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Eduardwww [97]
3 years ago
10

Methane gas at 25°C, 1 atm enters a reactor operating at steady-state and burns with 80% theoretical air entering at 227°C, 1 at

m. An equilibrium mixture of CO2, CO, H2O(g), H2, and N2 exits at 1427°C, 1 atm. Assume N2 is inert.
Determine, per kmol of methane entering;
(a) the composition of the exiting mixture.
(b) the heat transfer between the reactor and its surroundings, in kJ.
Neglect kinetic and potential energy effects.

Engineering
1 answer:
Damm [24]3 years ago
5 0

Answer:

The composition of existing misture are,

a.

Co2 =0.5679

Co= 0.9326

H2O=1.6326

H2 = 0.3679

N2 = 0.6016

b.

K= 3.3878

Explanation:

Please see attachment for the solving.

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Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the
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Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$P(h

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore,   $x_1=368 \ veh/h$

                        $=\frac{368}{1000} = 0.368$

Given,   $t_1=2+x_1$

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, $t_2=t_1$

∴  $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

    = 1368 veh/h

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8 0
3 years ago
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The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)
xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

\sigma = n|e|\mu_e+p|e|\mu_h (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

\mu_e= mobility of electron

\mu_h=mobility of holes

\sigma = electrical conductivity

Making the substitution in (S)

Mobility of electron

8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)

0.639=\mu_e+\mu_h

Mobility of hole in (S)

2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))

0.1436 = 7.6*10^{-3}\mu_e+\mu_h

Then, solving the equation:

0.639=\mu_e+\mu_h (1)

0.1436 = 7.6*10^{-3}\mu_e+\mu_h (2)

We have,

Mobility of electron \mu_e = 0.5m^2/V.s

Mobility of hole is \mu_h = 0.14m^2/V.s

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Which of the following can minimize engine effort and
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