newt. law cooling 2 diff. equations!!!!

**zena** · September 7th 2007, 03:49 AM

I apologise.. this question may be a little convoluted

this problem is based on the temperature of a room....
the variable (t) is the time in hours after 4pm eg. at 5pm t=1 .... for the purpose of the question we are considering from 4pm until midnight t=0... till t=8

there are to differential equations to consider

1. dT/dt= -k(T-8) without heater And

2. dT/dt= -k(T-8)+H .... where H=2 with heater

where k is a constant relating to insulation k is approx= 0.1732867951
when t=0, T=16 and..... in d.e number 2. the H, which =2, is relating to the power of the heating system

the temperature in the room throughout the 8 hour period Cannot fall below T=15 if it does the heater automatically comes on until the temperature rises to T=16... then the heater goes off... if it goes below T=15 then the heater comes on again until T=16 and this cycle continues throughout the 8 hour period...
When the heater is ON, you use differ. equation 2
When the heater is OFF, you use D.E 1

I'm really sorry i hope this is clear!!!!!!

a. graph the fluctuation in Temperature against time for the 8 hour period and find the FIRST time, to the nearest minute, that the heater First turns OFF

b. As a percentage, find the proportion of time that the heater is ON during the 8 hour period

(i have a suspicion that the graph may have a zig zag shape ..??!!)

I am sorry if it is confusing... i have no idea how work this out because you have to use both differential equations.... someone PLEASE help me!!!!!

Thank you for your time... if you want to contact me to clarify what i have asked please fell free to email me Thanks again,
Zena

**topsquark** · September 7th 2007, 10:01 AM

Originally Posted by zena

I apologise.. this question may be a little convoluted

this problem is based on the temperature of a room....
the variable (t) is the time in hours after 4pm eg. at 5pm t=1 .... for the purpose of the question we are considering from 4pm until midnight t=0... till t=8

there are to differential equations to consider

1. dT/dt= -k(T-8) without heater And

2. dT/dt= -k(T-8)+H .... where H=2 with heater

where k is a constant relating to insulation k is approx= 0.1732867951
when t=0, T=16 and..... in d.e number 2. the H, which =2, is relating to the power of the heating system

the temperature in the room throughout the 8 hour period Cannot fall below T=15 if it does the heater automatically comes on until the temperature rises to T=16... then the heater goes off... if it goes below T=15 then the heater comes on again until T=16 and this cycle continues throughout the 8 hour period...
When the heater is ON, you use differ. equation 2
When the heater is OFF, you use D.E 1

I'm really sorry i hope this is clear!!!!!!

a. graph the fluctuation in Temperature against time for the 8 hour period and find the FIRST time, to the nearest minute, that the heater First turns OFF

b. As a percentage, find the proportion of time that the heater is ON during the 8 hour period

(i have a suspicion that the graph may have a zig zag shape ..??!!)

I am sorry if it is confusing... i have no idea how work this out because you have to use both differential equations.... someone PLEASE help me!!!!!

Thank you for your time... if you want to contact me to clarify what i have asked please fell free to email me Thanks again,
Zena

You are just going to have to sit down and work with the solution for the time period before the heater comes on, to the point where it turns off again, and to the point where it comes on a second time. Yes, the graph is going to be "saw-toothed"-like.

Are you having a problem with the solutions to the two differential equations?

-Dan

**zena** · September 7th 2007, 07:25 PM

Thank you for your assistance but i am still confused. i don't know what to do

**zena** · September 8th 2007, 04:44 PM

hello,

for part a. i get t= 0.4090489288

for part b i allowed the temperature to equal 15 and i got t= .4766551919 then if you allow T to =16 you get t=o so i am supposed to work out the time taken for heater to turn off.. i though i might have to minus the T=16 time from the T=15 time but now i don't think that is the case. so there is no equation for the graph... i just have to keep alternating between the 2 differential equations? but.. how do i keep alternating between the two equations as for every temperature you only get one value for t. as when i substituted T=16 i got t=o, the initial temperature.. how can i get more than 1 value for t?

**zena** · September 9th 2007, 01:46 PM

Thank you for your help i do understand now....
Would it be possible for you to show me how you actually solved the differential equations... as when i solved them i got something different.. how do you get the T0 initial temp into the equation... when i solve it this is what i get
1. T= 8e^(-Kt)+8 and
2. T= (0.614e^(-kt)-3.38)/ (-k)

**topsquark** · September 9th 2007, 07:33 PM

Originally Posted by zena

Thank you for your help i do understand now....
Would it be possible for you to show me how you actually solved the differential equations... as when i solved them i got something different.. how do you get the T0 initial temp into the equation... when i solve it this is what i get
1. T= 8e^(-Kt)+8 and
2. T= (0.614e^(-kt)-3.38)/ (-k)

Hmmm...My notation here is slightly different than it was in my previous posts, but I did notice that I made an error in solving the heating equation.

After I finish solving this I'll rework the problem to see how much it changes my previous answers.

I will solve the equation
$T^{\prime} = -k(T - T_a) + H$
which can be adapted to either case. $T_a$ is the bath temperature. (For the cooling phase $T_a = 8$ and $H = 0$ , whereas for the heating phase $T_a = 8$ and $H = 2$ .)

So.
$T^{\prime} + kT = kT_0 + H$

The solution to the homogenous version of the equation
$T_h^{\prime} + kT_h = 0$
is
$T_h(t) = Ae^{-kt}$
where A is a constant.

The particular solution to
$T_p^{\prime} + kT_p = kT_a + H$
is
$T_p(t) = B$
where B is a constant. (Notice that the RHS of the differential equation is simply a constant.)

So putting the particular solution into the equation I get:
$0 + kB = kT_a + H$

Thus
$B = T_a + \frac{H}{k}$

So the particular solution is
$T_p(t) = T_a + \frac{H}{k}$

Thus the solution to the overall equation is
$T(t) = T_h(t) + T_p(t) = Ae^{-kt} + \left ( T_a + \frac{H}{k} \right )$

So when t = 0, the temperature will be $T_0$ . Thus
$T_0 = Ae^{-k \cdot 0} + \left ( T_a + \frac{H}{k} \right )$

$T_0 = A + T_a + \frac{H}{k}$

Thus
$A = T_0 - T_a - \frac{H}{k}$

Thus
$T(t) = \left ( T_0 - T_a - \frac{H}{k} \right )e^{-kt} + \left ( T_a + \frac{H}{k} \right )$

-Dan

**topsquark** · September 9th 2007, 08:01 PM

Originally Posted by zena

this problem is based on the temperature of a room....
the variable (t) is the time in hours after 4pm eg. at 5pm t=1 .... for the purpose of the question we are considering from 4pm until midnight t=0... till t=8

there are to differential equations to consider

1. dT/dt= -k(T-8) without heater And

2. dT/dt= -k(T-8)+H .... where H=2 with heater

where k is a constant relating to insulation k is approx= 0.1732867951
when t=0, T=16 and..... in d.e number 2. the H, which =2, is relating to the power of the heating system

the temperature in the room throughout the 8 hour period Cannot fall below T=15 if it does the heater automatically comes on until the temperature rises to T=16... then the heater goes off... if it goes below T=15 then the heater comes on again until T=16 and this cycle continues throughout the 8 hour period...
When the heater is ON, you use differ. equation 2
When the heater is OFF, you use D.E 1

The solution to the differential equation
$T^{\prime} = -k(T - T_a) + H$
is
$T(t) = \left ( T_0 - T_a - \frac{H}{k} \right )e^{-kt} + \left ( T_a + \frac{H}{k} \right )$

So for the cooling phase we have the equation
$\frac{dT}{dt} = -k(T - 8)$
and $T_0 = 16$ meaning that $T_a = 8$ and $H = 0$ . So the solution is:
$T(t) = ( 16 - 8 )e^{-kt} + 8 = 8e^{-kt} + 8$
(This is different from my initial solution.

)

So it takes a time t to drop the temperature to T = 15:
$15 = 8e^{-kt} + 8$

$7 = 8e^{-kt}$

$e^{-kt} = \frac{7}{8}$

$-kt = ln \left ( \frac{7}{8} \right )$

$t = -\frac{1}{k} ln \left ( \frac{7}{8} \right ) \approx 0.77058~hr$
(You didn't give me units for k but I assume t is measured in hours.)

For the heating phase we have the equation
$\frac{dT}{dt}= -k(T - 8) + 2$
with an initial temperature $T_0 = 15$ . So we have $T_a = 8$ and $H = 2$

The solution is
$T(t) = \left ( 15 - 8 - \frac{2}{k} \right )e^{-kt} + \left ( 8 + \frac{2}{k} \right ) = \left ( 7 - \frac{2}{k} \right )e^{-kt} + \left ( 8 + \frac{2}{k} \right )$
(This is significantly different from my initial solution.

)

So it takes a time t for the temperature to come back up to 16:
$16 = \left ( 7 - \frac{2}{k} \right )e^{-kt} + \left ( 8 + \frac{2}{k} \right )$

$16 - \left ( 8 + \frac{2}{k} \right ) = \left ( 7 - \frac{2}{k} \right )e^{-kt}$

$e^{-kt} = \frac{16 - \left ( 8 + \frac{2}{k} \right )}{\left ( 7 - \frac{2}{k} \right )} = \frac{16k - 8k - 2}{7k - 2} = \frac{8k - 2}{7k - 2}$

$-kt = ln \left ( \frac{8k - 2}{7k - 2} \right )$

$t = -\frac{1}{k} ln \left ( \frac{8k - 2}{7k - 2} \right ) \approx 1.43521~hr$

The temperature is now back to T = 16 and we go back to the cooling equation. It takes $t = 0.77058~hr + <br /> 1.43521~hr = 2.20579~hr$ for the temperature to cycle completely. Below I have included a graph of one temperature cycle. (Remember that to graph the heating part of the cycle we need to shift the time variable.)

-Dan

**zena** · September 9th 2007, 11:22 PM

Thank you ever so much!!!!!!

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