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Debora [2.8K]
2 years ago
12

It is possible to design a reactor where the scy conductor and the nitrogen/ammonia electrode operate at different temperatures.

Which combination of temperatures is expected to give the best results?.
Engineering
1 answer:
schepotkina [342]2 years ago
4 0

The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.

<h3>Why the case above?</h3>

The answer is option A because the proton conductivity of SCY is known to be one that often goes up along with increasing temperature, and the favorability of reaction goes down or reduces with total temperature.

Hence, It is said to be good for one to to keep the SCY conductor at a higher temperature and as such, The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.

See options below

. Which combination of temperatures is expected to give the best results?

A.

SCY temperature higher than electrode temperature

B.

SCY temperature lower than electrode temperature

C.

SCY temperature the same as electrode temperature

D.

The temperature of the components does not make a difference.

Learn more about reactor from

brainly.com/question/15051406

#SPJ1

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a)  Δh = 2 cm,  b) Δh = 0.4 cm

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Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = \sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}

in this expression the densities are constant

        v = A  √h

       A =\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }

 

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = \sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = \frac{1}{454.55^2} \ v^2

       Δh = \frac{dh}{dv}   Δv

       \frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}

Suppose we have a height reading of h = 20 cm = 0.20 m

             

a) uncertainty 2.5 m / s ( 0.05)

        \frac{\delta v}{v} = 0.05

       \frac{\Delta h}{h} = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        \frac{\Delta h}{h} =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

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3 years ago
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