Consider a mixture of hydrocarbons that consists of 60 percent (by volume) methane, 30 percent ethane, and 10 percent propane. After passing through a separator, the mole fraction of the propane is reduced to 1 percent. the mixture pressure before and after the separation is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture.

Answers

Answer 1
Answer:

Answer:

\Delta P_m=6\text{kPa}\n\Delta P_e=3\text{kPa}\n\Delta P_p=-9\text{kPa}

Explanation:

mole fraction of propane after passing through the separator is \beta_p

(\beta)/(0.6+0.3+\beta)=0.01

\beta =9.09* 10^-^3

mole fractions of ethane \beta _e and methane \beta_m after passing through separator are:

\beta_e =(0.3)/(0.3+0.6+0.00909)=0.66\n\beta_m=(0.6)/(0.3+0.6+0.00909)=0.33

Change in partial pressures then can be written as:

\Delta P=(y_2-y_1)\cdot P  where y_2 and y_1 are mole fractions after and before passing through the separator

Hence,

\Delta P_m=(0.66-0.6)\cdot 100\text{k}=6\text{kPa}\n\Delta P_e=(0.33-0.3)\cdot 100\text{k}=3\text{kPa}\n\Delta P_p=(0.01-0.1)\cdot 100\text{k}=-9\text{kPa}


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The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.

Answers

Answer

Given,

Reservoir pressure of a supersonic wind tunnel = 5 atm

Local Mach number = ?

Area ration = ?

a) 4 atm.

  Pressure ratio =(4)/(5)

                         = 0.8

From Isentropic Flow Tables

  M = 0.58   A/A* = 1.213

b) 2.64 atm

  Pressure ratio =(2.64)/(5)

                         = 0.528

From Isentropic Flow Tables

  M = 1  A/A* = 1

c)  0.5 atm

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From Isentropic Flow Tables

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8. Removing damage by tapping it out with a hammerand dolly is often referred to as
(A) contouring
(B) roughing out
(C) forming
(D) tapping out

Answers

Answer:

tapping out I think sorry if I'm wrong

Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car

Answers

Answer:

3 m/s²

Explanation:

Acceleration is calculated as :

a= Δv/ t

where ;

Δv = change in velocity

Δv = 45 - 0 = 45  m/s

t= 15 s

a= 45 /15

a= 3 m/s²

At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?

Answers

Answer and Explanation:

The computation is shown below

Total average cost + total variable cost = total cost

Let number of output be x

So,

Total fixed average cost = x × $30

Total variable cost = x × $15

Total cost = $2,500

Therefore,

$20 × x + $30 × x = $2,500

50 × x = $2,500

x = 50

Now the total variable cost is

= 50 × $20

= $1,000

And, the fixed cost is

= 50 × $30

= $1,500

Answer:

12 4

Explanation:

because the production average is variable

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

Answers

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

(d_(2) )/(d_(1) ) =(1)/(2) (\sqrt{1+8F^(2) } -1)

10*2=\sqrt{1+8F^(2) } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_(1) }

Here, g = gravity = 32.2 ft/s²

v=7.4162*√(32.2*1) =42.0833ft/s

b) The flow rate:

q=v*L*d_(1) =42.0833*80*1=3366.664ft^(3) /s

c) The Froude number:

v_(2) =(q)/(L*d_(2) ) =(3366.664)/(80*10) =4.2083ft/s

F=\frac{v_(2)}{\sqrt{gd_(2) } } =(4.2083)/(√(32.2*10) ) =0.2345

d) The flow energy dissipated:

E=((d_(2)-d_(1))^(3) )/(4d_(1)d_(2)) =((10-1)^(3) )/(4*1*10) =18.225ft

e) The critical depth:

d_(c) =((((q)/(L))^(2)  )/(g) )^(1/3) =((((3366.664)/(80))^(2)  )/(32.2) )^(1/3) =3.8030ft

Why do we care about a material's ability to resist torsional deformation?(A) Because the angle of twist of a material is often used to predict its shear toughness
(B) Because rotating shafts are used in engineering applications
(C) We don't care - simply an academic exercise.
(D) Because it can determine G and inform us of a materials ability to resist shear deformation

Answers

Answer:

(A) Because the angle of twist of a material is often used to predict its shear toughness

Explanation:

In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.

The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.

The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:

1- Tangential tensions appear parallel to the cross section.

2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.

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