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# Tutor profile: Michael S.

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Michael S.
PhD student in Chemical Engineering
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## Questions

### Subject:Physics (Thermodynamics)

TutorMe
Question:

Consider 2 air streams that are mixed. The first stream is 5 mol/s at 30 C while the the second one is at 3 mol/s 90 C. 1. Deduce the final temperature assuming that no heat is lost to the surrounding 2. What is the total change of entropy of the universe (or entropy generated)

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Michael S.

1. Then we can perform an energy balance Accumulation= energy from inlet flow - energy from outlet flow + Heat - Work Q=0 (no heat lost to the surrounding) W=0 ( No shaft Work) 0=n1h1+n2h2-(n1+n2)h3 (note that n3=n1+n2) 0=n1(h1-h3)-n2(h2-h3)=n1*cp(T1-T3)+n2Cp(T2-T3)=0 Assume cp constant=29 J/mol.K (cp will cancel out) 0=5*(30-T)+3(90-T) T=52.5 C 2. Delta S(sys)= Ds(gen) -Q/T Since no heat transfer to the surrounding, Q=0 Ds(generated)= Ds(system)= n1cpln(T3/T1)+ n2cpln(T3/T1) (entropy change for an ideal gas assuming isobaric conditions) Ds(generated)=10.386-9.487=0.899 J/K

### Subject:Pre-Algebra

TutorMe
Question:

Solve for X and Y 25^(x+y)= 5^(x)*25 (x/3)+(y/2)=3x-y

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Michael S.

5^(2x+2y)=5^x*5^2 2x+2y=x+2 x+2y=2 (1) 2x+3y=18x-6y -16x+9y=0(2) substitute 1 in 2 -16(2-2y)+9y=0 -32+32y+9y=0 -32+41y=0 y=0.78 X=0.439

### Subject:Chemical Engineering

TutorMe
Question:

1. Consider a mixture of acetone and water initially at 20 Degrees Celsius and atmospheric pressure. The mole fraction of Acetone is 0.4. Assume that the mixture is ideal. (2-Component Phase Equilibrium) A. Is the mixture a liquid, vapor, or a mixture of liquid and vapor? B. The pressure is reduced to 0.1 bar, what is the composition of both the vapor and liquid phase. C. Is the given assumption valid?

Inactive
Michael S.

A. At 20 Degrees Celsius, we can find the saturation pressures of both components using the Antoine's equation. Acetone: A: 4.424 B: 1312 C: -32.45 Psat(1)= 10^(A-(B/(t+C))/ P in bars and t in K Water: A:8.141 B:1811 C:-32.45 Psat(2)= 10^(A-(B/(t+C))/ P in mmHg and t in C Psat(1)=0.2446 bars Psat(2)=0.0259 bars Using Raoult's Law to find bubble point: yiP=xiPsat Summation of yi=1 Pb=x1Psat(1)+x2Psat(2)= 0.1134 bars Since P>Pbubble, Mixture is liquid B.Calculate Pdew summation(xi=1) P=1/(y1/p1sat+y2/p2sat)= 0.04 bars Since Pdew<P<Pb, then the the system is a mixture of both liquid and vapor. y1P1=x1P1sat(1) y2P2=x2P2sat(2) y1+y2=1(3) x1+x2=1(4) 4 equations 4 unknowns(replace 1 and 2 in 3 and solve for x) x1=0.3388 y1=0.829 x2=0.6612 y2=0.171

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