3 replies to this topic

[eugenelys]

I have been tasked recently to investigate the amount of heat energy that can be recovered from humid air in a dryer. As this is something relatively new for me, I would like to request your kind advice towards pointing me in the right direction.

 

The dryer current utilises steam from the steam boiler for the drying processes and the measured outlet has air leaving at estimated 10,000 Nm3/hr @ 90C, RH of 100%.

 

Before using designing a HX for heat recovery, how would I go about estimating the amount of energy available? Would it be correct to use psychrometric charts to obtain the enthalpy of humid air and estimate the amount of energy using Q=mcdT for a quick estimation?

 

I have attached the drawing for references. Would also want to learn more regarding this so I would be greatful if you can direct me towards what resources that I can read up on.

Attached Files

•  Heat Recovery Dryer.pdf   805.09KB   202 downloads

[breizh]

Hi ,

Let you find an Air humid  enthalpy diagram . This should be your starting point.

 

http://www.newbooks-...Excerpt_001.pdf

 

Breizh

Edited by breizh, 21 October 2016 - 09:45 PM.

[sgkim]

To calculate the heat by Q=mcdT is only to calculate the sensible heat of the exhausting wet air.  But the latent heat of moisture would be much larger than the sensible heat if fully (100%RH) saturated air drops in temperature down to a certain temperature and some part of moisture condenses.       

 

For every 1kg of hot dry air,  H1 kg-moisture will be fed at inlet and H2 kg-moisture discharged at outlet, Then the differential moisture (H1-H2)kg of water per kg-of-dry-air will condense at the outlet, so the latent heat of the condensate can be utilized 'additionally' to the sensible heat of condensing vapor.          

 

Therefore the total heat to be extracted('utilized") by the cooling medium can be calculated from the overall energy balance as:    

Q = (Dry Air Mass Rate, m)*{(Specific Enthalpy Difference of Wet Air, h1-h2) - (Absolute Humidity Difference, H1-H2)*(Specific Enthalpy of Liquid Water at t2, hw2)} ......(1)

 

where, H1 and H2 are absolute humidities of air at inlet and outlet respectively,  [kg-moisture/kg-dry air]  

            h1 and h2  are specific enthalpies of wet air at inlet and outlet, [kJ/kg-dry air]

            hw2 is specific enthalpy of pure water at outlet temperature, t2    

 

Absolute humidities and specific enthalpies can be found from psychrometric charts.  But be sure to calculate the mass rate of dry air correctly from the figure, '10,000 Nm3/h', of 90℃ saturated air.  It is not clear whether the volume rate by Nm3/h is based on dry air or wet.   And also check if the figure "RH 100%" of wet air stream from the dryer is correct or not: most often the case dryer outlet air is not fully 'saturated'.

 

Stefano    

Edited by sgkim, 21 October 2016 - 04:31 AM.

[eugenelys]

Thank you for your replies towards the problem I stated. Noted regarding the relative humidity of outlet dryer.

 

@breizh Greatly appreciate the link that you posted. Always happy to learn