Temperature change due to mixing liquids, heating and heat losses

[tjosan]

TL;DR Summary

I need help to combine these equations in order to get one expression that governs the temperature over time. Mixing liquids with different temperature, at the same heating the mixture, at the same time there are heat losses.

Hello,
I am stuck how to proceed with the equations below.

The problem:
A tank containing $m_0$ mass, with a temperature of $T_0$, is to be filled with a total mass of $m_1$, with the constant mass flow rate $G$ and temperature $T_1$ during $t_p$ seconds. $T_0>T_1$. Other than the temperature, the physical properties of the liquids are the same.

While all this is happening, the mixture in the tank is being heated with the effect $Q_H$, and there is heat loss with heat transfer coefficient $h$.

Looking at all of these events independently (only considering the time interval $0\le t \le t_p$):

a. Temperature due to filling the tank with a lower temperature liquid. No heating, no heatloss.

The temperature can be given as the weighted average. To make it time-dependent, I divide by $t_p$ and multiply with $t$ (since the liquids are identical, the $C_p$ value cancels, which otherwise is present.)
$$T_a(t)=\frac{m_0T_0+m_1T_1}{(m_0+m_1)t_p}t$$
__

b. Heat loss. No heating, no filling.
The heat loss is given by:
$$\dot{Q}=hA(T(t)-T_S)$$
Since $\dot{Q}=C_pm\frac{dT}{dt}$, I get:
$$\frac{dT_b}{dt}=\frac{hA}{C_pm_0}(T(t)-T_S)$$

Since there is no filling in this case, the mass is set to $m_0$ rather than $Gt$.
__

c. Heating. No heat loss, no filling.
$$T_c(t)=\frac{\dot{Q_H}}{C_pm_0}t$$

Since there is no filling in this case, the mass is set to $m_0$ rather than $Gt$.
__

I want one expression, $T(t)$ that governs the temperature over time, but I can't seem to figure out how I can combine these equtions.

For example, to calculate the heat loss, I need to factor in the temperature change from case a and the heating from case c. But since the mass change with time, the heating in case c is not constant (case b is also dependent on the mass). This is just one of the problems I came across. I suspect I might need to solve a system of equations maybe?

Any help would be appreciated. Thank you!

 

[Philip Koeck]

TL;DR Summary: I need help to combine these equations in order to get one expression that governs the temperature over time. Mixing liquids with different temperature, at the same heating the mixture, at the same time there are heat losses.

Hello,
I am stuck how to proceed with the equations below.

The problem:
A tank containing $m_0$ mass, with a temperature of $T_0$, is to be filled with a total mass of $m_1$, with the constant mass flow rate $G$ and temperature $T_1$ during $t_p$ seconds. $T_0>T_1$. Other than the temperature, the physical properties of the liquids are the same.

While all this is happening, the mixture in the tank is being heated with the effect $Q_H$, and there is heat loss with heat transfer coefficient $h$.

Looking at all of these events independently (only considering the time interval $0\le t \le t_p$):

a. Temperature due to filling the tank with a lower temperature liquid. No heating, no heatloss.

The temperature can be given as the weighted average. To make it time-dependent, I divide by $t_p$ and multiply with $t$ (since the liquids are identical, the $C_p$ value cancels, which otherwise is present.)
$$T_a(t)=\frac{m_0T_0+m_1T_1}{(m_0+m_1)t_p}t$$
__

b. Heat loss. No heating, no filling.
The heat loss is given by:
$$\dot{Q}=hA(T(t)-T_S)$$
Since $\dot{Q}=C_pm\frac{dT}{dt}$, I get:
$$\frac{dT_b}{dt}=\frac{hA}{C_pm_0}(T(t)-T_S)$$

Since there is no filling in this case, the mass is set to $m_0$ rather than $Gt$.
__

c. Heating. No heat loss, no filling.
$$T_c(t)=\frac{\dot{Q_H}}{C_pm_0}t$$

Since there is no filling in this case, the mass is set to $m_0$ rather than $Gt$.
__

I want one expression, $T(t)$ that governs the temperature over time, but I can't seem to figure out how I can combine these equtions.

For example, to calculate the heat loss, I need to factor in the temperature change from case a and the heating from case c. But since the mass change with time, the heating in case c is not constant (case b is also dependent on the mass). This is just one of the problems I came across. I suspect I might need to solve a system of equations maybe?

Any help would be appreciated. Thank you!

Try to write an expression for the change of temperature at any time between 0 and t_p.
This should lead to a differential equation for T(t) which you then solve with the given starting value T(0).
Don't split it up the way you do. The T(t) you are looking for is most likely not a combination of the expressions for the individual processes.
For a few simpler examples you can look at the first two problems in https://www.researchgate.net/publication/333479286_Differential_equations_for_thermal_processes

 

[Chestermiller]

What is the exact word-for-word statement of the problem?

 

[Chestermiller]

The mass balance equation is : $$\frac{dm}{dt}=G$$subject to the initial condition m = $m_0$ at t = 0. Assuming perfect mixing, the heat balance equation is $$\frac{d(mu)}{dt}=Gu_I+\dot{Q}-hA(T-T_S)$$where $u=C(T-T_R)$ is the internal energy per unit mass, $u=C(T_I-T_R)$ is the entering internal energy per unit mass, and $T_R$ is the reference temperature for zero u. If we combine the heat- and mass balances, we then obtain: $$mC\frac{dT}{dt}=GC(T_i-T)+\dot{Q}-hA(T-T_S)$$with $m=m_0+Gt$.

 

[tjosan]

Thank you very much. I don't have a word-to-word statement, as this problem is an invention of my own. :)

 

[tjosan]

The solution is far from pretty :)
https://www.wolframalpha.com/input?i=(m+G*t)*C*dT/dt=GC(T_0-T)+Q-h*A*(T-T_1),+T(0)=56

 

[Chestermiller]

The solution is far from pretty :)
https://www.wolframalpha.com/input?i=(m+G*t)*C*dT/dt=GC(T_0-T)+Q-h*A*(T-T_1),+T(0)=56

This is what happens when you let Wolfram Alpha brute force it for you. On the other hand,
$$\frac{dT}{(T-T_{\infty})}=-\left(1+\frac{hA}{GC}\right)\frac{Gdt}{m_0+Gt}$$where $$T_{\infty}=\frac{GCT_I+\dot{Q}(GC+hA)+hAT_S}{(GC+hA)}$$So $$\frac{T-T_{\infty}}{T_0-T_{\infty}}=\left(\frac{m_0}{m}\right)^{\left(1+\frac{hA}{GC}\right)}$$

 

[tjosan]

I just did the algebra and arrived at the same result, thanks!

Although I only used $\dot{Q}$ in $T_\infty$, and not $\dot{Q}(GC+hA)$, which I assume was a mistake.

 

[Chestermiller]

I just did the algebra and arrived at the same result, thanks!

Although I only used $\dot{Q}$ in $T_\infty$, and not $\dot{Q}(GC+hA)$, which I assume was a mistake.

Yes, that’s a mistake. The correct equation is $$T_{\infty}=\frac{GCT_I+\dot{Q}+hAT_S}{(GC+hA)}$$