Heating and Cooling of BUilding

My name is Erna Apriliana, usually called Erna and number ID 07305141011. I want to tell about my experience with my friend. My friend is Zulvina Tri Susanti. I had explained about Heating and Cooling of Building.
I choose topic “Heating and Cooling of Building”, because I was inspirited by study about Linear Equation in Differential Equation. From that topic, we can know to determine the building temperature.
Three main factors affecting the temperature inside the building. The first factor is the heat produced by people, lights, and machines inside the building. This causes a rate of increase in temperature that we will denote by H(t). The second factor is the heating or cooling supplied by the furnace (or air conditioning). This rate of increase or decrease in temperature we will by U(t).
In general, the additional heating rate H(t) and the furnace (or air conditioning) rate U(t) are described in terms of energy per unit time. The third factor is the effect of the outside temperature M(t) on the temperature inside the building. This is known as Newton’s Law of cooling, which states that there is a rate of change in temperature T(t) that is proportional to the difference between the outside temperature M(t) and the inside temperature T(t). That is, the rate of change in the building temperature due to M(t) is :
dT(t) / dt = K [ M(t) – T(t) ]
the positive constant K depends on the physical properties of the building, but K does not depends on M, T, or t. When the outside temperature is grater than the inside temperature, then M(t) - H(t) > 0, and there is an increase in the rate of change of the building temperature due to M(t). When the outside temperature is less than the inside temperature, then M(t) - H(t) < 0, and there is an decrease in the rate of change.
Summarizing, we find
(1) dT(t) / dt = K [ M(t) – T(t) ] + H(t) +U(t)
Where the additional heating rate H(t)
is always non negative and U(t is positive for nace heating and negative for air conditioner cooling.
Equation (1) is linear, it can be solved using the method with standards form
(2) dT(t) / dt + P(t)T(t) = Q(t)
where P(t) = K
(3) Q(t) = KM(t + H(t) + U(t)
We find that the integrating factor is
Μ(t) = exp ( ∫ K dt ) = e^Kt
To solve (2), multiply each side by e^Kt and integrate:
e^Kt dT(t) / dt + K e^Kt P(t)T(t) = e^Kt Q(t)
e^Kt T(t) = ∫ e^Kt Q(t) dt + C
solving for T(t) gives
(4) T(t) = e^-Kt ∫ e^Kt Q(t) dt + C e^-Kt
= e^-Kt {∫ e^Kt [ KM(t) + H(t) + U(t) ] dt + C}
In this process, I don’t have any troubles to explain that. Santi could understand what I had explained. To make sure that she has understood, I gave her question like this :
For example, at the end of the day (at time t0), when people leave the building, the outside temperature stays constant at M0, the additional heating rate H inside the building is zero, and the air conditioner rate U is zero. Determine T(t), the initial condition T(t0) = T0
And her result is :
M = M0, H = 0, U = 0
From equqtion (4) she found :
T(t) = e^-Kt {∫ e^Kt [ KM(t) + H(t) + U(t) ] dt + C}
= e^-Kt {∫ e^Kt [ KM(t) + 0+ 0 ] dt + C}
= e^-Kt {∫ e^Kt [ KM(t) ] dt + C}
= e^-Kt [ M0 e^Kt + C ]
= M0 + C e^-Kt
That was my experience of studying English 2 with the lecturer is Mr. Marsigit, MA. I’m feel so exciting.