How long does it take to heat a vessel?

In summary, the conversation discusses finding the time it takes to heat a vessel of water with specific heat capacity, density, volume, heat capacity of the vessel, and power rating of the element provided. The initial attempt at a solution uses the formula t=C(ΔT)/power and the Newton's law of cooling equation to calculate the time with and without taking into account ambient temperature. However, it is pointed out that the analysis is incorrect due to not considering the thermal inertia of water. The correct solution is requested before attempting to solve the problem with heat loss to the surrounding air included.
  • #1
Modrisco89

Homework Statement


I need to find the time it takes to heat a vessel of water which is closed.

Here is what I have:

Specific heat capacity of water: 4184 J kg^-1 K^-1
Density of water: 1000 kg m^-3
Volume of water added and volume of vessel: 500 L or 0.5 m^3
Heat capacity of the vessel: 91000 J K^-1
Power rating of the element: 25 kW

This is just a rough time estimator for a program I'm designing

Homework Equations


I'm thinking the formula to use without taking into account ambient temperature is:

t = C(ΔT)/power

t = time, C = to heat capacity, ΔT = final temp - initial temp and power = power rating of element.

The Attempt at a Solution


So I calculate C = (1000)×(0.5)x(4184) + 91000
And I plug in what ever temperature difference and the power of the heating element and calculate the time that way!

Now I want take into consideration of the ambient temperature outside the tank so it will take longer to heat up the tank.

Im looking Newton's law of cooling and noticed the equation:

T(t) = Ta + (T0 - Ta)e^(t/t0)

T(t) is the temperature at time t
Ta is the ambient temperature
T0 is the temperature at time 0
t0 is the time constant in seconds

Can I take my original equation for t above I'm the relevant equation and sub it into t0? And then solve for t in the solved DE to calculate the time it would take to heat the vessel with regards to ambient temperature

Hope I've been clear
 
  • Like
Likes SirMarx01
Physics news on Phys.org
  • #2
Can anybody assist me with this?
 
  • #3
Modrisco89 said:
Can I take my original equation for t above I'm the relevant equation and sub it into t0?
No.
Suppose that at time t the temperature is T(t). What gains and losses will occur in the next short period dt?
 
  • #4
haruspex said:
No.
Suppose that at time t the temperature is T(t). What gains and losses will occur in the next short period dt?

I actually don't know, I can understand how the equation t works but when it comes to taking into account the ambient temperature I'm lost, because the tank is heating while it'd losing a small bit of heat to the air outside the tank, so there's a gain and some loss at the same time, how do I model this? I spent 14 hours yesterday trying to do this
 
  • #5
Modrisco89 said:
I actually don't know, I can understand how the equation t works but when it comes to taking into account the ambient temperature I'm lost, because the tank is heating while it'd losing a small bit of heat to the air outside the tank, so there's a gain and some loss at the same time, how do I model this? I spent 14 hours yesterday trying to do this
If the current temperature is T, and the ambient is T0, at what rate is heat being lost? (You will need to assume some constant here, so it's just the form of the equation that matters.)
 
  • #6
haruspex said:
If the current temperature is T, and the ambient is T0, at what rate is heat being lost? (You will need to assume some constant here, so it's just the form of the equation that
haruspex said:
If the current temperature is T, and the ambient is T0, at what rate is heat being lost? (You will need to assume some constant here, so it's just the form of the equation that matters.)
haruspex said:
If the current temperature is T, and the ambient is T0, at what rate is heat being lost? (You will need to assume some constant here, so it's just the form of the equation that matters.)

That's a good question, so I have to assume a constant for the rate at which its lost? Is the solved DE useful to solve my problem?
 
  • #7
Modrisco89 said:
That's a good question, so I have to assume a constant for the rate at which its lost? Is the solved DE useful to solve my problem?
To get an actual answer you will need to know the value of the constant, of course, but we can get the right form of DE and its solution with it as an unknown.
 
  • #8
So this constant were talking about, this is the decay constant yes? So shall I scrap the original solved DE I have and form a new DE?
 
  • #9
Modrisco89 said:
So this constant were talking about, this is the decay constant yes? So shall I scrap the original solved DE I have and form a new DE?
Yes, except that I do not see a DE in this thread yet.
 
  • #10
haruspex said:
Yes, except that I do not see a DE in this thread yet.

Should it start with dQ/dt = C(T-Ta)?
 
  • #11
Modrisco89 said:
Should it start with dQ/dt = C(T-Ta)?
Yes, but now bring in the heat gained from the heating element.
 
  • #12
In this problem, you're supposed to assume that the heat lost to the surrounding atmosphere is negligible. Even with this assumption, your analysis in the original post is incorrect. You forgot to take into account the thermal inertia of the water. What do you get when you correct this? Let's see the number for the required time.

Before you start trying to solve the problem with heat loss to the air included, you first better know how to solve it without ( the additional thermal mechanism) of heat loss.
 
Last edited:
  • #13
Chestermiller said:
In this problem, you're supposed to assume that the heat lost to the surrounding atmosphere is negligible. Even with this assumption, your analysis in the original post is incorrect. You forgot to take into account the thermal inertia of the water. What do you get when you correct this? Let's see the number for the required time.

Before you start trying to solve the problem with heat loss to the air included, you first better know how to solve it without ( the additional thermal mechanism) of heat loss.

I'm able to calculate the time it takes to heat it from 1 K to 89 K and that works out to be 7684 seconds without heat loss to air surrounding the tank, that's when I use my t = C(Tf-Ti)/power as I said in my 1st post its just a rough estimate, I have to work with what I'm given...how do I work out the constant of heat loss to surrounding air?
 
Last edited by a moderator:
  • #14
haruspex said:
Yes, but now bring in the heat gained from the heating element.
So like dQ/dt = C(dT/dt)/power
 
Last edited by a moderator:
  • #15
Since Temperature changes with time I write it like this dQ/dt = C(T(t) - Ta)

Where C is the heat capacity, and T(t) is the temperature at time t, and Ta is the ambient temperature.

Do I write it like this:
dQ = C(T(t) - Ta)dt
Q(t) = C(Tf - Ta)t + Constant
Where Tf is the final temperature
Constant could be considered as intitial temperature by heat capacity C

So this is what I have
Q(t) = C(Tf - Ta)t + CTi

Does that sound right?
 
  • #16
Modrisco89 said:
Since Temperature changes with time I write it like this dQ/dt = C(T(t) - Ta)

Where C is the heat capacity, and T(t) is the temperature at time t, and Ta is the ambient temperature.

Do I write it like this:
dQ = C(T(t) - Ta)dt
Q(t) = C(Tf - Ta)t + Constant
Where Tf is the final temperature
Constant could be considered as intitial temperature by heat capacity C

So this is what I have
Q(t) = C(Tf - Ta)t + CTi

Does that sound right?
ac
Actually that's completely wrong, if someone can assist me I'd be grateful no one seems to have the answer to this problem
 
  • #17
Modrisco89 said:
Since Temperature changes with time I write it like this dQ/dt = C(T(t) - Ta)

Where C is the heat capacity, and T(t) is the temperature at time t, and Ta is the ambient temperature.

Do I write it like this:
dQ = C(T(t) - Ta)dt
Q(t) = C(Tf - Ta)t + Constant
Where Tf is the final temperature
Constant could be considered as intitial temperature by heat capacity C

So this is what I have
Q(t) = C(Tf - Ta)t + CTi

Does that sound right?
No. Not even close. The correct differential equation is $$(\rho V c+C)\frac{dT}{dt}=\dot{Q}-k(T-T_a)$$
where ##\rho## is the density of the water, V is the volume of water, c is the specific heat capacity of the water, C is the heat capacity of the vessel, ##\dot{Q}## is the rate of heat addition from the heating element, and k is proportionality coefficient for the rate of heat loss to the air.

Did you really mean for the starting temperature to be 1 K (since the water is well below the freezing point at that temperature)? Maybe you mean 1 C?
 
  • #18
Chestermiller said:
No. Not even close. The correct differential equation is $$(\rho V c+C)\frac{dT}{dt}=\dot{Q}-k(T-T_a)$$
where ##\rho## is the density of the water, V is the volume of water, c is the specific heat capacity of the water, C is the heat capacity of the vessel, ##\dot{Q}## is the rate of heat addition from the heating element, and k is proportionality coefficient for the rate of heat loss to the air.

Did you really mean for the starting temperature to be 1 K (since the water is well below the freezing point at that temperature)? Maybe you mean 1 C?
Thank you so much, I'll figure out how that equation works and I'll come back if I have questions, I Actually meant 1 degree celcius sorry
 
  • #19
Modrisco89 said:
Thank you so much, I'll figure out how that equation works and I'll come back if I have questions, I Actually meant 1 degree celcius sorry
If you meant 1 C, then you realize that this is lower than the room air temperature, so heat will be flowing from the room air to the water-filled vessel, rather than the other way around, correct?
 
  • #20
Chestermiller said:
If you meant 1 C, then you realize that this is lower than the room air temperature, so heat will be flowing from the room air to the water-filled vessel, rather than the other way around, correct?

I was thinking of that, would that mean the vessel will heat quicker when the initial temperature is lower than the ambient?
 
  • #21
Modrisco89 said:
I was thinking of that, would that mean the vessel will heat quicker when the initial temperature is lower than the ambient?
Sure, at least initially. Wouldn't you expect that?
 
  • Like
Likes Modrisco89
  • #22
Chestermiller said:
Sure, at least initially. Wouldn't you expect that?
Yes I would, it's starting to make more sense now,
Anyway, I solved your differentail equation using the laplace transform method and assumed initial condition to be T(0)

All I'm missing is the k value
 
  • #23
Modrisco89 said:
Yes I would, it's starting to make more sense now,
Anyway, I solved your differentail equation using the laplace transform method and assumed initial condition to be T(0)

All I'm missing is the k value
The k value depends on the specific geometry, the area available for heat transfer, and the motion of the air currents in the room.
 
  • #24
Chestermiller said:
The k value depends on the specific geometry, the area available for heat transfer, and the motion of the air currents in the room.
I'll try figure out the k value...

Here's my the solved DE,
$$T(t) =\frac {\dot{Q}+kT_a+T_0}{C}e^{\frac{kt}{C}}$$

Where C = density × volume x specific heat + heat capacity

I'm assuming the k value to be thermal conductance?
 
Last edited by a moderator:
  • #25
Modrisco89 said:
I'll try figure out the k value...

Here's my the solved DE,
$$T(t) =\frac {\dot{Q}+kT_a+T_0}{C}e^{\frac{kt}{C}}$$Where C = density × volume x specific heat + heat capacity

I'm assuming the k value to be thermal conductance?
This solution is incorrect. The parameter k is a thermal conductance of sorts; it is the heat transfer coefficient multiplied by heat transfer area.
 
  • #26
Chestermiller said:
This solution is incorrect. The parameter k is a thermal conductance of sorts; it is the heat transfer coefficient multiplied by heat transfer area.

$$T(t) =\frac {\dot{Q}+kT_a+CT_0}{C}e^{\frac{-kt}{C}}$$

Sorry made a mistake, though when I solve it for time, assuming the value of k to be the surface area of the tank (assumed spherical) x thermal conductivity of air. And T0 to be in tial temp...and Tf to be final temp, and I get a minus answer
 
  • #27
Modrisco89 said:
$$T(t) =\frac {\dot{Q}+kT_a+CT_0}{C}e^{\frac{-kt}{C}}$$

Sorry made a mistake, though when I solve it for time, assuming the value of k to be the surface area of the tank (assumed spherical) x thermal conductivity of air. And T0 to be in tial temp...and Tf to be final temp, and I get a minus answer
The solution is still in correct. It is not even close.
 
  • #28
Chestermiller said:
The solution is still in correct. It is not even close.
Really? I assumed all values to be constant except T and dT/dt, the rest are constants, I used the laplace transform method get it...while assuming initial conditions to be T_0...what am I doing wrong?
 
  • #29
Modrisco89 said:
Really? I assumed all values to be constant except T and dT/dt, the rest are constants, I used the laplace transform method get it...while assuming initial conditions to be T_0...what am I doing wrong?
Your solution doesn't even satisfy the initial condition. This is a variable-separable first order linear ordinary differential equation with constant coefficients. You can solve it either by separation of variables or by use of an integrating factor.
 
  • #30
Chestermiller said:
Your solution doesn't even satisfy the initial condition. This is a variable-separable first order linear ordinary differential equation with constant coefficients. You can solve it either by separation of variables or by use of an integrating factor.
I think I finally got it, I'll shoot myself if this is wrong:

$$T(t) = \frac{\dot{Q}+kT_a+CT_0}{k}(1-e^{\frac{-kt}{C}}$$
 
  • #31
Modrisco89 said:
I think I finally got it, I'll shoot myself if this is wrong:

$$T(t) = \frac{\dot{Q}+kT_a+CT_0}{k}(1-e^{\frac{-kt}{C}}$$
Get your gun out. I get $$T=T_0+\left[\frac{\dot{Q}}{k}+(T_a-T_0)\right](1-e^{-\frac{kt}{C}})$$
 
  • #32
Modrisco89 said:
I think I finally got it, I'll shoot myself if this is wrong:

$$T(t) = \frac{\dot{Q}+kT_a+CT_0}{k}(1-e^{\frac{-kt}{C}}$$
Chestermiller said:
Get your gun out. I get $$T=T_0+\left[\frac{\dot{Q}}{k}+(T_a-T_0)\right](1-e^{-\frac{kt}{C}})$$
Oh man, definitely getting the gun, assuming that the laplace transform method is illegal for that DE? That's what I used perhaps that wasn't good practice

I was told by lecturer that It can solve any DE
 
  • #33
Modrisco89 said:
Oh man, definitely getting the gun, assuming that the laplace transform method is illegal for that DE? That's what I used perhaps that wasn't good practice

I was told by lecturer that It can solve any DE
Certainly Laplae Transform can handle that equation.
 
  • #34
Chestermiller said:
Certainly Laplae Transform can handle that equation.

Must be my maths is the major issue here, listen thanks so much for your help, I'm coming with realistic values on excel so it's looking good, I took the surface area of the tank and multiplied it by the thermal conductility of air and it's working out alright
 
  • #35
Modrisco89 said:
Must be my maths is the major issue here, listen thanks so much for your help, I'm coming with realistic values on excel so it's looking good, I took the surface area of the tank and multiplied it by the thermal conductility of air and it's working out alright
The surface area times the thermal conductivity of air does not have the proper units.
 

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
514
  • Introductory Physics Homework Help
Replies
23
Views
1K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
781
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
867
Back
Top