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sertanlavr [38]
3 years ago
9

What is the composition of the two phases that form when a stream of 40% A, 39% B, and 21% C separates into two phases? Label th

e phases as A-rich or C-rich. Is there more of the A-rich or C-rich phase?
Engineering
1 answer:
irga5000 [103]3 years ago
8 0

Answer:

vapor fraction = 0.4 and 0.08

Explanation:

At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

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A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can
Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a)   ln(T2/T1)=μβ         β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π  radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π  

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb     T1=80 lb   μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

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3 years ago
What is a voltage divider circuit and how do you calculate the voltage across one element in a series
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Army people are good people right
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You are investigating surface hardening in iron using nitrogen gas. Two 5 mm thick slabs of iron are separately exposed to nitro
Luden [163]

Answer and Explanation:

The explanation is attached below

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3 years ago
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin
Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$  ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

       $=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

       $=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

       $=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0
3 years ago
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