Thermodynamics 1 PDF

Preview

Summary

PPE PROBLEM SET Thermodynamics 2 At 180c the entropy of a substance is 5 KJ/K, the quality of the substance is: At 180 ° c:Ssat=6. KJkg−kSact<Ssat:wet¿get x:Sact=sf+xsfg5.2=2+x(4)x=0¿68% 2. At 10m3 vessel initially contains 5m3 of saturated water vapor at 1000kPa Vv= 5 m 3 VL= 5 m 3 Vv=mvVg v...

Description

PPE PROBLEM SET Thermodynamics 2

1. At 180c the entropy of a substance is 5.2 KJ/K, the quality of the substance is:

At 180 ° c :

KJ Sact < Ssat :wet ¿ get x : kg − k x=0.688333 ¿ 68.833 %

S act=sf + xsfg

Ssat=6.5857

5.2=2.1396 + x ( 4.4461 )

2. At 10m3 vessel initially contains 5m3 of saturated water vapor at 1000kPa 3 3 vL=mL vF 5=mv ( 1.694 ) Vv=5 m VL =5 m Vv =mv Vg mL =4793.86385 kg 5=mL(0.001043 )mv=2.95159 kg mv 2.9515 = mT=mV + mL ¿ 2.95 + 4793.8 mT =4796.8154 kg ¿ get x : x= mT 4796.8 x=0.0006153236 ¿ get u : u=uf + xufg ¿ 417.3+ ( 0.006153236 )( 2506 )

u=418.842

KJ ( 4796.81 kg ) kg

6

u=2009107.781 KJ

¿ 2 X 10 KJ

3. A vessel with a volume of 1 m3 contains liquid water and vapor in equilibrium at 600 kPa. The liquid water has a mass of 1kg. Using steam tables, calculate the mass of water vapor at 600 kPa: vf= 0.001101 m3/kg, vg= 0.3157 m3/kg

vT =Vv + vL

¿ mV ( .3157 )+1 ( 0.001101 )

¿ mV Vg+mLVf

mV =3.1641 kg

4. A steam has a condition of 2 MPa and 250C undergoes a constant pressure process until its quality is 50%. What is the heat rejected by steam?

at 2 Mpa

tsat=212.42 ° c

h 1=2902.5

KJ kg

P=2 mPa hf =908.79

for pt 2

x=0.50 hfg=1890.7 Q=h 2−h 1

tsat hact : wet hf =1134.37, hfg=1662.50 , kg KG 1861=1134.37 + x ( 1662.50) x=0.4370706 v 1=vf + xvfg

h sat=2796.9

at 260 ℃

h=hf 1+ xhfg 1

¿ 0.0012755 + 0.437 (0.042 ) u 1=uf 1+ xufg1

3

m kg

¿ 0.0191667

1128.39+ (0,437 )( 1470.03)

vg 0.019294 0.0191667 0.018975

u 1=1771.146137

KJ kg

for pt 2

ug 2551.8 U2 2550.1

¿ 1 ( 2551.1219 −1771.1461 )

Q=u 2−u 1

Q=779.9758 KJ

8. One kg of steam at 121C and 10% moisture undergoes a constant volume process until the pressure becomes 0.28MPa. Determine the final temperature in C.

at 121 ℃ , x 1=90 %

v 1=vf + x vfg

vf =0.0010612

¿ 0.00106 + 0.9 ( 0.8659 )

at tablw 3

VsatVact : wet

¿ find x

x=29.4725 %

10. There are 2.27 kg/min of steam undergoing an isothermal process from 27.5 bar, 316C to 6.8bar. Determine the change entropy, KJ/min.

for pt 1

p=2.75 Mpa , t=316 ℃

tsat=229.11 ℃

tsat...