Piping Heat Loss in Closed Loop System 1

[092961]

I've got a system with a 7000 gal 316SS tank in a closed loop with 120 feet of 12 inch diameter 316SS piping. The system is located inside with ambient air temperature of 60 F. The water is heated in a separate heating tank to 110 F. The process cycle involves filling the 7000 gal tank with 110 F water, then circulating in the piping loop at a velocity of 8 ft/sec. The process runs for 1 hour with no additional heating, but the process must remain above 95 F for the entire hour. The circulation pump has a 30 Hp motor operating at about 90% efficiency. I have calculated the heat input from the pump to be 72531 BTU/hr. I've looked at the 3E Plus software (mentioned on this site) and found that with Bare stainless steel pipe, 12 inches in diameter, no wind, etc. the loss is 280 BTU/hr/ft. This tells me that I have a net heat INPUT to the system of 38931 BTU/hr. Unless I missed it, I didn't see anything in the manual about the 3E software calculating combined heat loss of convection and radiation, however the inputs imply that it does.

(1) Can someone who uses the software regularly confirm that it calculates the heat loss from both convection and radiation.
(2) If I ignore the addition of energy from the pump, I calculate a temperature drop over the course of 1 hour of less than 1 degree F. I am using the equation q=m(dot) x cp x dT and the heat loss from the program.
(3) If I ignore the addition of energy from the pump, my calculations indicate that my temperature drop over the course of an hour is less than 1 F. Can someone confirm that for me and let me know if this is because of the volume of fluid in the system and the heat capacity of that volume of fluid? If I am off, perhaps someone can point out the equation I SHOULD be using.
(4) If I wanted to manually calculate the BTU/hr loss, it looks like a convoluted process of Nusselt and Prandlt numbers based on the fluid and the Reynolds number. Any helpful hints on this approach?

I should know this but it's been a while since heat transfer and I would like reassurance that my approach is acceptable and I don't like using a canned program without being able to confirm it…at least get myself in the ballpark.

Thanks to whomever responds.

 

[sailoday28]

Heat Transmission-McAdams 1954 gives the following equation for horizontal pipes with Grashoff numbers between 10^3 and ~!0^9 with ordinary air.
h(natural convection)=0.27(delta T /od)^0.25
od units of ft
h(convection)=0.72
Text further states the above Grashoff numbers are usual case. Using radiation to a room with emissivity=1
q/A=.171(5.7^4-5.2^4)= 55.5 B/H/sq ft
h(radiation)=55.5/(110-60)= 1.11
h(combined) outside=1.83
Heat flux q/a = 1.83(50) 91.5 B/H/sq/ft
For nominal 1 ft od q/ft approx 287.5
This seems to be in agreement with your numbers.

 

[sailoday28]

In my previous response, I did not take into account the energy that will be lost from the fluid to heat the stainless steel piping. (Mass Cp delta t of stainless) Say from 60F to at most 110F.

 

[CRG]

092969, you said, "The circulation pump has a 30 Hp motor operating at about 90% efficiency."

This seems unusually large for this application. Is this correct? Quick rule of thumb, gpm*(head in feet)/(3960*eff)=bhp Do you need a pump with 283 feet of head for the closed loop?

 

[25362]

CRG: 75,281 Btu/h is the motor's 30 metric HP converted totally into heat. I assume 092961 meant that the 72,531 Btu/h would represent 96.3% of that, taken as the motor efficiency as if the total BHP pump only added heat to the water.

092961: Can you tell us, please, why the heat lost by the tank wasn't taken into account, is it out of the room or efficiently insulated ? Kindly comment.

 

[CRG]

oops, 25362 & 092961, I goofed when punching the numbers into my calculator. I did not convert ft^3/min into gal/min; hence my error asking if a pump with 283 feet of head was required. Rule of thumb using the correct units results in a head of 38 feet which could be correct for this application.

12" pipe flowing water at 8 ft/sec will have a pressure drop of 1.5 feet per 100 feet of pipe. Question stated that loop consisted of 120 feet of pipe. I don’t think entrance and exit losses add up to 36 feet (I did not calculate them). There must be something else not stated to require 38 feet of head.

092961, let me rephrase my question, did you calculate head loss for the 120 foot closed loop circuit? Was the head loss used in conjunction with the pump curve to come up with 30 hp?

 

[092961]

Thanks for all the feedback...here are a few responses:

Sailoday28:
This is important to me...yes. Was that piping material mass*material specific heat (cp)*temperature rise(dT)? If so, I believe the light might be coming on!

CRG:
The 90% is not the hydraulic efficiency, it's the motor efficiency (slip, mechanical, etc.). The hydraulic efficiency of this pump is ~55%. I sort of mis-typed. What I was really getting at was the heat input from the inefficiency of the motor would add 72531 BTU/hr (although there is some heat generate by disc friction in the pump itself...which I've neglected). I'm not sure if this is correct though...should it be the heat added from the difference between the motor BHP and the Hydrualic HP (WHP)?

25362:
The tank is FRP material which has a much lower thermal conductivity than the piping. On my second iteration, I did add the tank and found the heat loss to be about 19700 BTU/hr, about 1/2 the piping.

CRG Second post:
The pump is sized correctly based on the system losses. There is a pressure drop "device" in the loop...sorry, that's all I can say.

THANK YOU all again for your kind assistance!

 

[sailoday28]

The problem can be approximated as quasi steady state
In addition to energy convected by radiation and convection from the system. WHICH YOU HAVE A HANDLE ON
There is an enery loss of the fluid also required to heat the mass of metal piping and tank.
The energy loss of the fluid is m*C*delta t Here the mass is the mass of the fluid, specific heat of the fluid and temperature drop of the fluid.
The energy loss of the fluid that heats the metal up is
m*C*temperature rise of the metal Here the m and C are the mass and specific heata of the metal.
The temp rise of the metal is approximately (110-60),