Question

Data for carbon dioxide:

Molecular weight:44.0

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

Viscosity            Pr         k (W/m K)

180 oC    2.13 x10-5 kg/(m s)            0.721        0.029

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

 80oC    1.72x10-5 kg/(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)

180 oC    139 x10-6 kg/(m s)            0.94        0.665

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

 80oC    472x10-6 kg/(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.

Heat transfer coefficient of the water in the shell: 500 W/m2 oC

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.

Calculate the following (show work here):

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

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

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

mass flow rate of the CO2 in each tube, in kg/s:_____________

overall heat transfer coefficient, Uo___________units of:____________

area needed for heat transfer, Ao, m2________________

Number of tubes needed, (round the number to nearest integer):__________

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

Answer 1

IF YOU HAVE ANY DOUBTS PLEASE COMMENT BELOW

RATE THUMBSUP PLEASE

ANS:

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/mole^oC

              Cp2 = 0.036+ 4.23 x10^-5(80) =0.039384 kJ/mole^oC

Thus Cp (avg) = 0.041499 kJ/mol^oC

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

RATE THUMBSUP