Day 2 (2)

Problem Set for Class number 6:

Calc 1 & 2: 6. A stoichiometric problem was solved on the basis of 100 moles dry flue gas (DFG). The given condition 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: Given: 100 Moles DFG P=780 mmHg T= 970 K PH20= 24 mmHg Reqd: Total Volume Solution:

nHO = P nToal+nH PA nHO = 4 Hg  ol+nH 8 Hg nH20= 3.1746 mol (PV = nRT)total

  x 9 K  ( ) +.  46) 46 ol x .  8  V=  x = 8.0010 m ≈ 8 m   a   L   3

3

Transpo: 29. In a natural gas pipeline at Station 1 the pipe diameter is 2 ft and the flow conditions are 800 psia, 60o F and 50 ft/s velocity. At station 2, the pipe diameter is 3 ft and the flow conditions are 500 psia, 60 o F. What is the mass flowrate in kg/s Given: Condition 1: Diameter: 2ft Pressure: 800 psia Temperature= 60o F Velocity:50 ft/s

Condition 2: Diameter: 3ft Pressure: 500 psia Temperature= 60o Velocity:?

Solution: Simple solution 1: Reqd: Mass flowrate, m

   x 6  8 psia x lb .    = .    x R  = 2.2932   m1= m2 m1=



1 A1 V1

m1= 2.2932

lb x π (2ft) x 50  x  kg  = 163.4369 kg/s  4 s .4 lb

Trial and error solution namalapit sa sagotni Anti :

   x 6  (8−4.  )psia x .    = 2.2511 lb  = .      x R      x 6  (−4.  )psia x .    = 1.3911 lb  = .    x R   1

2

Using Bernoulli equation:

(8−4.) x 44    (−4.) x 44    ( )−V  = 0  .  .9   x .4      V  = 49.1645 s lb π   kg  = 160.4365 kg/s m1= = 2.2511  x (2ft  ) x 50  x  4 s .4 lb lb π   kg  = 219.3465 kg/s m2= = 1.3911  x (3ft  ) x 49.1645  x  4 s .4 lb 6.46+9.46 kg/s = 189.8915 kg/s ≈ 184 kg/s mave =  2

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= 500 oC



 = 24 kg/m3

Reqd: Pressure Solution:

 P x 8 kg  24  =  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 350 o  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

350oC H1MPa = 3160.9078 KJ/kg

H3MPa = 3114.3190 KJ/ kg

H5MPa = 3067. 7301 KJ/kg S1MPa = 7.3043 KJ/kg K S5MPa = 6.4449 KJ/kg K

S3MPa = 6.8746 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 = 350 o C

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

  x   x    MW x     r  = 70.8764 Mg/ hr ≈ 70 Mg/ hr Mass flowrate = J  9.69  x  x   Heat transfer: 41 Calculate the temperature of its inner surface for a day during which the room is maintained at 20 oC while the temperature of the outdoors is -10oC. h1 = 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 ≈. C   /+./ o

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

 x r . 4 4 r   = 0.01  U=  .4  s  x 8  .    x .   Re = . pa s   = 4  (−) ρu ∆ P = h{[150u μ(−) 1.75][ ∅ ∅ ]}  x . (−.)  x . pa s x (−.)  8 .      ∆ P = 1 m150 . x . 1.75 .x . 

∆ -

 P = 6560 Pa