# ME444-Assignment #2-Sol

```ME444-Assignment #2-Sol
September 16, 2014
1. Your classroom has dimensions of 3m x 8m x 12m. On a given day the relative
humidity is 70%, pressure is 0.1 MPa and the temperature is 30C. Calculate
the humidity ratio, the dew point, the mass of air and the mass of the water
vapor. If the water vapor were condensed into liquid, how many liters of
liquid would one expect?
Solution
The volume of the room is:
3𝑚 &times; 8𝑚 &times; 12𝑚 = 288 𝑚2
𝑝𝑣
𝑃𝑔
→ 𝑝𝑣 = 0.7(4.246) = 2.972 𝑘𝑃𝑎
∅=
Dew point is the sat. temperature at =𝑃𝑣 = 2.972 𝑘𝑃𝑎.
We can get the temperature by interpolation as:
2.972 − 2.339
𝑇 − 20
=
3.169 − 2.339 25 − 20
𝑇 = 23.81 &deg;𝐶
Also partial pressure of air can be calculated from 𝑃 = 𝑃𝑣 + 𝑃𝑎 .
𝑃𝑎 = 100 − 2.972 = 97.028 𝑘𝑃𝑎
Humidity ratio is:
𝑃𝑣
𝑤 = 0.622
𝑃𝑎
2.972
𝑤 = 0.622
= 0.0191
97.028
Therefore mass of air can be calculated as below:
𝑃𝑎 𝑉 97.028 (288)
=
= 321 𝑘𝑔
𝑅𝑎 𝑇
0.287(303)
𝑚𝑣 = 𝑤 𝑚𝑎 = 0.0191(321 𝑘𝑔) = 6.13 𝑘𝑔
𝑚𝑎 =
1
ME444-Assignment #2-Sol
September 16, 2014
𝑉𝑣 = 𝑚𝑣 𝑣𝑓 @𝑝𝑣
𝑣𝑓 − 0.001002
2.972 − 2.339
=
3.169 − 2.339 0.001003 − 0.001002
𝑣𝑓 = 0.0010027
𝑘𝑔
𝑉𝑣 = 6.146𝐸 − 3 3 = 6.146 𝐿
𝑚
2. Calculate the theoretical air–fuel ratio for the combustion of octane, C8H18.
Solution
The combustion equation is
C8H18 + 12.5O2 + 12.5(3.76) N2 → 8 CO2 + 9H2O + 47.0N2
The air–fuel ratio on a mole basis is
AF = (12.5 + 47.0)/1= 59.5 kmol air/kmol fuel
The theoretical air–fuel ratio on a mass basis is found by introducing the molecular
mass of the air and fuel.
AF = 59.5(28.97)/114.2= 15.0 kg air/kg fuel
3. Determine the molal analysis of the products of combustion when octane,
C8H18, is burned with 200% theoretical air, and determine the dew point of
the products if the pressure is 0.1 MPa.
Solution
The equation for the combustion of octane with 200% theoretical air is
C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 → 8 CO2 + 9H2O + 12.5O2 + 94.0N2
Total kmols of product = 8 + 9 + 12.5 + 94.0 = 123.5
Molal analysis of products:
CO2 = 8/123.5 = 6.47%
H2O = 9/123.5 = 7.29%
O2 = 12.5/123.5 = 10.12%
N2 = 94/123.5 = 76.12%
100.00%
2
ME444-Assignment #2-Sol
September 16, 2014
The partial pressure of the water is 100(0.0729) = 7.29 kPa, so the saturation
temperature corresponding to this pressure is 39.7 &deg;𝐶, which is also the dew-point
temperature.
3
```