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sveticcg [70]
3 years ago
11

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

:
O+ N2 → NO + N
N + O2 → NO+ O

Write out expressions for d[NO] / dt and d[N] / dt.
Engineering
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer: hello attached below is the properly written chain reaction  to your question

answer :

d[NO] / dt = k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] + K_{3f}[N][OH]

d[N] / dt = k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] - K_{3f}[N][OH]

Explanation:

<u>write out expressions for  d[NO] / dt and d[N] / dt</u>

Given :

properly written chain reaction ( attached below)

Expression for d[NO] / dt can be written as

k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] + K_{3f}[N][OH]

Expression for d[N] / dt can be written as

k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] - K_{3f}[N][OH]

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3 years ago
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A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla
creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed =  0.5 m

cohesion = 25 kPa

internal friction angle =  5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

and we know formula for N_{\gamma } is

N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

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7 0
4 years ago
Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is
ryzh [129]

Answer:

a) \dot W = 0.978\,kW, b) I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}

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The reversible work input is:

\dot W = \frac{\dot Q_{H}}{COP_{HP}}

\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

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b) The irreversibility is given by the difference between real work and ideal work inputs:

I = \dot W_{real} - \dot W_{ideal}

I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW

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