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]](https://tex.z-dn.net/?f=k_%7B1f%7D%20%5BO%5D%20%5BN_%7B2%7D%5D%20%2B%20K_%7B2f%7D%20%20%5BN%5D%5BO_%7B2%7D%20%5D%20%2B%20K_%7B3f%7D%5BN%5D%5BOH%5D)
d[N] / dt = ![k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH]](https://tex.z-dn.net/?f=k_%7B1f%7D%20%5BO%5D%20%5BN_%7B2%7D%5D%20%2B%20K_%7B2f%7D%20%20%5BN%5D%5BO_%7B2%7D%20%5D%20-%20K_%7B3f%7D%5BN%5D%5BOH%5D)
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
Expression for d[N] / dt can be written as
Parallel hookups increase amp hour capacity but voltage remains the same... 1.5 volts
Series the two and voltage is 3.0 volts.
Answer: heat flux into the fun is 21.714 mW/m^2
Explanation:
Heat flux Q = q/A
q = heat transfer rate W
A = area m^2
q = area * conductivity * temperature gradient
Temperature gradient = difference in temperature of the metal faces divided by the thickness.
Therefore Q = k * ( temp. gradient)
Q = 200 * ((400-20)/3.5*10^-2)
Q = 21714285.71 = 21.714 mW/m^2
Answer 2: convective heat transfer flux between fin and air
is 3800W/m^2
Explanation :
q = hA*(Ts-Ta)
h = convective heat transfer coefficient
Ts = temperature of fin
Ta = temperature of air
Q = q/A
Q = h(Ts-Ta)
Q = 10(400 - 20)
Q = 3800 W/m^2
