Dry Bulb Temperature Lab Report
To begin the analysis, the properties of the air at the base of the tower were examined. The wet and dry bulb temperatures of the air at the inlet were used to find the humidity, humid volume, and enthalpy of dry air. For example, in the first trial the dry bulb temperature was 22.5⁰C and the wet bulb was 16.5⁰C, and the properties found from the psychrometric chart are listed below.
H_bottom=0.0095 kg/(kg dry air)
V_bottom=0.85 m^3/(kg dry air) h_bottom=47 kg/(kg dry air)
Then, the cross-sectional area of the tower outlet of air was found to be 0.005 m^2. With the values from the psychrometric chart and cross-sectional area, the mass flow rates of the dry air and moist air were calculated, as shown below for Trial 1. m ̇_( dry air,bottom)=(0.005 m^2*2.01 m/s)/(0.85 m^3/(kg dry air))=0.0118 (kg dry air)/sec m ̇_( …show more content…
Rate of Saturation of Air
Trial Mass Flow Rate of Moist Air at Inlet
(kg H20/s) Mass Flow Rate of Moist Air at Outlet
(kg H20/s) Mass Flow Rate (kg/s)
1 0.000112 0.000219 0.000107
2 0.000101 0.000210 0.000109
3 0.000117 0.000256 0.000139
4 0.000118 0.000188 0.000689
5 0.000118 0.000159 0.000412
With the mass flow rate of water leaving the make-up tank and water gained by air calculated, the two values was compared to observe the closure of the mass balance. The percent difference was calculated for each trial. An example for Trial 1 is shown below.
% difference=|0.000157 kg/s-0.000107 kg/s|/[(0.000157 kg/s+0.000107 kg/s)/2] *100%=31.9%
Table 10 summarizes the mass flow rates of the water leaving the make-up tank and the water gained by the air and the percent difference between the values.
Table 10. Water Mass Balance Percent Difference Calculations
Trial Mass Flow Rate Leaving the Tank (kg/s) Mass Flow Rate Gained in the Air (kg/s) % Difference
1 0.000157 0.000107 31.9
2 0.000212 0.000109 48.6
3 0.000202 0.000139 31.3
4 0.000164 0.000689 57.4
5 0.000187 0.000412
H_bottom=0.0095 kg/(kg dry air)
V_bottom=0.85 m^3/(kg dry air) h_bottom=47 kg/(kg dry air)
Then, the cross-sectional area of the tower outlet of air was found to be 0.005 m^2. With the values from the psychrometric chart and cross-sectional area, the mass flow rates of the dry air and moist air were calculated, as shown below for Trial 1. m ̇_( dry air,bottom)=(0.005 m^2*2.01 m/s)/(0.85 m^3/(kg dry air))=0.0118 (kg dry air)/sec m ̇_( …show more content…
Rate of Saturation of Air
Trial Mass Flow Rate of Moist Air at Inlet
(kg H20/s) Mass Flow Rate of Moist Air at Outlet
(kg H20/s) Mass Flow Rate (kg/s)
1 0.000112 0.000219 0.000107
2 0.000101 0.000210 0.000109
3 0.000117 0.000256 0.000139
4 0.000118 0.000188 0.000689
5 0.000118 0.000159 0.000412
With the mass flow rate of water leaving the make-up tank and water gained by air calculated, the two values was compared to observe the closure of the mass balance. The percent difference was calculated for each trial. An example for Trial 1 is shown below.
% difference=|0.000157 kg/s-0.000107 kg/s|/[(0.000157 kg/s+0.000107 kg/s)/2] *100%=31.9%
Table 10 summarizes the mass flow rates of the water leaving the make-up tank and the water gained by the air and the percent difference between the values.
Table 10. Water Mass Balance Percent Difference Calculations
Trial Mass Flow Rate Leaving the Tank (kg/s) Mass Flow Rate Gained in the Air (kg/s) % Difference
1 0.000157 0.000107 31.9
2 0.000212 0.000109 48.6
3 0.000202 0.000139 31.3
4 0.000164 0.000689 57.4
5 0.000187 0.000412