Answer:
0.5°c
Explanation:
Humidity ratio by mass can be expressed as
the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air
Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.
Humidity ratio expressed by mass:
x = mw / ma (1)
where
x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)
mw = mass of water vapor (kg, lb)
ma = mass of dry air (kg, lb)
It can be as:
x = 0.005 (100) / [(100 - 100)]
x = 0.005 x 100 / (100 - 100)
x = 0.005 x 100 / 0
x = 0.5°c
So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c
Answer:
1.96 kg/s.
Explanation:
So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;
=> Superheated water vapor at a pressure = 20 MPa,
=> temperature = 500°C,
=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."
=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."
K1 = 3241.18, k2 = 93.28 and 2725.47.
Therefore, m1 + m2= m3.
10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.
=> 1.96 kg/s.
