Change in temperature of water as it flows through a pipe.

  • #1
Dear Physicists,
I'm not sure if my question qualifies as coursework, but here it is anyway:
I'm taking ENGR-101 this term, and for a group project my group chose the design of a solar water heater. I'm trying to find a formula to calculate our design's efficiency, and when I search for thermodynamic equations I can't find what I'm looking for.

Homework Statement



An insulated 30 ga. water tank has a 3/4" dia. , 40' long copper pipe connected to it (the pipe leaves the tank and returns to it). A 1 ga./min. pump moves the water through the pipe. The air around the pipe is hotter than the water, so the water in the tank gets hotter as it circulates around the pipe. If the initial temperature of the water in the tank is 45 degrees fahrenheit, and the air temperature around the pipe remains constant at 100 degrees fahrenheit, how hot will the water in the tank be after 1 hour?

thermal conductivity (k) of copper: 400
wall thickness of pipe (l): 1 mm = 0.001 m
Surface area of pipe (A): (2.54 cm * 0.75 * pi)/100 * 12.2 m = 0.73 m^2
Heat capacity of water (Cp): 4.2
mass flow rate (m-dot): 1 ga/min = 3.8 l/min = 3.8 kg/min = 0.06 kg/s
Mass of water: 30 ga. = 113.6 l = 113.6 kg (mass of water in pipe negligible)
Initial difference in temp: 37.8 C - 7.2 C = 30.56 C = 30.56 K


Homework Equations



The only equation I have found, which deals only with the change in temp between the ends of the pipe, is:

Tf - Ti = q * A / m-dot * Cp
q = k * (T outside - T water) / l

The Attempt at a Solution



When I punched in my values in the equation above, I got for an answer millions of kelvins difference between Tf and Ti!

I'm also taking Calculus II this term, so I'm prepared (hopefully) to deal with the integrals that should pop up as the water in the tank gets gradually hotter as a function of time.

Thanks!
 

Answers and Replies

  • #2
389
0
Hi! Welcome to PF!

For your question, i think this equation will be the one you need to work with:

[tex] \frac{dQ}{dt} = KA\frac{dT}{dr} [/tex]

K is you thermal conductivity, A is surface area. dQ/dt is the power transfer, and dT/dx is the temperature gradient across the pipe.

If i were to make certain assumptions to make integration easier, like the power is a constant value and that it is at a steady state. We will get some textbook answer that goes like this:

[tex] \int_{r-x}^{r} \frac{1}{r} dr = \int_{T_w}^{T_s} \frac{k2\pi l}{P} dT [/tex]

[tex] ln(\frac{r}{r-x}) = \frac{k2\pi l}{P} (T_s - T_w) [/tex]

Which i am sure that would't work in reality. Since as power will not be constant and depends on the temperature gradient. This equation also assume that the water in pipe can keep taking in energy with no temperature gain.

Maybe the PF mentors can help.
 
  • #3
4,239
1
I can't help you a lot. You have a lot of units conversions to do. 40 inches should be closer to 1.22 meters.
 
  • #4
51
0
The first thing that you need to do is an energy balance for the tank (i.e., what goes in minus what goes out equals to accumulation). Of course, you make the assumption that the tank has a uniform temperature.
Then you need to calculate the amount of heat that is transferred from the hot air to the cold water, as the latter moves in the pipe. For that, you need to use an equation that is similar to the one you wrote: q=h*(Tw-Ta), where h is the heat transfer coefficient (look for "Convective heat transfer" or "Newton's law of cooling"). Once you know q, you can compute the new temperature of the water. Pay attention to the fact that q depends on the water temperature. Thus you have a coupled system of equations.
 
  • #5
20
0
just make sure that all of your numbers are in the correct units and double check your calculations, other wise you should be correct.
 
  • #6
Hi! Welcome to PF!

For your question, i think this equation will be the one you need to work with:

[tex] \frac{dQ}{dt} = KA\frac{dT}{dr} [/tex]

K is you thermal conductivity, A is surface area. dQ/dt is the power transfer, and dT/dx is the temperature gradient across the pipe.

If i were to make certain assumptions to make integration easier, like the power is a constant value and that it is at a steady state. We will get some textbook answer that goes like this:

[tex] \int_{r-x}^{r} \frac{1}{r} dr = \int_{T_w}^{T_s} \frac{k2\pi l}{P} dT [/tex]

[tex] ln(\frac{r}{r-x}) = \frac{k2\pi l}{P} (T_s - T_w) [/tex]

Which i am sure that would't work in reality. Since as power will not be constant and depends on the temperature gradient. This equation also assume that the water in pipe can keep taking in energy with no temperature gain.

Maybe the PF mentors can help.

Dear Delzac,
Thank you for your welcome and your help. I have looked carefully at your equations, and I'm afraid that I don't know how to put them to use. I read the first equation as "the instant rate of change in power with respect to time equals k times A times the instant rate of change in temperature with respect to r". (I don't know what r stands for.)

In your second equation, I don't know what P or r stand for.

In my mind, I need an equation that contains the variables I know, whose area beneath the curve between t-final and t-initial would tell me how much energy I have in that water, and another equation that translates that energy into a difference in temperature.

Please bear in mind that I haven't studied college physics yet, and that my problem has not come out of a textbook that contains what I need to solve it.
 
  • #7
I can't help you a lot. You have a lot of units conversions to do. 40 inches should be closer to 1.22 meters.

Thank you for looking into my problem, dear Phrak. The following reads "forty feet": 40'. The following reads "forty inches": 40".

Can you see any other problems with my units or my conversions?
 
  • #8
The first thing that you need to do is an energy balance for the tank (i.e., what goes in minus what goes out equals to accumulation). Of course, you make the assumption that the tank has a uniform temperature.
Then you need to calculate the amount of heat that is transferred from the hot air to the cold water, as the latter moves in the pipe. For that, you need to use an equation that is similar to the one you wrote: q=h*(Tw-Ta), where h is the heat transfer coefficient (look for "Convective heat transfer" or "Newton's law of cooling"). Once you know q, you can compute the new temperature of the water. Pay attention to the fact that q depends on the water temperature. Thus you have a coupled system of equations.

Dear MasterX, thank you for your help. I'm afraid that I don't know how to do an energy balance. I have a formula to calculate the energy in the tank, but I still don't know how to calculate how much energy the system gains per unit of time.

There's something I don't understand about the formula for q you gave me: it changes the variables k and l for the variable h, which depends not on the pipe's material but on the substance flowing through it. That seems to imply that the water will gain the same heat going through a thin silver pipe or a thick plastic one.

Am I to understand that there isn't any simple formula to calculate either the rate of change or the total energy gained as a function of time?

I've checked my units and my conversions 10 times, and they seem (to me) to be correct. I'm still getting over 35 million kelvins difference from one end of the pipe to the other. Would someone point out specifically whether I'm doing something wrong or whether my formulas are incorrect? You don't need to be so mysterious!

Thank you.
 
  • #9
51
0
There's something I don't understand about the formula for q you gave me: it changes the variables k and l for the variable h, which depends not on the pipe's material but on the substance flowing through it. That seems to imply that the water will gain the same heat going through a thin silver pipe or a thick plastic one.

This is correct!

One thing that I missed is that in the 2nd eqn you have the term "k/l". This term has the units of "h", and some people do suggest to set "h=k/l". So, the 2nd eqn is OK.

When you write an eqn make sure that the units are in balance.
Your first eqn is wrong:

Tf - Ti = q * A / m-dot * Cp

The units on the left side are "K". On the right side you have

(joule/s)*(m^2/(kg/s))*(joule/kg*K)

Based on the units, this eqn should be

Tf - Ti = q / (m-dot* Cp)

or else q=m-dot*Cp*(Tf-Ti)

On the Internet I found that Cp=4.2kj/(kg * K)
Also the thermal conductivity of copper is 400 (units?) W/mK = kcal/(hr*m*C)

1cal=4.183 joule So, k=0,465kj/(s*m*K) (convert hours to sec)
 
Last edited:
  • #10
just make sure that all of your numbers are in the correct units and double check your calculations, other wise you should be correct.

dear donutz610,
Thanks for looking into my problem and checking my formulas. I've checked my units and my calculations many times, and my answers still don't make any sense.
 
  • #11
This is correct!

One thing that I missed is that in the 2nd eqn you have the term "k/l". This term has the units of "h", and some people do suggest to set "h=k/l". So, the 2nd eqn is OK.

When you write an eqn make sure that the units are in balance.
Your first eqn is wrong:

Tf - Ti = q * A / m-dot * Cp

The units on the left side are "K". On the right side you have

(joule/s)*(m^2/(kg/s))*(joule/kg*K)

Based on the units, this eqn should be

Tf - Ti = q / (m-dot* Cp)

or else q=m-dot*Cp*(Tf-Ti)

On the Internet I found that Cp=4.2kj/(kg * K)
Also the thermal conductivity of copper is 400 (units?) W/mK = kcal/(hr*m*C)

1cal=4.183 joule So, k=0,465kj/(s*m*K) (convert hours to sec)

Dear MasterX,
Thank you taking the time to help me out again with this problem. I've checked the units in the equation Tf - Ti = q * A / m-dot * Cp, and they are correct. On the left of the equation we have kelvins; on the right we have W/m^2 (q) times m^2 (A) divided by kg/s (m-dot) times J/kg*K (Cp). That's W*m^2*s*kg*K/m^2*kg*J = W*s*K/J; since W=J/s, we have J*s*K/s*J = K

The thermal conductivity of copper is 400 W/m*K (already a SI unit). 1 W/m*K = 0.86 kcal/hr*m*degC.

I changed my heat capacity Cp of water from 4.2 kJ/kg*K to 0.0042 J/kg*K, but since Cp is in the denominator of the equation, now my answer (Tf-Ti) is in the billions of kelvins instead of millions! I can't figure out what I'm doing wrong!

This is what I have so far:

Tf - Ti = q*A/m-dot*Cp
q = k*(Tair - Twater)/l
therefore, Tf - Ti = k*(Tair - Twater)*A/l*m-dot*Cp

l = thickness of pipe = 1 mm = 0.001 m
A = surface area of pipe 3/4" dia. * 40 ft. = 0.73 m^2
m-dot = mass flow rate = 1 ga/min = 3.8 l/min = 3.8 kg/min = 0.06 kg/s
Cp = heat capacity of water = 4.2 kJ/kg*K = 0.0042 J/kg*K
k = thermal conductivity of copper = 400 W/m*K
Tair-Twater = 100degF - 50degF/2 = 28 K/2 = 14 K (I'm dividing by 2 because I believe I'm supposed to take the average.)

Tf - Ti = 400*14*0.73/0.001*0.06*0.0042 = 16.2 billion K

That's a powerful solar water heater!
 
  • #12
51
0
Dear MasterX,
Thank you taking the time to help me out again with this problem. I've checked the units in the equation Tf - Ti = q * A / m-dot * Cp, and they are correct. On the left of the equation we have kelvins; on the right we have W/m^2 (q) times m^2 (A) divided by kg/s (m-dot) times J/kg*K (Cp). That's W*m^2*s*kg*K/m^2*kg*J = W*s*K/J; since W=J/s, we have J*s*K/s*J = K

You made one mistake: You wrote W*m^2*s*kg*K/m^2*kg*J = W*s*K/J
On the left hand side you have W*j, but on the right hand side you have W/j. How did you do that?

Also, your second eqn should be
q = A* k * (T outside - T water) / l

so that q has units of W (or j/s), although, that is not really important.
 
Last edited:

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