Fall’04

1.044J, 2.66J, 4.42J

Solutions for HW Set # 3

1.

(a) From 8:00 am to 8:00 pm, there is an average rate of electrical and solar energy into the room (totals 500 watts over the 12 hours). So the room temperature vs. time for the 12 hours period between 8 am and 8 pm is:

(b) To obtain the maximum room temperature, we can define a system that includes both room air and concrete floor slab. The total electrical and solar energy into the room over 12 hours is:

|Q| + |W| = 500 J/sec x 12 hours x 3600 sec/hour = 2.16 x 107 J

Thus the increase of internal energy of both room air and the slab during 12 hours is:

Uair + Uslab = cair mairT + cslab mslabT = |Q| + |W|

As we know

cair = 700 J/kg K, air = 1.22 kg/m3 at 18 oC, Vair = 12.54 m2 x 2.79 m = 35 m3

cslab = 880 J/kg K, slab = 2100 kg/m3, Vslab = 12.54 m2 x 0.1 m = 1.254 m3

So the increased temperature is

T = 2.16 x 107 /(700x1.22x35 + 880x2100x1.254) = 9.2 oC

So the maximum room air temperature is

Tmax = To + T = 18 + 9.2 = 27.2 oC

2. Given: room air remains at 18 oC; Air starts out dry.

Find: the maximum amount of liquid water evaporated to the air

When the air reaches equilibrium, it becomes saturated with water vapor. We need to find out the specific volume of saturated vapor at 18 oC. Given on Page 353 (copies of the water property table: constant temperature table), we know that

 = 77.93 m3/kg at 15 oC

 = 57.79 m3/kg at 20 oC

Interpolate:

So we get:  = 65.85 m3/kg at 18 oC

As we know that the room volume is 35 m3, so the maximum amount of liquid water evaporated to the air is:

Mwater = V/ = (35 m3) / (65.8 m3/kg) = 0.53 kg

3. We can define a system that includes the floor slab, the water, and the air. So the net energy change of the system is

U = |Q| + |W| = 2.16 x 107 J (from problem 1)

Note, |W| is only electrical work. Since the room volume is constant, pdV work is zero.

If we take state 1 as water, floor slab and air at 18 oC, the conditions at 8 am, and take state 2 as the slab and saturated air at 8 pm, there are three processes going on:

1) the water evaporates

2) the slab changes in temperature

3) the air changes in temperature

So we can get

Uevaporate + Uslab + Uair = 2.16 x 107 J

We can look at each process individually. For process 1, we know that

Uevaporate = (2396.1 – 62.99) kJ/kg = 2333.11 kJ/kg at 15 oC

Uevaporate = (2402.9 – 83.95) kJ/kg = 2318.95 kJ/kg at 20 oC

Interpolate to find Uevaporate at 18 oC

U18oC = 2324.6 kJ/kg

Since we have 0.53 kg water evaporated,

Uevaporate = 2324.6 x 0.53 = 1232.05 kJ

Now look at the slab and air

Uslab = cslab mslabT = 880x(2100x1.254) T

Uair = cair mairT = 700x(1.22x35) T

So we can get

Uevaporate + Uslab + Uair = (1.232 x 106) J + (2.3 x 106T) J + (3.0 x 104T) J

= 2.16 x 107 J

T = 8.7 oC

So the room temperature at 8:00 pm is

T = 18 + 8.7 = 26.7 oC

4. When using the night cooling method, the daytime temperature is increased by 9.2 oC, and reaches 27.2 oC, which is not too bad. But if night temperature is not very cool, then increasing of 9 Celsius degrees in the daytime will lead to a high room temperature.

The supplement method to spray liquid water droplets into the air helps to lower room temperature by 0.5 oC. However, increasing the water vapor to saturation is not going to provide comfortable conditions for the room because the room air will be too humid.