An empty building\'s temperature on a cold morning drops to 48 degree at 1: 00 a

Question

An empty building's temperature on a cold morning drops to 48 degree at 1: 00 a.m. At that time the furnace is turned on with the thermostat set at 66 degree. The time constant for the building is 2 hours and for the building along with its heating system is 1/4 hour. Assuming that the outside temperature varies as a cosine wave with a maximum 54 degree at 1: 00 p.m. and a minimum of 27 degree at 1: 00 a.m., set up the initial value problem (with initial condition) that describes this problem. Do not solve this IVP.

Explanation / Answer

time constant of the building with the heating system = T'
time constant of the building without the heating system = T
at time t = to ( 1 am ) , temperature, Ko of the building is 48 F
Thermostat termperature, K' = 66 F
temperature of surroundings at time t = L

so, after t = to
d(K - Ko)/dt = -(T')*(K -L)

now, K is temperature of building at time t
Ko = 48 F
T' = 0.25 hours

L = (54 + 27)/2 - (54 - 27)cos(w(t - to)/2
at t = 1 am, t - to = 0, L = 27 F
at t = 1 pm, t - to = 12 hrs, L = 54 F, so cos(w(t - to)) = -1
w(t - to) = pi
w = pi/12 per hour

so, L = 40.5 - 13.5cos(pi(t - to)/12) [ where to is 1 am and t is in hours]
so

d(K - 48)/dt = -(0.25)*(K - 40.5 - 13.5cos(pi(t - to)/12))
d(K)/dt = -(0.25)*(K - 40.5 - 13.5cos(pi(t - to)/12))

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