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# Data for carbon dioxide: Molecular weight:44.0 Heat capacity of gas phase: 0.036+4.23 x10-5T   (in kJ/moleoC, T is...

Data for carbon dioxide:

Molecular weight:44.0

Heat capacity of gas phase: 0.036+4.23 x10-5T   (in kJ/moleoC, T is in oC)

Viscosity                                 Pr                    k (W/m K)

180oC2.13 x10-5kg/(m s)                  0.721               0.029

130oC  1.93x10-5kg/(m s)                  0.738               0.025

80oC  1.72x10-5kg/(m s)                  0.755               0.020

Approximate density: 2.0 kg/m3

Data for water:

Molecular weight:18.0

Heat capacity for liquid water: 0.0754 kJ/mole oC

Viscosity:

Viscosity                                 Pr                    k (W/m K)

180oC139 x10-6kg/(m s)                   0.94                 0.665

130oC  278x10-6kg/(m s)                   1.72                 0.682

80oC  472x10-6kg/(m s)                   3.00                 0.658

Ideal gas constant:

0.08206 L atm/(mol K)

Carbon dioxide, with flow rate of 0.10 kg/s, is to be cooled from 180 oC to 80 oC.  The utility stream is cold water, which enters the shell at 15 oC and leaves at 60 oC.    The carbon dioxide flows through the tubes and the water flows through the shell. The heat exchanger is a 1-1 shell and tube design (i.e. one shell pass, one tube pass).  The flow is counter-current.  The number of tubes is unknown.

Data (additional data on previous pages):

Heat transfer coefficient of the water in the shell: 500 W/m2oC

Re in the tubes is 15,000

Assume the resistance to heat transfer of the pipe wall is negligible.

tube dimensions:  0.716 cm ID,   1.271 cm OD

tube pitch/diameter ratio=1.5

Pressure everywhere is approximately 1.5 atm

Assume carbon dioxide is an ideal gas.

Note—you do not need to interpolate physical properties; use values at the closest temperature given.

a. expected rate of energy transfer from the hot to the cold stream, in W:______________

b. mass flow rate of the water, in kg/s:______________________

c. heat transfer coefficient in the tubes (hint, tube Re=15,000):___________units of:____________

d. mass flow rate of the CO2in each tube, in kg/s:_____________

e. overall heat transfer coefficient, Uo___________units of:____________

f. area needed for heat transfer, Ao, m2________________

g. Number of tubes needed, (round the number to nearest integer):_________

h. Length of each tube (using integer number of tubes, from part (g), m:___________

1. Expected rate of energy transfer from the hot to the cold stream, in W:______________

For this we will use, mCpdT for CO2 side as we know, Q= mCpdT (hot) = mCpdT (Cold)

Step1 :For CO2 , T1 = 180 C and T2 = 80, so dT = 100C

Step2 : Cp= 0.036+4.23 x10-5T

Cp1 = 0.036+ 4.23 x10-5(180) =0.043614 kJ/moleoC

Cp2 = 0.036+ 4.23 x10-5(80) =0.039384 kJ/moleoC

Thus Cp (avg) = 0.041499 kJ/moloC

Step3 : m = 0.10 kg/s * 3600 = 360 kg/hr =8.18 kmol/hr =8180mol/hr

Step 4 : Q = mCpdT = 8180* 0.0415* 100 = 33497 kJ/hrC = 9304 W /c

2. mass flow rate of the water, in kg/s:

As we know, Q= mCpdT (hot) = mCpdT (Cold)

Thus , 33497 kJ/hr C = m (water ) Cp( water ) dt(water )

m = 9.872 kmol/hr = 0.049 kg/s

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