ENGINEERING COLLEGE

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 process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal

Answers

Answer:

Isobaric process

Explanation:

The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.

At constant pressure, the work done is given by :

$W=p* \Delta V$

Where

W is the work done by the system

p is the constant pressure

$\Delta V$ is the change in volume

So, the correct option is (c) " isobaric process ".

Use Euler’s Method: ????????????????=???? ????????????????=−????????−????3+????cos(????) ????(0)=1.0 ????(0)=1.0 ????=0.4 ????=20.0 ℎ=0.01 ????=10000 ▪ Write the data (y1, y2) to a file named "LASTNAME_Prob1.dat" Example: If your name is John Doe – file name would be "DOE_Prob1.dat" ▪ Plot the result with lines using GNUPLOT (Hint: see lecture 08) ▪ Submit full code (copy and paste). Plot must be on a separate page. ▪ Run the code again and plot the result for: ????=0.1 ????=11.0

Answers

Answer:

Too many question marks

Explanation:

One kind of SS-3xX steel alloy has a melting point of 1450°c. Its specific heat = 0.46 J/g°C, and its heat of fusion 270 J/g. For a 200 kg block of this steel, determine how much heat is required to (a) raise its temperature from 25°C to its melting point and (b) transform it from solid to liquid phase.

Answers

Answer:

a)Q=131.1 MJ

b)Q=54 MJ

Explanation:

Given that

Mass ,m=200 kg

Specific heat Cp=0.46 J/g°C

Cp=0.46 KJ/kg°C

Heat of fusion = 270 J/g

Heat of fusion = 270 KJ/kg

Melting point temperature = 1450°C

Initial temperature = 25°C

Final temperature=1450°C

Heat required to rise temperature from 25°C to 1450°C.

Q= m CpΔT

Q=200 x 0.46 x (1450-25) KJ

Q=131,100 KJ

Q=131.1 MJ

Heat required to transform from solid phase to liquid phase

Q= Mass x heat of fusion

Q=200 x 27 KJ

Q=54,000 KJ

Q=54 MJ

Will mark brainliest if correctWhen a tractor is driving on a road, it must have a SMV sign prominently displayed.

True
False

Answers

Answer: true

Explanation:

A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.

Answers

Answer:

V₂=1.76 m³

P=222.03 KPa

Explanation:

Given that

For tank 1

V₁=1 m³

T₁= 10°C = 283 K

P₁=350 KPa

For tank 2

m₂=3 kg

T₂=35°C = 308 K

P₂=150 KPa

We know that for air

P V = m R T

P=pressure ,V= Volume,R= gas constant ,T= temperature ,m =mass

for tank 2

P₂ V₂ = m₂ R T₂

By putting the values

150 x V₂ = 3 x 0.287 x 308

V₂=1.76 m³

Final mass = m₁+m₂

m =m₁+m₂

The final volume V= V₂+V₁

V= 1.76 + 1 m³

V= 2.76 m³

The final temperature T= 19.5°C

T= 292.5 K

$m=(PV)/(RT)$

$m_1=(P_1V_1)/(RT_1)$

$m_1=(350* 1)/(0.287* 283)$

$m_1=4.3\ kg$

m =m₁+m₂

m =4.3 + 3 = 7.3 kg

Now at final state

P V = m R T

P x 2.76 = 7.3 x 0.287 x 292.5

P=222.03 KPa

The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?

Answers

Answer:

891.027 lbm/ft³

Explanation:

See it in the pic

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