shinglai
26-02-2009, 09:24 AM
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
---
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
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
---
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