1 reply to this topic

[sgkim]

Hi,

Present "Approach"(temperature difference between the cold water and dew point) is 19-17 or 2 ℃ and "Range"(temperature differenct between hot water and cold water) is 25.1-19 or 6.1 ℃ in summmer.

Now you are going to increase the COLD cooling water temperature(T1) from 19 to 23 ℃. Then the HOT cooling water temparature(T2) shall be 23+6.1= 29.1℃ assuming the cooling duty in the "tempered water system" remain unchanged. The "Approach" will become 23-17= 6℃.

The governing equation for the cooling tower calculation was given by Merkel expressed as:
K*a*V/L=Integral(T1 to T2)dT/(hs-hw) ------(1)
where, K*a*V/L= characteristic factor of cooling tower
K=Mass-transfer coefficient, a=contact area per tower volume, V=active cooling volume per plan area, L=water rate per plan area, T1=COLD water temperature, T2=HOT water temperature, hs=enthalpy of saturated air at the same local temperaure T of water. hw=enthalpy of hot air/moisture mixture in equlibrium with water at local temperature T.

Refer to Perry's CEH 6th ed. pp12-13~14 for details and the sketch attacked.

The term (hs-hw) makes a kind of "driving force" in the cooling tower. As the "Approach" increases, the driving force (hs-hw) increases, but the value K*a*L/G will not be changed. The idea that the factor will not change could be employed for the estimation of variables such as Hot Air Temperature and L/G ratio.

You must first check the present hot air temperature which is not given in your question. The point "D" shall be determined from the present "Hot Air Temperature". Calculate the area of the shape "A-B-C-D" in the attached drawing, then move C to C", A to A", but the "Range" is the still the same 6.1℃. Find A"-B"-C"-D" to get the same area as A-B-C-D.
L/G Ratio can be greatly increased by decreasing G(Air Rate). L/G ratio is the slope of the Operating Line.

It seems to be a quite complicate job.