Liquid Cooled Heat Exchanger Verification

[Raddy13]

My friend and I are trying to design a simple liquid-cooled heat exchanger and I'm getting some odd results so I wanted to get another set of eyes on my math here. We're running four 1.5m sections of nylon tube in parallel through the heat source, which was a thermal input of 265W. The cooling loop will be supplied by a tank of cold water and pump. As the system runs, the water will gradually warm up, and once the the surface temperature of the nylon tube reaches 305K, we will consider that the point at which the water needs to be replaced/chilled, so I was trying to determine what the cold tank temperature would be at that point.

My assumptions:
*All thermal input is absorbed by the exchanger tube
*Heat is transferred to the water and tube by conduction and convection
*Neglect radiation and contact resistance
*The thermal input is equally divided among the four tubes

Known:
\dot{V}_{pump}=0.114 L/s
\dot{V}_{tube}=\frac{0.114}{4 tubes}=0.0285 L/s
OD_{tube}=0.953 cm
ID_{tube}=0.699 cm
L_{tube}=1.5m
k_{tube}=0.25 W/m \cdot K
T_H=305K
\dot{Q}_{per tube}=265W/4 tubes=66.25W

Using the overall heat transfer equation:
\dot{Q}=UA\Delta T

Where:
\frac{1}{UA}=\frac{1}{hA_{conv}}+\frac{d}{kA_{cond}}

For my convection coefficient, I got the following (check attached document for more complete work):
Re=2900
Pr=12.56
f=0.046
Nu=26.14
h=10.37 W/m^{2}\cdot K
A_{conv}=\pi (0.699cm)(1.5m)=0.033m^2

For the conduction term:
k_{tube}=0.25 W/m\cdot K
d=r_o - r_i =\frac{0.953cm-0.699cm}{2}=0.127cm
A_{cond}=\pi (0.953cm)(1.5m)=0.045m^2

Plugging it in:
\frac{1}{UA}=\frac{1}{(10.37)(0.033)} + \frac{0.127cm}{(0.25)(0.045)}=3.035
UA=0.33W/K

Substituting into the heat transfer equation:
66.25W=(0.33W/K)\Delta T
\Delta T = 200.8K
200.8K=T_H - T_C=305K-T_C
T_C=104.2K

Obviously a ridiculously low temperature. I thought that was odd, so I checked how much heat could be handled by conduction alone:
\dot{Q}=k\frac{2\pi L\Delta T}{ln(r_o/r_i)}
66.25W=(0.25 W/m\cdot K)\frac{2\pi (1.5m)\Delta T}{ln(0.477/0.350)}=\frac{2.36\Delta T}{0.309}
\Delta T=\frac{(66.25)(0.309)}{2.36}=8.67K

So according to conduction alone, the system will be effective until the water is 296K, but with conduction AND convection, it's only good until the water is 104.2K, which would be ice. Any idea where I'm screwing up?

 

[256bits]

... I thought that was odd, so I checked how much heat could be handled by conduction alone:
\dot{Q}=k\frac{2\pi L\Delta T}{ln(r_o/r_i)}
66.25W=(0.25 W/m\cdot K)\frac{2\pi (1.5m)\Delta T}{ln(0.477/0.350)}=\frac{2.36\Delta T}{0.309}
\Delta T=\frac{(66.25)(0.309)}{2.36}=8.67K

So according to conduction alone, the system will be effective until the water is 296K, but with conduction AND convection, it's only good until the water is 104.2K, which would be ice. Any idea where I'm screwing up?

The gives you the temperature ( 296K ) of the inner surface of the tube, not the water temp.