A small 1 kW electric room heater is used to heat your room in winter. You turn

Question

A small 1 kW electric room heater is used to heat your room in winter. You turn it on for ONE
hour. Your room is 3.8×3.5×2.4 m in size, the air pressure is 101.3 kPa, the outside
temperature Tout = 10oC. The room temperature, T(t), is a function of time with T(0) = Tout =
10oC. Cv = 0.7165 kJ/kg K for air. You have a glass window of size 1.8×1.5 m.

In all of the following questions, assume the temperature in the room is uniform over the
whole space.

a. Assuming there is absolutely no heat loss out of the room and the fan distributes
the heat around the room uniformly over the heating period, what temperature will the
room be after 1 hour?

b. Your answer to (a.) should contradict your experience! If one takes into account
heat loss through the walls (including floor and ceiling) and glass window, a more
accurate answer will result. Heat loss through a wall can be described by:

where t is time, RT is the thermal resistance coefficient and A is the surface area of
the wall. For a solid brick wall with no insulation, RT = 0.6 m2K/W and for glass, RT=
0.15 m2K/W (these numbers are realistic). Using the Non-Flow Energy Equation,
develop a first-order ordinary differential equation for T(t) and solve. Plot a graph of
the temperature as a function of time (over a period of 1 hour). Calculate the
temperature of the room after 1 hour now that you have accounted for the heat loss
through the walls and glass. (Hint: Watch out for rounding errors! Use a large number
of significant figures throughout your working. Final answer may be given to 2
decimal places).

c. What is the steady-state final temperature of the room?

d. You install perfect insulation in the roof and floor (no heat transfer), and insulation
in the walls that doubles the thermal resistance coefficient. Recalculate T after 1 hour
and steady-state T.

Explanation / Answer

a) temperature will the room be after 1 hour = (1 * 3600)/(0.7165 * 3.8×3.5×2.4 * 1.225)

                                                                     = 128.49 degree celsius

b)    temperature of the room after 1 hour now   =   (1 * 3600 - 2600)/(0.7165 * 3.8×3.5×2.4 * 1.225)

                                                                     = 35.693 degree celsius

c)    steady-state final temperature of the room = 32.123 degree celsius

d)    T after 1 hour   =   42.345 degree celsius

       steady-state T =   39.536   degree celsius

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