Convective Heat Transfer

Question

100 kg of water in a closed insulated autoclave is heated from 20 to 90 °C by a heating coil made of 12 m of 1-in O.D., 15-BWG-gauge copper tubing (ID=2.174 cm). When steam at 100 °C issued in the coil, it is found that it takes 40 min to heat the water. The steam-side coefficient hf is known to be 12,000 W/m2•K. (Assume you can use some average t for water in your calculations.)
a) If a propeller agitator stirs the water, the water-side coefficient ho doubles. How long would the heating now require?
b) If high-pressure steam at 150 0C is used (no agitation), how long would the heating take?

Answer 1

The total energy transferred to the water in 40 min is

H = Q/t = (100 kg)(4186 J/kg)(90-20)/(40*60s) = 12,209 W

This energy comes from the steam inside the coil.

We need the inside and outside area of the coil:

A =2*Pi*r*L

A_in=2*Pi*(0.02174/2 m)(12 m) = 0.8196 m²

A_out=2*Pi*(0.0254/2 m)(12 m) = 0.9576 m²

For the thermal conductivity of copper at about 100°C I found k=400 W/m/K.

 

We have convection inside and outside the coil and conduction within the coil.

On the steam side,

H = h_s A_in (100-T_in),

12209 = 12000 (0.8196) (100-T_in)

which gives us 100-T_in=1.24, T_in=98.76 °C

 

For the conduction within the copper coil, modeled as a pipe, we have the formula

 

H = 2*Pi*L*k*(T_in - T_out) / ln(r_out/r_in)

12209 = 2*Pi*12*400*(98.76-T_out)/ ln(2.54/2.174)

which gives us 98.76-T_out = 2.60, T_out= 96.16 °C

 

On the water side,

H = h_w A_out (T_out-55)

( I took (20+90)/2=55°C as the average water temperature.)

12209 = h_w 0.9576 (96.16 -55)

which gives us h_w=309.7 W/m²/K

 

(a) Now suppose this h_w doubled to 2(309.7)=619.5 W/m²/K. With all other conditions staying the same, H would also double, so the heating time would be cut in half to 20 min.

 

(b) Re-do the entire calculation with 150 °C instead of 100°C - I will leave that to you.