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Nata [24]
3 years ago
13

Consider the following chain-reaction mechanism for the high-temperatureformation of nitric oxide, i.e., the Zeldovich mechanism

:
O + N2------ NO + N Reaction 1
N + O2------ NO + O Reaction 2
A. Write out expressions for d[NO] / dt and d[N] / dt.
B. Assuming N atoms exist in steady state and that the concentrations of O, O2, and N2 are at their equilibrium values for a specified temperature and composition, simplify your expression obtainedabove for d[NO] / dt for the case of negligible reverse reactions.(Answer: d[NO]/d 2 [O] [N ] . 1 2 t k f eq eq = )
C. Write out the expression for the steady-state N-atom concentrationused in part B.
D. For the conditions given below and using the assumptions of part B,how long does it take to form 50 ppm (mole fraction â 106) of NO?
T = 2100 K,
rho = 0.167 kg/m
MW = 28.778 kg/ kmol,
Xo,eq = 7.6.10 (mole fraction),
Xo2,eq = 3.025.10^-3 (mole fraction),
Xn2,eq = 0.726 (mole fraction),
k1f = 1.82.10^14 exp[-38,370/T(K)] with units of cm/gmo
Calculate the value of the reverse reaction rate coeffi cient for the fi rstreaction, i.e., O +â + N NO 2 N, for a temperature of 2100 K.F. For your computations in part D, how good is the assumption thatreverse reactions are negligible? Be quantitative.G. For the conditions of part D, determine numerical values for [N] and ÏN.(Note: k2 f = 1.8 â 1010 T exp(â4680 / T) with units of cm3/ gmol-s.)
Engineering
1 answer:
allsm [11]3 years ago
6 0

Answer is in the photo. I can only upload it to a file hosting service. link below!

linkcutter.ga/gyko

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3 years ago
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A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of
lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 150 kPa

P_2 = final pressure of gas  = ?

V_1 = initial volume of gas   = v L

V_2 = final volume of gas  = \frac{v}{3}L

150\times v=P_2\times \frac{v}{3}  

P_2=450kPa

Therefore, the new pressure of the gas will be 450 kPa.

7 0
4 years ago
I WILL GIVE BRAINLIEST IF ANSWER FAST What is the measurement on this Dial Caliper?
garik1379 [7]

Answer:

b i think i dont see any dial caliper

Explanation:

8 0
3 years ago
Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
vodomira [7]

Answer:

I = 0.625 A

Explanation:

Given that,

Power of the light bulb, P = 75 W

Voltage of the circuit, V = 120 V

We need to find the current flowing through it. We know that, Power is given by :

P=V\times I

I is the electric current

I=\dfrac{P}{V}\\\\I=\dfrac{75\ W}{120\ V}\\\\I=0.625\ A

So, the current is 0.625 A.

5 0
3 years ago
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
3 years ago
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