Drinking Fountain / Water Cooler

[shinglai]

Good day! RE people, i believe everyone or if not most of the people here knows what a Drinking Fountain is. Im making one and I have made a few calculations already and i don't know if i'm on the right path i just wanna hear some side comments and/or a few suggestions in obtaining the following:

1. Compressor specification and capacity (I have two suppliers ready: Danfoss and Tecumseh)

2. Air-Cooled Condenser specification (Im planning of using 5/16 copper tube, how much FPI, Fin Length, Number of tube face, No. of rows etc. to use?). Fan Selection

3. Evaporator Length (im using 5/16 copper tube too)

4. Capillary Tube Length (im using 0.036 in)



Here it is:

This is the part were water is stored to be cooled regulated by a solenoid valve:


Design Conditions


Ambient Temp: 32 ºC
Inlet Water Temp (T1): 29 ºC
Output Water Temp (T2): 5 ºC


Cooling Tank Selection

Outside diameter (do): 4-in
Tank length (L): 12-in
Critical pressure (Pcr): 150 psi
Thickness (t): ?
Capacity of stainless tube (V): ?
Inside diameter (di): ?


Pcr = (50,200,000 (t/do)^3) / 3

Pcr = (50,200,000 (t/4)^3) / 3

t = 0.0831 in or 2.111 mm


V = π(di)2 L / 4

V = π(3.8338)2 (12) / 4

V = 138.53 cu. in or 2.27 Li



Cooling System Design

Materials


Stainless tube: 0.0831 in x 4-in dia. X 12 in
Copper tube: 5/16 in dia. x L?
Styropore insulation: 1.0 in x 6-in dia. X 14 in


Data

Specific heat of water (Cp): 4.187 KJ/Kg-ºC
Specific heat of SS (Cs): 0.450 KJ/Kg-ºC
Water density: 1000 kg/cu.mtr
SS density: 8.03 g/cu.cm = 0.1316 kg/cu.in
Water Inlet temp


Volume conversion

138.53 cu.in (1 cu.ft)(1cu.mtr)
(1728 cu.in)(35.29 cu. cu.ft)

= 2.27 x 10 –3 cu mtr




Mass of water (Mw)

Mw = 2.27 x 10 –3 cu.mtr (1000 kg/cu.mtr)

Mw = 2.27 kg


Load from water

Qw = Mw Cp (T1 – T2)

Qw = (2.27)(4.187)(29 – 5)



Qw = 228.11 KJ





========================================

Volume of stainless steel

Vs1 = 0.26 cu.in


Vs2 = 2 {[π (4)2 /4] (0.0831)}

Vs2 = 2.1 cu.in


Vs = Vs1 + Vs2

Vs = 0.26 + 2.1

Vs = 2.36 cu.in



Mass of stainless Steel (Mss)



Ms = Density of SS x Volume of SS



Ms = (0.1316)(2.36)

Ms = 0.311 Kg




Load from stainless tube



Qs = Ms Cs (T1 – T2)



Qs = (0.31)(0.450)(29 - 5)



Qs = 3.348 KJ

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Calculated load from temperature difference between ambient and water condition inside the tank:

Q1 = 4.7116 Watts

Q1 = 4.7166 J/s

Q1 = 0.0047166 KJ/s


@ 60 secs


Q1 = 0.3 KJ



Total Load


Q = Qwater + Qstainless + Q1

Q = 228.11 + 3.348 + 0.3

Q = 231.8 KJ = 219.72 BTU : Is this correct or i am missing something here? i have a feeling that my tabulated total load is somewhat small

[icemakerman]

what i think you have missed is the flow rate of the water passing through the drinking fountain. The calculations are fine if the water in the heat exchanger is static so the next thing to calculate (guestimate) is how many litres /min flow rate ...how many water changes inside heat exchanger /hr and also how quickly is the water to be cooled.hope this helps a bit

[shinglai]

hi there! thanks for you comment, i was wondering on that part too considering that my reference(s) books are stating the flow rate. and yes, you are right, the water is not static, i think it is even safer to assume that from time to time cold water is drawn from the drinking fountain. anyway, i don't really know what to put in here because of the varying flow rate of water here in our country (any suggestions?) or perhaps it would be best to assume a certain flow rate for this part?

I think the cooling rate can be calculated from the specifications given by the compressor manufacturer, 30 - 60 mins cooling time perhaps?

[Brian_UK]

If you want a fixed water flow rate then install flow regulation devices to maintain your desired flow rate.