Day 2 (2)

Problem Set for Class number 6: Problem Set for Class number 6: Calc 1 & 2:

Calc 1 & 2:

6. A stoichiometric problem was

6. A stoichiometric problem was solved on the basis of 100 solved on the basis of 100 moles dry flue gas (DFG). The moles dry flue gas (DFG). The given conditiongiven condition at the stack outlet are as follows 780 mmHg, 970 K and the partial pressure is 24 mm Hg. The volume of at the stack outlet are as follows 780 mmHg, 970 K and the partial pressure is 24 mm Hg. The volume of the wet flue gas is:

the wet flue gas is: Given:

Given:

Reqd: Total Volume Reqd: Total Volume Solution: Solution:

nHO

nHO

nToal+nH

nToal+nH

==

PP





PP

A

A

nHO

nHO

 ol+nH

 ol+nH

==

4 Hg

4 Hg

8 Hg

8 Hg

nH20= 3.1746 mol nH20= 3.1746 mol (PV = nRT) (PV = nRT)totaltotal V= V=

((+.46

+.46))ol x .8

_{ }

_{ }

ol x .8

 

 

 

 

 x 9 K

 x 9 K

 

 

 a

 a

 x x







 L

 L

= 8.0010 m= 8.0010 m33

≈≈

_{8 m8 m}33 Transpo: Transpo:

29. In a natural gas pipeline at Station 1 the pipe diameter is 2 ft and the flow conditions are 800 psia, 29. In a natural gas pipeline at Station 1 the pipe diameter is 2 ft and the flow conditions are 800 psia, 60

60oo_{F and 50 ft/s velocity. At station 2, the pipe diameter is 3 ft and the flow conditions are 500 psia, 60F and 50 ft/s velocity. At station 2, the pipe diameter is 3 ft and the flow conditions are 500 psia, 60}oo

F. What is the mass flowrate

F. What is the mass flowrate in kg/sin kg/s Given: Given: 100 Moles DFG 100 Moles DFG P=780 mmHg P=780 mmHg T= 970 K T= 970 K P PH20H20= 24 mmHg= 24 mmHg Condition

Condition 1: 1: Condition 2:Condition 2: Diameter:

Diameter: 2ft 2ft Diameter: Diameter: 3ft3ft Pressure:

Pressure: 800 800 psia psia Pressure: Pressure: 500 500 psiapsia Temperature= 60

Temperature= 60oo_{F F Temperature= Temperature= 6060}oo

Velocity:50

Solution:

Simple solution 1: Reqd: Mass flowrate, m



=

8 psia x

. 

 

 x 6





.

_{ }



 x R

 = 2.2932



lb



m1= m2 m1=



1 A1 V1 m1= 2.2932

lb





 x

π

4

(2ft



)

 x 50



s

 x

 kg

.4 lb

 = 163.4369 kg/s Trial and error solution namalapit sa sagotni Anti:



1=

(8−4.)psia x

 

. 

 x 6





.

_{ }



 x R

 = 2.2511



lb





2=

(−4.)psia x

 

. 

 x 6





.

_{ }



 x R

 = 1.3911



lb



Using Bernoulli equation:

(8−4.) x 44

 

_

.

_





(−4.) x 44

 



.9

_





(



_

₎

_

_−V

_

 x .4

 

_{ } 

 = 0 V2 = 49.1645



s

m1= = 2.2511

lb





x

π

4

(2ft



)

 x 50



s

 x

 kg

.4 lb

 = 160.4365 kg/s m2= = 1.3911

lb





x

π

4

(3ft



)

 x 49.1645



s

 x

 kg

.4 lb

 = 219.3465 kg/s mave =

6.46+9.46



 kg/s =189.8915 kg/s

≈

 184 kg/s

Thermo 1:

12. Calculate the pressure in KPa of steam at a temperature of 500 o _{C and a density of 24 kg/m}3 _using

the ideal gas law: Given: T= 500o_C



 = 24 kg/m3 Reqd: Pressure Solution: 24

kg





=

P x 8





8.4

_{ }

J

 x . k

P= 8570.6255 KPa

 ≈

 8570 KPa Thermo 2:

45. An ideal Rankine cycle with reheat operates the boiler at 3 MPa, the reheater at 1 MPa and the condenser at 50 KPa. The temperature at the boiler and reheater outlets is 350o  _{C. The boiler and}

reheaterare fired with a fuel that releases 9000 kJ/kg of heat as it is burned. What is the mass flowrate of the fuel for such a cycle when sized to produce 50 MW of network?

Solution: Wnet= 50 MW

Thermodynamics of water using Handbook: Condition 1: 3Mpa SHS, Tsat= 506.8554 K 350o_C H1MPa= 3160.9078 KJ/kg H3MPa= 3114.3190 KJ/ kg H5MPa = 3067. 7301 KJ/kg S1MPa= 7.3043 KJ/kg K S3MPa = 6.8746 KJ/kg K S5MPa = 6.4449 KJ/kg K

Condition 2: P2= P3= 1 MPa H2= 2190.4960 KK/kg S1 = S2 = 6.8749 KJ/kg K Condition 3: T1 = T3 = 350oC SHS H3 =3285.9911 KJ/kg P2 = P3 = 1 MPa , Tsat = 184.3772 oC S3 = 7.3043 KJ/kg K Condition 4: P4=50 KPa Saturated SL(1.0869) < 7.3043 < SV (7.6065) S3 = S4 = 7.3043 KJ/kg K 7.3043 = 1.0869(1-X) + 7.6065 X X = 0.9536 HL= 339.1120 KJ/kg H4= 339.1120 (1-0.9536) + 2646.7031(0.9536) HV= 2646.7031 KJ/kg H4 = 2539.6309 KJ/kg Mass flowrate =

 MW x

  



x

 

 

x

 

 r

9.69

_

J

 x

_{ }



x

  

_{ }

 = 70.8764 Mg/ hr

≈

70 Mg/ hr Heat transfer:

41 Calculate the temperature of its inner surface for a day during which the room is maintained at 20 o_C

while the temperature of the outdoors is -10o_{C. h}

1 = 10 W/m2oC ; h2 = 40 W/m2oC

Solution:

From previous problem:

Q=

_

.(.8)(− −) C

/

+

/



+

./

   

+

./

 

Q= 69.2478 W 69.2478 W=

.  x .8  (−T

_



)

/

+

./

 

= 13.9376

≈.

o_C

Momentum Transfer:

14. Calculate the frictional pressure drop across the bed when the volume flow rate of liquid is 1.44 m3_{/hr. Use Ergun Equation}

Solution



= 0.5 U=

.44



r

 x

 

r

.4 



 = 0.01



s

Re =

.  x .





 x 8





. pa s

 = 4

∆

 P =

h{[150

u μ(−)



∅



_



1.75][

ρu



_

(−)



_∅

]}

∆

 P =

1 m150

.





 x . pa s x (−.)



.



 _{x .}



1.75

8





 x .







_{(−.)}

.



_{x .}



∆

 P = 6560 Pa