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VikaD [51]
2 years ago
9

10.0 kmol of a 40.0 mol% methanol and 60.0 mol% water mixture is processed in a normal batch distillation system with a still po

t that acts as an equilibrium stage and a column with two equilibrium stages (total of three equilibrium contacts). The column has a total condenser, and reflux is a saturated liquid. The column is operated with a varying reflux ratio so that xD is held constant. We desire a final still-pot concentration of 8.0 mol% methanol, and distillate should be 84.0 mol% methanol. Pressure is 1 atm, and CMO is valid.
Required:
a. What initial external reflux ratio, LD, must be used?
b. What final external reflux ratio must be used?
c. How much distillate product is withdrawn, and what is the value of Wfinal?
Engineering
1 answer:
Serggg [28]2 years ago
3 0

Answer:

[a]. 0.49.

[2]. 0.536

[c]. 4.15 kmol; 5.84 kmol.

Explanation:

Without mincing words let's dive straight into the solution to the question above.

                                                            [a].

The initial external reflux ratio, LD that must be used = [(0.85 - 0.57)/ 0.85 - 0]/ [ 1 - (0.85 - 0.57)/ 0.85 - 0].

The initial external reflux ratio, LD that must be used = 0.329/ 1- 0.329 = 0.49.

                                                             [b].

The final external reflux ratio that must be used = [ 0.85 - 0.13/ 0.85 - 0]/ [ 1 - 0.85 - 0.13/ 0.85 - 0].

Hence, the final external reflux ratio that must be used =0.847/ 1 - 0.847 = 5.536.

                                                              [c].

The amount of distillate product that is withdrawn:

4 = 0.85 H(t) + 0.8 - 0.08.

H(t) = 4.15 kmol, and the value of Wfinal = 5.84 kmol.

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3 years ago
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3 years ago
P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si
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complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

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A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

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      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

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       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

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Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

4 0
3 years ago
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Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

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