞ , I for A:F of 1:1, 2:1 and 4:1; ρcoal = 1300 kg/m3, air
density 1.1 kg/m3. Assume that T∞,I =

Problem 13,5


Most explosive conditions are presumed to exist in the pulverizer mills at 0.5 g per liter [Power
Engineering, 1993].a) What is the probable ignition temperature for a cylindrical cloud of dia 20
cm? b)What is the temperature for 20 cm dia spherical cloud? Assume other dara similar to
Problem 13.4
Problem 13.7

 *= hc*/[1+{ exp(1/θ)/ DIII,mod ] where θ is particle temperature of carbon and h*c =
If Qgen

(Sh/ Nu ) B ( R hc/[E Le cp]); DaIII,mod=[Ahet νO2 d/Sh Dw]. (seeText) . Ignoring change in

concentrations of fuel and oxidizer on Q gen
 vs θ occurs such that
(a) show that maximum slope of Qgen
 = exp(1/θ)/ DIII,mod
{1/θ} = 2 (1+x)/(x-1) where Qgen

(b) If heat loss QL*= θ-θ∞, show that at the limiting DIII,mod,
{1/θ} = 2 {(1/hc*) +1}1/2, and limiting DIII,mod is given by
DIII,mod = [{1/(2θ)}-1] exp (1/θ)/[ 1+ {1/(2θ)}]

Problem 15.7.

Consider the solution in text for non-dimensional speed for any order of reaction with fuel
Λ = {YF,02hc* )/2} /{∫ θ0 1 ( 1-θ) nF exp(-E*/θ) θα1-n1 dθ } (A)
Where Λ = AY {ρD / m  " 2 } YO2 nO2 T∞ α1 (p /RT∞) n1[1 /hc*]nF How does this equation simplify
if nF=1 and α1=n1 ( Hint Chapter 05 for exponential integral).

Problem 15.8

The premixed gas mixture C3H8 and 50 % excess air flows at an average velocity of 25 cm/s
through Bunsen burner of dia 1 cm. Flow is assumed to be laminar. Determine the flash back
and blow off limits. Let T0=298 K; hc=46357 kJ/kg ( Lower HV); p= 1bar
assume λ=7.63E-05 kW/(m K); cp= 1.2 kJ(kg K); then ρD = λ/cp; A, PRE EXPONENTIAL
FACTOR (1/s):- 4.84E+09 for kmole/m^3 s of fuel which could be converted into AY
as 1.3e09; E= 125604 kJ/kmole, nF= 10.1; nO2= 1.65; alpa1= n1=0; d quench= 3.9 mm
Problem 15.9


5

Les avantages d'acheter des résumés chez Stuvia:

Qualité garantie par les avis des clients

Qualité garantie par les avis des clients

Les clients de Stuvia ont évalués plus de 700 000 résumés. C'est comme ça que vous savez que vous achetez les meilleurs documents.

L’achat facile et rapide

L’achat facile et rapide

Vous pouvez payer rapidement avec iDeal, carte de crédit ou Stuvia-crédit pour les résumés. Il n'y a pas d'adhésion nécessaire.

Focus sur l’essentiel

Focus sur l’essentiel

Vos camarades écrivent eux-mêmes les notes d’étude, c’est pourquoi les documents sont toujours fiables et à jour. Cela garantit que vous arrivez rapidement au coeur du matériel.

Foire aux questions

Qu'est-ce que j'obtiens en achetant ce document ?

Vous obtenez un PDF, disponible immédiatement après votre achat. Le document acheté est accessible à tout moment, n'importe où et indéfiniment via votre profil.

Garantie de remboursement : comment ça marche ?

Notre garantie de satisfaction garantit que vous trouverez toujours un document d'étude qui vous convient. Vous remplissez un formulaire et notre équipe du service client s'occupe du reste.

Auprès de qui est-ce que j'achète ce résumé ?

Stuvia est une place de marché. Alors, vous n'achetez donc pas ce document chez nous, mais auprès du vendeur mizhouubcca. Stuvia facilite les paiements au vendeur.

Est-ce que j'aurai un abonnement?

Non, vous n'achetez ce résumé que pour €28,84. Vous n'êtes lié à rien après votre achat.

Peut-on faire confiance à Stuvia ?

4.6 étoiles sur Google & Trustpilot (+1000 avis)

77858 résumés ont été vendus ces 30 derniers jours

Fondée en 2010, la référence pour acheter des résumés depuis déjà 14 ans

Commencez à vendre!
€28,84
  • (0)
  Ajouter