Second order differential - Tanks in series cooling coil

[gmaverick2k]

Homework Statement

Second order differential - Tanks in series cooling coil

Relevant Equations

Second order differential eq's

I'm stuck on a problem:
T1 = dT2/dt + xT2 - y
T2 = (Ae^(-4.26t))+(Be^(-1.82t))+39.9

I'm unsure how to proceed

 

[gmaverick2k]

 

[gmaverick2k]

 

[gmaverick2k]

I was able to get XXV by setting T2 = 94.9 @ t = 0 and putting this into XXIV, but I'm stumped with the other boundary condition result in XXVI

 

[Chestermiller]

Can you please provide the original problem statement?

 

[gmaverick2k]

Problem statement below. Final two pages in posts #2 & 3

 

[gmaverick2k]

Note for XXIII, I got 159.6 for the right hand sided term instead of 309 (see post #2) although the coeffcients remained unchanged

 

[Chestermiller]

I would never have solved this problem the way that they solved it. The first thing I would have done would have been to solve Eqns. 19 and 20 for the final steady state temperatures in the two tanks and large $\theta$ ($T_1(\infty)$ and $T_2(\infty)$), when the two time derivatives approach zero. Then I would write $$T_1=T_1^*+T_1(\infty)$$and$$T_2=T_2^*+T_2(\infty)$$What do you get for the final steady state temperatures, and what do you get when you substitute my Eqns. 1 and 2 into their Eqns. 19 and 20?

 

[gmaverick2k]

I would never have solved this problem the way that they solved it. The first thing I would have done would have been to solve Eqns. 19 and 20 for the final steady state temperatures in the two tanks and large $\theta$ ($T_1(\infty)$ and $T_2(\infty)$), when the two time derivatives approach zero. Then I would write $$T_1=T_1^*+T_1(\infty)$$and$$T_2=T_2^*+T_2(\infty)$$What do you get for the final steady state temperatures, and what do you get when you substitute my Eqns. 1 and 2 into their Eqns. 19 and 20?

I'm most likely wrong here, setting the differentials to 0 in both eq.'s [19] and [20], I get for both eq.'s:

4486.2*T2 - 1500*T1 = 59724

as expected because eq.'s [20] is rearranged form of [19]

 

[Chestermiller]

I'm most likely wrong here, setting the differentials to 0 in both eq.'s [19] and [20], I get for both eq.'s:

4486.2*T2 - 1500*T1 = 59724

as expected because eq.'s [20] is rearranged form of [19]

Sorry, I meant Eqns. 18 and 19.

 

[gmaverick2k]

Sorry, I meant Eqns. 18 and 19.

I get:
T1 (∞) = 79.3°C
T2 (∞) = 39.8°C

 

[Chestermiller]

I get:
T1 (∞) = 79.3°C
T2 (∞) = 39.8°C

I didn't check them, but these values look like they are probably right. Now, what do you get when you substitute my Eqns. 1 and 2 into their Eqns. 18 and 19 (in terms of the T*'s)?

 

[gmaverick2k]

I didn't check them, but these values look like they are probably right. Now, what do you get when you substitute my Eqns. 1 and 2 into their Eqns. 18 and 19 (in terms of the T*'s)?

I get:
T1* = 0.01°C
T2* = 0.03°C
Most likely incorrect because of rounding error

 

[gmaverick2k]

I get:
T1* = 0.01°C
T2* = 0.03°C
Most likely incorrect because of rounding error. Differentials again set to zero

 

[Chestermiller]

I get $$C\frac{dT_1^*}{d\theta}=-\left[C_W(1-\alpha)+C\right]T_1^*+C_W(1-\alpha)(1-\beta)T_2^*$$ and $$C\frac{T_2^*}{d\theta}=CT_1^*-[C_W(1-\beta)+C]T_2^*$$
Note, that, in terms of the asterisk parameters, there are no constant terms in these equations. Initial conditions are $$T_1^*=142.4-79.3=63.1$$and $$T_2^*=94.9-39.8=55.1$$

Does this make sense to you so far?

 

[gmaverick2k]

I get $$C\frac{dT_1^*}{d\theta}=-\left[C_W(1-\alpha)+C\right]T_1^*+C_W(1-\alpha)(1-\beta)T_2^*$$ and $$C\frac{T_2^*}{d\theta}=CT_1^*-[C_W(1-\beta)+C]T_2^*$$
Note, that, in terms of the asterisk parameters, there are no constant terms in these equations. Initial conditions are $$T_1^*=142.4-79.3=63.1$$and $$T_2^*=94.9-39.8=55.1$$

Does this make sense to you so far?

So T1 is T1,i (after cooling water is put back on) and T1 (∞) was T1 at inifinite time by setting differential to zero
T1* is T1,i+1
Subbed T1* into the differential term in place of T1 as it will vary with time and T (∞) is constant differential is now dT1*/dt

 

[Chestermiller]

So T1 is T1,i (after cooling water is put back on) and T1 (∞) was T1 at inifinite time by setting differential to zero
T1* is T1,i+1
Subbed T1* into the differential term in place of T1 as it will vary with time and T (∞) is constant differential is now dT1*/dt

If I understand you correctly, yes. Can you figure out how to solve the T* differential equations?

 

[gmaverick2k]

If I understand you correctly, yes. Can you figure out how to solve the T* differential equations?

so when T1* is subbed into the differential the terms such as Cw(1-alpha)beta*t3 and CT0 disappear as they are not a function of time

 

[gmaverick2k]

dT1*/dt = 1.489T2* - 3.094T1*
dT2*/dt = T1* - 2.991T2*

I think this is where eq's. [3] to [7] kick in for simultaneous 1st ODE.

 

[gmaverick2k]

If I understand you correctly, yes. Can you figure out how to solve the T* differential equations?

I may be wrong. Changed the initial temps

 

[Chestermiller]

I can't read what you did. From the initial values of the T*'s and the two differential equations, what are the initial values of the T*'s?

If you solve the first differential equation for T2* (in terms of T1* and its time derivative) and substitute that into the second differential equation, what do you get?

 

[gmaverick2k]

I initially used @ t = 0, T2* = 55.1°C as per post #15, which gave T1* = 39.97°C. I thought thats pretty low as the outlet temperature of tank 1 when the water is switched back on shouldn't be that low. So I changed the initial conditions as follows:
@ t = 0; T1* = 94.9°C ; T2* = 81.3°C calculated from the 1st ODE as the input into the dT2*/dt

 

[Chestermiller]

I initially used @ t = 0, T2* = 55.1°C as per post #15, which gave T1* = 39.97°C. I thought thats pretty low as the outlet temperature of tank 1 when the water is switched back on shouldn't be that low. So I changed the initial conditions as follows:
@ t = 0; T1* = 94.9°C ; T2* = 81.3°C calculated from the 1st ODE as the input into the dT2*/dt

This is not correct.

 

[gmaverick2k]

This is not correct.

I'm stumped. Tried the T* method but I'm hitting a wall..
where t > 0
T1*(t) = T1(0) - T1(t) = 142.4 - T1(t)
T1*(∞) = T1(0) - T1(∞) = 142.4 - 79.3 = 63.1°C
T1*(1) = T1(0) - T1(1) = 142.4 - 23.3 = 119.1°C

T2*(t) = T2(0) - T2(t) = 94.9 - T2(t)
T2*(∞) = T2(0) - T2(∞) = 94.9 - 39.8 = 55.1°C
T2*(1) = T2(0) - T2(1) = 94.9 - 20.3 = 74.6°C

I calculate:
T1(1) = 119.1°C
T2(1) = 74.6°C

Apologies, post #'s 19 & 20 are incorrect and should be ignored

 

[Chestermiller]

I'm stumped. Tried the T* method but I'm hitting a wall..
where t > 0
T1*(t) = T1(0) - T1(t) = 142.4 - T1(t)
T1*(∞) = T1(0) - T1(∞) = 142.4 - 79.3 = 63.1°C
T1*(1) = T1(0) - T1(1) = 142.4 - 23.3 = 119.1°C

T2*(t) = T2(0) - T2(t) = 94.9 - T2(t)
T2*(∞) = T2(0) - T2(∞) = 94.9 - 39.8 = 55.1°C
T2*(1) = T2(0) - T2(1) = 94.9 - 20.3 = 74.6°C

I calculate:
T1(1) = 119.1°C
T2(1) = 74.6°C

Apologies, post #'s 19 & 20 are incorrect and should be ignored

We are going. to first solve the two homogeneous coupled differential equation in post #19 subject to the initial conditions: $$T_1^*(0)=63.1$$ and $$T_2^*(0)=55.1$$If we substitute these values into the two differential equations, we can obtain the initial conditions on the first derivatives of $T_1^*$ and $T_2^*$ at time t = 0: $$\left[\frac{dT_1^*}{dt}\right]_{t=0}=-3.094T_1^*(0)+1.489T_2^*(0)=-113.2$$ and $$\left[\frac{dT_2^*}{dt}\right]_{t=0}=T_1^*(0)-2.991T_2^*(0)=-101.7$$

The next step is to solve the second equation post #19 for $T_1^*$ in the 2nd differential equation in terms of $T_2^*$ and its first derivative: $$T_1^*=\frac{dT_2^*} {dt}+2.991 T_2^*$$We can then eliminate $T_1^*$ by substituting this into the first differential equation in post #19: $$\frac{d^2T_2^*}{dt^2}+2.991\frac{dT_2^*}{dt}=-3.094\left(\frac{dT_2^*}{dt}+2.991 T_2^*\right)+1.489T_2^*$$or $$\frac{d^2T_2^*}{dt^2}+6.085\frac{dT_2^*}{dt}-7.7652T_2^*=0$$The solution to this homogeneous linear ODE is of the form $e^{\lambda t}$, with the two exponential terms satisfying $$\lambda^2+6.085\lambda+7.7652=0$$The solution to this quadratic equation for $\lambda$ is $$\lambda=\frac{-6.0850\pm\sqrt{(6.0850)^2-4(7.7652)}}{2}=-4.2638\ and\ -1.8212$$So, $$T_2^*(t)=Ae^{-1.8212t}+Be^{-4.2638t}$$where A and B are constants to be determined from the initial conditions: $$T_2^*(0)=A+B=55.1$$and
$$\left[\frac{dT_2^*}{dt}\right]_{t=0}=-1.8212A-1.4638B=-101.7$$Solving these two linear algebraic equation for the constants A and B gives: $$A=54.3$$ and $$B=0.8$$So, we have $$T_2^*(t)=54.3e^{-1.8212t}+0.8e^{-4.2638t}$$The solution for $T_2$ is then $$T_2=T_2(\infty)+T_2^*=39.8+54.3e^{-1.8212t}+0.8e^{-4.2638t}$$

 

[gmaverick2k]

We are going. to first solve the two homogeneous coupled differential equation in post #19 subject to the initial conditions: $$T_1^*(0)=63.1$$ and $$T_2^*(0)=55.1$$If we substitute these values into the two differential equations, we can obtain the initial conditions on the first derivatives of $T_1^*$ and $T_2^*$ at time t = 0: $$\left[\frac{dT_1^*}{dt}\right]_{t=0}=-3.094T_1^*(0)+1.489T_2^*(0)=-113.2$$ and $$\left[\frac{dT_2^*}{dt}\right]_{t=0}=T_1^*(0)-2.991T_2^*(0)=-101.7$$

The next step is to solve the second equation post #19 for $T_1^*$ in the 2nd differential equation in terms of $T_2^*$ and its first derivative: $$T_1^*=\frac{dT_2^*} {dt}+2.991 T_2^*$$We can then eliminate $T_1^*$ by substituting this into the first differential equation in post #19: $$\frac{d^2T_2^*}{dt^2}+2.991\frac{dT_2^*}{dt}=-3.094\left(\frac{dT_2^*}{dt}+2.991 T_2^*\right)+1.489T_2^*$$or $$\frac{d^2T_2^*}{dt^2}+6.085\frac{dT_2^*}{dt}-7.7652T_2^*=0$$The solution to this homogeneous linear ODE is of the form $e^{\lambda t}$, with the two exponential terms satisfying $$\lambda^2+6.085\lambda+7.7652=0$$The solution to this quadratic equation for $\lambda$ is $$\lambda=\frac{-6.0850\pm\sqrt{(6.0850)^2-4(7.7652)}}{2}=-4.2638\ and\ -1.8212$$So, $$T_2^*(t)=Ae^{-1.8212t}+Be^{-4.2638t}$$where A and B are constants to be determined from the initial conditions: $$T_2^*(0)=A+B=55.1$$and
$$\left[\frac{dT_2^*}{dt}\right]_{t=0}=-1.8212A-1.4638B=-101.7$$Solving these two linear algebraic equation for the constants A and B gives: $$A=54.3$$ and $$B=0.8$$So, we have $$T_2^*(t)=54.3e^{-1.8212t}+0.8e^{-4.2638t}$$The solution for $T_2$ is then $$T_2=T_2(\infty)+T_2^*=39.8+54.3e^{-1.8212t}+0.8e^{-4.2638t}$$

Thank you. Apologies, had a family emergency last week so wasn't able to get back on this