Heat absorption by apples

  • Thread starter hilly1989
  • Start date
  • #1
6
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I have a problem here im trying to solve for work. Does not need to be 100% accurate.
Apple slices are pushed through a bath of hot water to heat the outside of them for 30 seconds.

- Apple slice has an area of 3500mm² = 0.035m²
- Apple temperature is 3 degree cel
- Hot water temp = 53 deg cel. (kept constance from heat exchanger)
- Apple slices are heated for 30 seconds
- 82,300 slices per hour are processed.
- Thermal Conductivity of Apple = 0.5 W/(m K)

Im trying to find the Heat (energy transfered to the apples) - in other words, heat required by the heat exchanger to keep the bath at 53 degrees.

Here is my workings.

82,300 slices / (60*60) = 22.8 slices per second.
30 seconds bath time * 22.8 slices = 686 slices in bath at any one time.

686 * 0.035 = 24m2 of surface area in the bath.

Q/t = K*A*(T1-T2)

Q/t= 0.5*24*50 = 600 J/s = 600W.

After looking on google, the formula I have above sometimes get divided with a distance. mainly when looking at heat losses through windows. distance = thickness. This makes sense as the units leaves me a with meter on the top. Q/t = W/(m K) * m2 * K,

So should I now divide the 600W by the average depth of a slice/2 (say 8mm)

giving me 75kW of heat being absorbed by the slices. Thats a lot of heat! (the Heat exchanger is capable of having upto 96kW provided to it, i think)

Thanks,
Anthony
 

Answers and Replies

  • #2
688
1
In your formula: Q/t = K*A*(T1-T2)

Note the T2 increases and Q varies with time.

SO, you could assume in 30 seconds the apples reach 50 C. if you want to perform a transient heat transfer analyis on the applies let me know!

BUT, we can get an upper limit of the energy required as follows:

The energy to heat one apple slice:

Q = mass_apple_slice X specific heat_apple X temperature rise of apple.

THEN, with the rate you pass apples, you can figure the total input power to maintain 50C !

Let me know if you need any help.
 
  • #3
1,506
18
I would follow edgepflow's idea. If you assume apples are more or less the same as water (SHC 4200J/kgC) and you know the mass of a slice you should get a reasonable upper limit for the power required.
 
  • #4
6
0
In your formula: Q/t = K*A*(T1-T2)
The energy to heat one apple slice:
Q = mass_apple_slice X specific heat_apple X temperature rise of apple.
750 kg of apples if processed per hour
750 (kg/hr) / 82,300 (slc/hr) = 9.11 grams per slice

Q = 0.00911 * 4,200 (J/kgC) * 50 (C) = 1913 J

22.8 slices per second;
1913 *22.8 = 43.62 KJ/s = 43.62 KW

This 43.62 kW would only be reached if the apples managed to reached 53 degree, correct? But is a maximum possible consumption.
 
  • #5
1,506
18
I agree with your calculation.....43kW to heat this quantity of apples to 53C in 30secs.
I don't know but I imagine that in practice you would make sure that your heater could supply about 60kW to give some lee way.
 
  • #6
6
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Is there a way for me to determine a more accurate final temperature of the apples?
I don't think the whole slice heats up to 53 degrees.
 
  • #7
688
1
750 kg of apples if processed per hour
750 (kg/hr) / 82,300 (slc/hr) = 9.11 grams per slice

Q = 0.00911 * 4,200 (J/kgC) * 50 (C) = 1913 J

22.8 slices per second;
1913 *22.8 = 43.62 KJ/s = 43.62 KW

This 43.62 kW would only be reached if the apples managed to reached 53 degree, correct? But is a maximum possible consumption.
I came up with 38 kW - so I think your numbers are good. That is a lot of power, but that is a lot of apples.
 
  • #8
688
1
Is there a way for me to determine a more accurate final temperature of the apples?
I don't think the whole slice heats up to 53 degrees.
Yes, you could start with a simple lumped capacity model if the Biot number < 0.1. Otherwise, we have to move to some 1D models.
 

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