Temperature loss along a duct

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  • #1
jordanb
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TL;DR Summary
A solution to a problem regarding temperature entering a length of ducting at which the exit temperature needs to be calculated
Hello,

I am looking for assistance regarding temperature losses along a length of pipe with hot air entering one. I need to know the temperature leaving the other end of the pipe.

I have been able to calculate the W/m-K (in theory but stopped a i knew the answer wasnt what i needed) losses but was hoping to be able to calculate the temperature loss.

Info:

Pipe OD = 105mm
Pipe ID = 100
Flow rate of air = 800L/min
Air temp entering pipe = 200°C
Ambient temp = 30°C
Flow type - Laminar
Material = Not yet decided but any ABS plastic can be assumed (open on values regarding material for now)

i have been reading 2009 ASHARE handbook but with no achieved results. Any help would be greatly received.

Jordan
 

Answers and Replies

  • #2
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TL;DR Summary: A solution to a problem regarding temperature entering a length of ducting at which the exit temperature needs to be calculated

Hello,

I am looking for assistance regarding temperature losses along a length of pipe with hot air entering one. I need to know the temperature leaving the other end of the pipe.

I have been able to calculate the W/m-K (in theory but stopped a i knew the answer wasnt what i needed) losses but was hoping to be able to calculate the temperature loss.

Info:

Pipe OD = 105mm
Pipe ID = 100
Flow rate of air = 800L/min
Air temp entering pipe = 200°C
Ambient temp = 30°C
Flow type - Laminar
Material = Not yet decided but any ABS plastic can be assumed (open on values regarding material for now)

i have been reading 2009 ASHARE handbook but with no achieved results. Any help would be greatly received.

Jordan
What is the air pressure, and what is the length of the pipe? What is the nature of the boundary conditions on the outside of the pipe (e.g., insulated, vertical natural convection, horizontal natural convection)?

To get some background on this, see Chapter 14 in Transport Phenomena by Bird, Stewart, and Lightfoot.
 
  • #3
jordanb
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Thank you for your reply. Air pressure can be assumed as 1bar. Length of the pipe being 5 meters (sorry i missed that off!) Horizontal natural convection.

I have researched both Thermodynmaic an engineering approach 4th edition and Transport Phenomena but i was hoping for some guidance as i don't fully understand.
 
  • #4
22,219
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OK. I'll help by asking some leading questions.

1. From the ideal gas law, what is the density of the entering air in g/l?
2. What is the viscosity of the entering air?
3. What is the Reynolds number of the entering air?
 
  • #5
jordanb
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1. 2.03 g/l
2. 1.90 x 10^-4 gm/cm-sec
3. 40000
 
  • #7
jordanb
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1)

p/RT

P = 101325 Pa

R = 287.058 J

T = 304.15 K

1.16 g/l


3) 46300

I have tried to copy my workings from a word doc with no prevail
 
  • #8
berkeman
Mentor
63,574
14,698
I have tried to copy my workings from a word doc with no prevail
Funny malapropism... :smile:

Please have a look at the "LaTeX Guide" link below the Edit window to see how to post math here at PF. Note that you will need to refresh your browser one extra time when first posting LaTeX in a thread.
 
  • #10
jordanb
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Understood. Im trying to understand so please bare-with my incompetence.

So with the current values what would i need to calculate next?
 
  • #11
22,219
5,120
Understood. Im trying to understand so please bare-with my incompetence.

So with the current values what would i need to calculate next?
You still haven't calculated the density correctly or the Reynolds number for the flow. We need to know the Reynolds number in order to determine the heat transfer coefficient between the gas and the inside wall of the pipe. The Reynolds number determines the heat transfer coefficient and the relationship for turbulent flow is different from their relationship for laminar flow. The Reynolds number also helps us determine whether the flow is turbulent or laminar. So please try again to calculate the Reynolds number correctly.
 
  • #12
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The game plan in this calculation is to first determine 3 things:

1. The forced convection heat transfer coefficient between the air and the inside surface of the tube
2. The equivalent heat transfer coefficient across the wall of the tube
3. The natural convection heat transfer coefficient between the outside wall of the tube and the outside air at 30 C.

Item 1 can be determined from the heat transfer correlation presented in Fig. 14.3-2 of Transport Phenomena. This expresses the Nusselt number (dimensionless heat transfer coefficient) as a function of the Reynolds number for the flow and the Prantdl number of the air. The correlation is supposed to use the air physical properties averaged over the bulk flow and the wall temperature (arithmetic averages at inlet and outlet). However, since these are initially unknown, they would have to be determined iteratively. So, as a first approximation, we would be using the properties evaluated at the inlet air temperature (200 C). From the ideal gas law, the density of the inlet air is $$\rho=\frac{PM}{RT}=\frac{(101325)(0.029)}{(8.314)(473)}$$$$=0.747\ \frac{kg}{m^3}$$and the mss flow rate is $$\dot{m}=\rho Q=(0.747)(0.8)=0.598\ kg/min=0.00996\ kg/sec$$The viscosity of air qt 200 C is $$\mu=2.58\times 10^{-5}\ \frac{kg}{m.sec}$$So, the Reynolds number at the inlet temperature is $$Re=\frac{\rho v D}{\mu}=\frac{4\dot(m)}{\pi D\mu}=\frac{(4)(0.00996)}{(3.14159)(2.58\times 10^{-5})(0.1)}=4920$$This is beyond the transition between laminar and turbulent flow, and the flow is definitely turbulent.

From the heat transfer = correlation in Fig. 14.3-2 of Transport Phenomena, at this Reynolds number, we have that the Nusselt number is given by $$Nu=\frac{hD}{k}=0.004(Re)(Pr)^{1/3}=(0.004)(4920)(0.7)^{1/3}=17.5$$The thermal conductivity of air at 200 C is ##0.0382\ \frac{W}{m.C}## so the initial estimate of the inside heat transfer coefficient is given by $$h_{inside}=\frac{k}{D}Nu=\frac{0.0382}{0.1}17.5=6.69\ \frac{W}{m^2.C}$$

CONTINUED IN NEXT REPLY
 
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  • #13
22,219
5,120
CONTINUED FROM PREVIOUS REPLY

For conduction through the ABS wall, the equivalent heat transfer coefficient is equal to the thermal conductivity of the plastic divided by the thickness of the tube wall (neglecting curvature). The thermal conductivity of ABS is about ##0.17\ W/m.C##. So, $$h_{wall}=\frac{0.17}{0.0025}=68.0\ W/m^2.C$$This is 10X the heat transfer coefficient for the air inside the tube, and is nearly negligible.

For natural convection heat transfer outside the tube, Bird et al (Transport Phenomena) give the following equation for the outside Nusselt Number ffrom a turrrbe with air on the outside : $$Nu_{outside}=0.365 (Gr)^{0.25}$$where Gr is the Grashoff number: $$Gr=\frac{g\beta |T_W-T_0|D^3}{\nu^2}$$where ##\beta## is the coefficient of volumetric thermal expansion (equal to 1/T for an ideal gas) nd ##\nu## is the cinematic viscosity (presumably evaluated at the wall temperature). The Nusselt number is going to be a function of the wall temperature itself, so the natural convection calculation is also going to be iterative. The highest the Nusselt number for the outside natural convection could be is if the outside wall temperature were equal to the inlet gas temperature, such that : $$Gr=\frac{(9.8)(1/473)(170)(0.105)^3}{(2.58\times 10^{-5}/0.747)^2}=3.42\times 10^6$$ $$NU_{outside, max}=0.365(3.42\times 10^6)^{0.25}=15.7$$This gives a maximum possible value of the outside heat transfer coefficient of $$h_{outside,max}=15.7\frac{k}{D}=\frac{(15.7)(0.0382)}{0.105}=5.7\ \frac{W}{m^2.C}$$This is consistent with the typical range of 3 - 20 ##\frac{W}{m^2.C}## for natural convection from tubes in Table 14.1-1 of Transport Phenomena.

The overall heat transfer coefficient U (neglecting wall curvature) is determined from the inside-, wall-, and outside heat transfer coefficients from the equation: $$\frac{1}{U}=\frac{1}{h_{inside}}+\frac{1}{h_{wall}}+\frac{1}{h_{outside}}$$Since, as noted previously, the values of ##h_{inside}## and ##h_{outside}## are upper bound estimates to the more accurate values obtained by further iteration, the value of U obtained from this equation will also be an upper bound: $$\frac{1}{U_{max}}=\frac{1}{6.69}+\frac{1}{68}+\frac{1}{5.7}$$This yields $$U_{max}=2.94$$

Solution of the differential heat balance equation for the pipe gives the following equation for the exit temperature of the air from the pipe: $$T_{exit}=30+(200-30)\exp{\left(-\frac{UA}{\dot{m}C_p}\right)}$$where A os the total heat transfer area in the pipe ##\pi DL##. Next, calculating the upper bound to the estimate of ##\frac{UA}{\dot{m}C_p}## in the above equation yields $$\frac{U_{max}A}{\dot{m}C_p}=\frac{(2.94)(1.61)}{(0.00996)(3.5)(8.314)\left(\frac{1000}{29}\right)}=0.474$$If we now substitute this into the equation for the exit temperature gives us a lower bound to what the exit temperature can be: $$T_{exit,min}=30+(200-30)\exp{(-0.474)}=136\ C$$
The exit temperature cannot be lower than this value. Any refinements (iterations) we make to the calculated exit temperature will result in a value higher than this.
 
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  • #14
jordanb
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Thank you for this. I have digested the information as best i can. I have tried to apply this to another example however, i cant see where the length of the ducting is factored into the formulas?
 
  • #15
jordanb
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No problem, ive found it!
 
  • #16
jordanb
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I have completed a different problem using the above as a template. Im not too happy with the results. Can anyone see what ive done wrong?
 

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  • #17
jordanb
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I have created a flow simulation to second back up calculations. The simulation is giving a result of 150°C ± 5°C
 
  • #18
22,219
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I have completed a different problem using the above as a template. Im not too happy with the results. Can anyone see what ive done wrong?
Where did you get 0.394 for U from?
 
  • #19
jordanb
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Thank you for spotting that, that is an error i have since noticed. I carried the wrong value down. However, this does not make the result any better.
 
  • #20
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Thank you for spotting that, that is an error i have since noticed. I carried the wrong value down. However, this does not make the result any better.
What does "male the result any better" mean?
 
  • #21
jordanb
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Im expecting to have a result of around 150°C. Im currently getting a result of 192°C which seems unrealistic that over 5meters of stainless steel, only 8°C of heat is lost?
 
  • #22
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Im expecting to have a result of around 150°C. Im currently getting a result of 192°C which seems unrealistic that over 5meters of stainless steel, only 8°C of heat is lost?
You used a value of U=4 ?
 
  • #23
jordanb
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Umax = 0.25
 
  • #25
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If I were you, I would be using the information in BSL to refine these rough estimates as much as possible.
 

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