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ME 607 – Leading-edge Thermodynamics – Homework Answers

Week 1 Answers
1.4 T3 = 100C

1.7b. DeltaS1-2 = 0.199 kJ/kg-K, Q1-2 = 49.7 kJ/kg, W1-2 = -49.7 kJ/kg
1.7c.

Q2-3 = 0, DeltaS2 = 0, W2-3 = 57.3kJ/kg
1.7d. W3 advanced thermodynamics homework 0, Q3-1 = -57.33 kJ/kg, S1-3 = -0.199 kJ/kg-K
1.7e. COPHP = 7.5

1.10a.

Advanced Thermodynamics Physics Investigation Help

v = 5.58 bars
1.10b. Queen = 1.71 kJ
1.10c. DeltaSTOTAL = 0.0091 kJ/K

1.16a. TF = hungry to get worship ebook assessment essay Chemical, DeltaSTOTAL = 2.402 kJ/K
1.16b.

Irreversible
1.16c.

Chapter 6 Answers

University of malta stockpile dissertations = 2.560 kJ/K

1.18a. (1) mdot = 18.03 kg/min
1.18a. (2) v3 = 0.060 m/s
1.18b. (1) mdot = 686.4 kg/min
1.18b.

ganeshotsav essay or dissertation throughout marathi language v3 = 227 m/s

1.26 TR = 416.7 C

1.28a.

TF = TA/(1-(R/2Cp)*(1-Po/Pa))
1.28b. TF = 2kTa/(k+1)
1.28c. TF = Ta

1.30 q = 383 KJ

Week advanced thermodynamics homework Answers
2.9a.

Thermodynamics

deltaSH2 = 0.6264 kJ/K, sigmaH2=0
advanced thermodynamics homework. sigmatotal = 0.1772 kJ/K
2.9d. sigmaH2 = 0.1273 kJ/K, sigmatotal = 0.1772 kJ/K
free business recovery program web theme essay. sigmaH2 = 0.6264 kJ/K

2.17a. p2 = 119 kPa
2.17b. 0.297

2.18a. sigmacv = 0.180 kJ/kg K
2.18b.

Thermodynamics I

1.0

2.21a. w = -625.1 kJ/kg
2.21b. x2 = 0.902
2.21c. qstream = -24 kJ/kg
2.21d. m = 618.2 kJ/kg

Week 3 Answers –

3.2a. f500 = -4.04 kJ

3.2b.f280 = 0.643 kJ

I = 11,500 kJ

3.8a.W = 3.576 kJ

3.8b.Won min = 0.57 kJ

3.8c.I =3.01 kJ

3.8d.e = 0.16

3.11a.

Advanced Thermodynamics

Watts = 43.17 kJ

3.11b. Wreversible on = 3.8 kJ

3.11c. I just = 39.4 kJ

3.11d.e = 0.088

3.16a.w = 271.9 kJ/kg

3.16b.Df = 97.8 kJ/kg

3.16c.i =174.1 kJ/kg

3.16d.e = 0.36

3.19 DERIVATION

 

Week 5 Answers

 

4.2 a) that i = 73.5 kJ/min, age = 0.706

b) irefrig = 78.69 kJ/min, iQH = 21.3 kJ/min, iQL = 23.5 kJ/min, o = 0.588

 

4.5 a) erinarians =0.20 kJ/K

b) 2009 hsc native english speakers marking critical elements for essay = 60kJ

c) –EQL=100kJ

d)e = 0.625

e) COPact = 3.125

f) land W not vs Q,W

 

4.16 a)y5 = 81.94 kJ/kg

b) i= 8.03 kJ/kg

 

4.25          a)Dy(tank, 1st) = -680.3 kJ/kg

Dy(tank, 2nd) = -785.8 kJ/kg

Dy(cond) = -135.2 kJ/kg

Dy(heater) = -60.7 kJ/kg

Dy(pump1) = 1.0 kJ/kg

Dy(pump2) = 14.2 kJ/kg

b) i(tank, 1st) = 62.1 kJ/kg

i(tank, 2nd) = 146.2 kJ/kg

i(cond) = 113.3 kJ/kg steam

i(heater) = 59.9 kJ/kg

i(pump1) = 0.3 kJ/kg

i(pump2) = 3.5 kJ/kg

c)      ecycle = 0.760

 

Chapter 6 Answers

6.7              a) cp = 2.601 kJ/kg-K, cp = 2.578 kJ/kg-K

b) cp = 2.775 kJ/kg-K, cp = 2.789 kJ/kg-K

6.11

thanksgiving for denmark essay 6.13 (dv/dT)P = 0.0001797 m3/kg-K, (ds/dP)T = -0.0001796 m3/kg-K

 


6.15 a)DuT,RK=(3a/(2T0.5))(-1/b ln(((v2+b)v1)/((v1+b)v2))

b)DuT = 4.93 kJ/kg

c) DuT = 5.64kJ/kg

 

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