Calculating the air venting flow rate needed during CIP of a tank

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Discussion Overview

The discussion revolves around calculating the air venting flow rate required during the cleaning-in-place (CIP) process of a tank, specifically addressing the impact of temperature changes when hot water is introduced to cool the tank. Participants explore various methods for calculating the venting flow rate using Python and the CoolProps module, focusing on the theoretical and practical implications of their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents three methods for calculating the venting flow rate, noting significant discrepancies in the results obtained from each method.
  • Another participant points out that the calculated "tank_volume" is not explicitly shown in METHOD 1 and suggests that METHOD 3 appears to use the volume calculated in METHOD 2, indicating potential issues in consistency.
  • Further contributions highlight the importance of correctly identifying parameters used in calculations, such as using C_v instead of C for specific heat in METHOD 1 and clarifying the use of enthalpy versus internal energy in METHOD 3.
  • Questions are raised about how the temperature differential is accounted for in METHOD 1, suggesting that this aspect may influence the results.
  • One participant emphasizes that the cold rinse water will heat the cold air, causing it to contract, which necessitates the venting system to accommodate the displaced volume.

Areas of Agreement / Disagreement

Participants express differing views on the methods used for calculations, with some questioning the validity of certain parameters and others pointing out inconsistencies in the calculations. No consensus is reached regarding the best approach or the reasons for the discrepancies in results.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the definitions of specific heat and enthalpy, as well as the implications of temperature changes on the air volume. The discussion does not resolve these limitations.

Remusco
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I have an application where I must calculate the venting flowrate of air required due to temperature change when hot tank is rinsed with cool water. The tank has spray nozzles inside so it can be cleaned in place. For the properties of the air in the tank I used the CoolProps module in Python.

I used three different methods to calculate the venting flowrate, as shown below:

METHOD 1:
Python: 
from CoolProp.HumidAirProp import HAPropsSI
import numpy as np
import matplotlib.pyplot as plt


tank_height=(251/12)/3.281[/ICODE]
tank_diam=(104/12)/3.281[/ICODE]
tank_volume=tank_height*np.pi*0.25*tank_diam**2[/ICODE]

#tank air state before cooling
initial_air_pressure=100594
initial_air_temp=335.928
initial_air_specific_vol=HAPropsSI('Vha','T',initial_air_temp,'P',initial_air_pressure,'R',1)

#tank air state after cooling
final_air_pressure=100594
final_air_temp=285.928
final_air_specific_vol=HAPropsSI('Vha','T',final_air_temp,'P',final_air_pressure,'R',1)

#calculating the volume displaced by the cooling
air_mass=tank_volume/initial_air_specific_vol
volume_displaced=air_mass*(final_air_specific_vol-initial_air_specific_vol)

#heat lost by the water = heat gained by the air
vdot_CIP=.0053
rho_CIP=998
C_p_water=4184
mdot_CIP=rho_CIP*vdot_CIP
deltaT=final_air_temp-initial_air_temp
Qdot=mdot_CIP*C_p_water*deltaT

#time to cool the air
C_v_air=HAPropsSI('C','T',initial_air_temp,'P',final_air_pressure,'R',1)
time=air_mass*C_v_air/(mdot_CIP*C_p_water)
print(time)

#venting flowrate
vdot_air=volume_displaced/time
print(vdot_air)

METHOD 2:
Python: 
from CoolProp.HumidAirProp import HAPropsSI
import numpy as np
import matplotlib.pyplot as plt

tank_height=(251/12)/3.281
tank_diam=(104/12)/3.281
tank_volume=tank_height*np.pi*0.25*tank_diam**2

#tank air state before cooling
initial_air_pressure=100594
initial_air_temp=335.928
initial_air_specific_vol=HAPropsSI('Vha','T',initial_air_temp,'P',initial_air_pressure,'R',1)

#tank air state after cooling
final_air_pressure=100594
final_air_temp=285.928
final_air_specific_vol=HAPropsSI('Vha','T',final_air_temp,'P',final_air_pressure,'R',1)

#calculating the volume displaced by the cooling
air_mass=tank_volume/initial_air_specific_vol
volume_displaced=air_mass*(final_air_specific_vol-initial_air_specific_vol)

#heat lost by the water = heat gained by the air
vdot_CIP=.0053
rho_CIP=998
C_p_water=4184
mdot_CIP=rho_CIP*vdot_CIP
deltaT=final_air_temp-initial_air_temp
Qdot=mdot_CIP*C_p_water*deltaT

#time to cool the air
initial_air_enthalpy=HAPropsSI('H','T',initial_air_temp,'P',initial_air_pressure,'R',1)
final_air_enthalpy=HAPropsSI('H','T',final_air_temp,'P',final_air_pressure,'R',1)
time=air_mass*(final_air_enthalpy-initial_air_enthalpy)/Qdot
print(time)

#venting flowrate
vdot_air=volume_displaced/time
print(vdot_air)

METHOD 3:
Python: 
from CoolProp.HumidAirProp import HAPropsSI
import numpy as np
import matplotlib.pyplot as plt

tank_height=(251/12)/3.281
tank_diam=(104/12)/3.281
tank_volume=tank_height*np.pi*0.25*tank_diam**2

#tank air state before cooling
initial_air_pressure=100594
initial_air_temp=335.928
initial_air_specific_vol=HAPropsSI('Vha','T',initial_air_temp,'P',initial_air_pressure,'R',1)

#tank air state after cooling
final_air_pressure=100594
final_air_temp=285.928
final_air_specific_vol=HAPropsSI('Vha','T',final_air_temp,'P',final_air_pressure,'R',1)

#calculating the volume displaced by the cooling
air_mass=tank_volume/initial_air_specific_vol
volume_displaced=air_mass*(final_air_specific_vol-initial_air_specific_vol)

#heat lost by the water = heat gained by the air
vdot_CIP=.0053
rho_CIP=998
C_p_water=4184
mdot_CIP=rho_CIP*vdot_CIP
deltaT=final_air_temp-initial_air_temp
Qdot=mdot_CIP*C_p_water*deltaT

#time to cool the air
initial_air_enthalpy=HAPropsSI('U','T',initial_air_temp,'P',initial_air_pressure,'R',1)
final_air_enthalpy=HAPropsSI('U','T',final_air_temp,'P',final_air_pressure,'R',1)
time=air_mass*(final_air_enthalpy-initial_air_enthalpy)/Qdot
print(time)

#venting flowrate
vdot_air=volume_displaced/time
print(vdot_air)

My result for the vdot_air on each method is as follows:

METHOD 1:
-3.6722 kg/sec

METHOD 2:
-0.4986 kg/sec

METHOD 3:
-0.5431 kg/sec

I'm not sure why I am getting such different values for each. I know the difference lies in how I am calculating my time value; however, I would think all the methods would provide similar results.
 
Last edited by a moderator:
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The calculated "tank_volume" is not shown in METHOD 1:.

METHOD 3: seems to use "tank_volume" as calculated in METHOD 2:

These may or may not be part of the problem.

Have Fun!
Tom
 
Tom.G said:
The calculated "tank_volume" is not shown in METHOD 1:.

METHOD 3: seems to use "tank_volume" as calculated in METHOD 2:

These may or may not be part of the problem.

Have Fun!
Tom
The tank volume is the same for all three.
 
Method 1:
Remusco said:
C_v_air=HAPropsSI('C','T',initial_air_temp,'P',final_air_pressure,'R',1)
This will give C_p_air. Use CV instead of C for C_v_air. (Reference)

Method 2:
Remusco said:
initial_air_enthalpy=HAPropsSI('H','T',initial_air_temp,'P',initial_air_pressure,'R',1)
final_air_enthalpy=HAPropsSI('H','T',final_air_temp,'P',final_air_pressure,'R',1)
You are using the correct parameter H to get the enthalpy here.

Method 3:
Remusco said:
initial_air_enthalpy=HAPropsSI('U','T',initial_air_temp,'P',initial_air_pressure,'R',1)
final_air_enthalpy=HAPropsSI('U','T',final_air_temp,'P',final_air_pressure,'R',1)
U is not listed in the parameter's list but I can only assume it means internal energy and not enthalpy.

You seem to want to evaluate heat lost by the water = heat gained by the air. What motivates you to use either of these methods for this purpose?
 
How was the temperature differential considered in method #1?

Specific-Heat.jpg
 
jack action said:
Method 1:

This will give C_p_air. Use CV instead of C for C_v_air. (Reference)

Method 2:

You are using the correct parameter H to get the enthalpy here.

Method 3:

U is not listed in the parameter's list but I can only assume it means internal energy and not enthalpy.

You seem to want to evaluate heat lost by the water = heat gained by the air. What motivates you to use either of these methods for this purpose?
The cold rinse water will heat the cold air and cause it to contract. The vent needs to be able to suck in that displaced volume.
 

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