Second order differential - Tanks in series cooling coil

Chestermiller said:

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 said:

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 said:

Sorry, I meant Eqns. 18 and 19.

gmaverick2k said:

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

Chestermiller said:

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 said:

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

Chestermiller said:

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 said:

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 said:

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

Chestermiller said:

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

gmaverick2k said:

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 said:

This is not correct.

gmaverick2k said:

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 said:

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 $$T_2=T_2(\infty)+T_2^*=39.8+54.3e^{-1.8212t}+0.8e^{-4.2638t}$$