PROBLEM SET IN OPTIMIZATION IN FLOW

OF HEAT

Submitted by: QTX Chemicals Inc.

Collado, Ellaine Vanezza

Dumandan, Florenze Jesse

Gecha, Andrea

Ulep, Michael

Problem 5:Determine the optimum economic thickness of insulation that should be used under the following conditions: saturated steam is being passed continuously in a vertical 1” Sch 40 steel pipe. The temperature of the steam is 180°C, and the steam is valued at $0.0035/kg. The pipe is to be insulated with a material that has a thermal conductivity of 0.06 W/mK. The cost of the installed insulation per m of pipe length is $ 200x, where x is the thickness of the insulation in meters. Life of the installation is 10 years with time value of money equal to 15%. The total length of the pipe is 250 m, and the average temperature of the surroundings is 27°C. Heat transfer resistances due to steam film, scale, pipe wall are negligible. The air film coefficient at the outside of the insulation may be assumed constant for all insulation thicknesses and given by the equation:

1

421.37 W/m K where Tw, Ts are in oC and L, mW ST T

hfL

− =

Solution:

Do 1.32 2.54⋅:= Do 3.353=

L 250:= To 180:=

bc 0.0035:=

K 0.06:=

ac200

100:= ac 2=

Hy 8760:=

λo 2013.5:=

Solution:

hr

5.672Tw 273+

100

4To 273+

100

4

−

Tw To−:=

hr 13.624=

Solution:To Tw−Dso Do−

200

K π⋅ L⋅Dso Do−( )

lnDso

Do

100⋅

⋅

Tw Tsur−1

hf hr+( ) π⋅Dso L⋅

100⋅

Tw Find Tw( ):=

Tw 29.959=

Problem 11:A condenser for a distillation unit must be designed to condense 2000 kg/h of vapor. The effective condensation temperature for the vapor is 132°C.The heat of condensation for the vapor is 350000 J/kg. Cooling water is available at 15°C.The cost of cooling water is $25/1000m3.The optimum Heat Transfer Coefficient is 300 W/m2K.The cost of the installed heat exchanger is $400/m2 and life of the equipment is 10 years. The heat capacity of water is 4184 J/kgK. If the condenser is to operate 7200 hrs/yr, determine the cooling water flow rate in kg/hr and exit temperature for optimum economic conditions.

Solution:

Solution:

Problem 17:An air cooler consisting of a bundle of 2” Sch 40 steel pipes enclosed in a well baffled shell is being built to cool 250 kg/hr of air from 82°C to 27°C. The air flows under pressure through the tubes of a 1-shell 2-tube pass heat exchanger. Cooling water at 21°C is under sufficient pressure to force it through at any desired rate, flows through the shell. The heat transfer coefficient based on the inside fluid is 50 W/m2K and for the outside fluid, 2100 W/m2K. Cooling water costs PhP 0.55/m3 and the annual fixed charges is 15% with cooler cost at PhP 18,000/m2. Calculate the Optimum exit temperature of cooling water and optimum length of tubes.

Solution:Ei 0:= Eo 0:= t1p 82:= t2p 27:= t1 21:= ∆t1 t2p t1−:=

Ui 50:= Di 2.067:= Do 2.375:= KfCao 2700:= assume t2 60:= Ft 1:=

Uo UiDi

Do⋅:= Hy 8760:= Cu 0.00055:= Cpu 1:= ∆t2 t1p t2−( ):=

∆t2m root 1t1p t2p−

∆t1 ∆t2−+

2

ln∆t2

∆t1

1−

∆t1

∆t2+

⋅

Ft Uo⋅ Hy⋅ Cu⋅KfCao Cpu⋅

− ∆t2,

:=

∆t2m 49.992=

∆t2 27.13975501:= t2 t1p ∆t2m−:= t2 32.008=

Rt1 t2−

t2p t1p−:= R 0.2= S

t2p t1p−t1 t1p−

:= S 0.902=

Solution:from chart

Ft = 0.53

Ft 0.53:=

∆t2m root 1t1p t2p−

∆t1 ∆t2−+

2

ln∆t2

∆t1

1−

∆t1

∆t2+

⋅

Ft Uo⋅ Hy⋅ Cu⋅KfCao Cpu⋅

− ∆t2,

:=

∆t2m 52.654=

∆t2 27.13975501:= t2 t1p ∆t2m−:= t2 29.346=

Rt1 t2−

t2p t1p−:= R 0.152= S

t2p t1p−t1 t1p−

:= S 0.902=

Solution:from chart

Ft = 0.75

Ft 0.75:=

∆t2m root 1t1p t2p−

∆t1 ∆t2−+

2

ln∆t2

∆t1

1−

∆t1

∆t2+

⋅

Ft Uo⋅ Hy⋅ Cu⋅KfCao Cpu⋅

− ∆t2,

:=

∆t2m 51.276=

∆t2 27.13975501:= t2 t1p ∆t2m−:= t2 30.724=

Rt1 t2−

t2p t1p−:= R 0.177= S

t2p t1p−t1 t1p−

:=

Solution:from chart

Ft = 0.70

Ft 0.70:= S 0.902=

∆t2m root 1t1p t2p−

∆t1 ∆t2−+

2

ln∆t2

∆t1

1−

∆t1

∆t2+

⋅

Ft Uo⋅ Hy⋅ Cu⋅KfCao Cpu⋅

− ∆t2,

:=

∆t2m 51.564= t2 t1p ∆t2m−:= t2 30.436=

∆t2 27.13975501:=R 0.177= S

t2p t1p−t1 t1p−

:=

Rt1 t2−

t2p t1p−:=

S 0.902=

Solution:from chart

Ft = 0.73

Ft 0.73:=

∆t2m root 1t1p t2p−

∆t1 ∆t2−+

2

ln∆t2

∆t1

1−

∆t1

∆t2+

⋅

Ft Uo⋅ Hy⋅ Cu⋅KfCao Cpu⋅

− ∆t2,

:=

∆t2m 51.39= t2 t1p ∆t2m−:= t2 30.61=

∆t2 27.13975501:=R 0.172= S

t2p t1p−t1 t1p−

:=

Rt1 t2−

t2p t1p−:=

S 0.902=Ft = 0.73 ok!

t2 30.61= o C optimum temperature

Solution:

q = UopDoL?Tlm = wCpu(t 2 - t1) w 250:=

Lw Cpu⋅ t2 t1−( )⋅

Uo π 0.060325⋅∆t2 ∆t1−

ln∆t2

∆t1

⋅

1000

3600⋅:=

L 5.778= m

Thank You!