Question

Your dorm room is heated by a steam radiator. The room is 10 ft x 10 ft and has an 8 ft. ceiling. The room has a single pane glass window with dimensions 2 ft. wide and 3 ft. high. If the room is 25 °C and the outside temperature is -5 °C, how much steam (in grams per minute) must condense to maintain this temperature at steady state? Assume the steam is saturated at 125 °C and transfers only its latent heat. (hint: You will need to look up a few things).

Answer 1

Assuming, the walls of the dorm room are insulated and no heat is lost through them and the heat is lost only from the single pane glass window.

Assuming the thickness of the glass to be of 1/8" (0.01041 ft) and looking up for the thermal conductivity of glass, we find,

K_glass = 0.8 W / (m K) .

dimension of the glass, 2 ft * 3 ft * 0.01041 ft

in metric units, 0.6096 m * 0.9144 m * (3.173 * 10^-3) m.

therefore, the width of the glass window, w =  2 ft = 0.6096 m

Height of the glass window, h = 3 ft = 0.9144 m

the thickness of the window, t =  0.01041 ft = 3.173*10^-3 m.

The rate of heat loss from the room through the window using thermal conduction equation,

Q = KA dT/dx

Q_loss = K_glass*w*h*[(T_room - T_outside)/t]

Temperature of room, T_room = 25 ^oC = (273 + 25) K = 298 K

Outside Temperature, T_outside = -5 ^oC = (273 - 5) K = 268 K

Q_loss = K_glass*w*h*[(T_room - T_outside)/t]

Q_loss = 0.8*0.6096*0.9144*[(298 - 268)/(3.173*10^-3]

Q_loss = 4216.211 J/s = 4.2162 kJ/s

Now to maintain the temperature of the room at steady state, the steam radiator should input the same amount of heat into the room as lost by the window.

Q_loss = Q_input

Now the heat is produced by the steam radiator by condensing the steam and transferring the latent heat of vaporisation.

Q_input = m_steam*L

The steam condenses at 125 ^oC.

Using steam table, the latent heat of vaporization of steam at 125 ^oC; L = h_fg = 2188.9 KJ/kg

Note : It is better to use interpolation to get the exact value of h_fg at exactly 125 ^oC.

Q_loss = m_steam*L

4.2162 = m_steam*2188.9

m_steam = 1.9262*10^-3 kg/s

Now, 1 Kg/s = (1000 g)/ (1/60 min) = (60 * 1000) g/min

m_steam = 1.9262*10^-3*60000 g/min

m_steam = 115.57 g/min