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…
moist air,bottom )=0.0095 kg/(kg dry air)*0.0118 (kg dry air)/sec=0.000112 kg/s
After the mass flow rate of dry air was found, the enthalpy value for the moist air was calculated by multiplying the enthalpy value for dry air by the mass flow rate of dry air. An example is shown below for Trial 1. h_( moist air,bottom)=47 kJ/(kg dry air)*0.0118 (kg dry air)/sec=0.556 kJ/s=0.556 kW
All of the values found from the psychrometric chart, calculated mass flow rates, and enthalpies of the moist air for each trial are presented in Table 3.
Table 3. Psychrometric Chart Readings for Air at the Base of the Tower
Trial Humidity
(kg/kg of dry air) Humid Volume (m^3/kg of dry air) Enthalpy of Dry Air
(kJ/kg of dry air) Mass Flow Rate Dry Air
(kg dry air/s) Mass Flow Rate of Moist Air
(kg H20/s) Enthalpy of Moist Air
(kW)
1 0.0095 0.85 47 0.0118 0.000112 0.556
2 0.0095 0.85 47 0.0106 0.000101 0.498
3 0.0098 0.86 54 0.0119 0.000117 0.644
4 0.0100 0.85 46 0.0118 0.000118 0.544
5 0.0110 0.86 57 0.0118 0.000118 0.613
The same calculations for the air at the base of the tower were performed for the air leaving the tower at the top. Thus, Table 4 summarizes the psychrometric chart values, the mass flow rates, and the enthalpy values of the moist air for each trial.
Table 4. Psychrometric Chart Readings for Air at the Top of the Tower
Trial Humidity
(kg/kg of dry air) Humid Volume (m^3/kg of dry air) Enthalpy of Dry Air
(kJ/kg of dry air) Mass Flow Rate Dry Air
(kg dry air/s) Mass Flow Rate of Moist Air
(kg H20/s) Enthalpy of Moist Air
(kW)
1 0.019 0.88 75 0.0114 0.000219 0.857
2 0.021 0.88 80 0.0090 0.000210 0.818
3 0.022 0.89 90 0.0103 0.000256 1.037
4 0.016 0.86 65 0.0101 0.000188 0.764
5 0.015 0.87 68 0.0093 0.000159 0.723
With the enthalpy values calculated for air at the top and bottom of the tower, the change in enthalpy of the air was calculated by subtracting the enthalpy values, as shown below for Trial 1.
〖∆h〗_(moist air)=0.857 kW-0.556 kW=0.301 kW
The change in enthalpy of the moist air for each trial is present in Table 5.
Table 5. Change in Enthalpy of Air over the Tower
Trial Δ Enthalpy of Air (kW)
1 0.301
2 0.321
3 0.393
4 0.220
5 0.110
For the water being cooled in the tower, the mass flow rate was determined to be a constant 21 grams per minute for all of the trials. Thus, the change in enthalpy of water was easily calculated using the specific heat capacity of water, 4.186 J/(g*⁰C) , as shown below for Trial 1.
〖∆h〗_water=4.186 J/(g*⁰C)*(〖27.3〗^0 C-〖24.5〗^0 C)*21 g/s*(1 kJ)/(1000 J)=0.246 kJ/s=0.246 kW
The change in enthalpy of the water for each trial is presented below in Table 6.
Table 6. Change in Enthalpy of Water over the Tower
Trial Mass Flow Rate (g/sec) Change in Temperature (⁰C) Δ Enthalpy of Water (kW)
1 21 2.8 0.246
2 2.5 0.219
3 1.7 0.149
4 1.6 0.140
5 2.0 0.176
After the enthalpy changes were calculated for the air and water in the system, the energy balance of the system was analyzed. The percent difference of the calculated enthalpy values was determined, as shown below for Trial 1.
% difference=|0.246 kW-0.301 kW|/[(0.246 kW+0.301 kW)/2] *100%=18.3%
Table 7 summarizes the change in enthalpy values for the water and air and the percent difference of the values for each trial.
Table 7. Percent Difference Calculations for Overall Energy Balance
Trial Δ Enthalpy of Air (kW) Δ Enthalpy of Water (kW) % Difference
1 0.301 0.246 18.3
2 0.321 0.219 31.5
3 0.393 0.149 62.0
4 0.220 0.140 36.2
5 0.110 0.176 59.7
From Table 7, the effect of each of the variables being tested on the enthalpy values, and in turn the temperature changes, can be observed. Trial 1, the base trial, has the largest change in temperature of the water at 2.8⁰C. Additionally, this trial has the best closure of the energy balance by approximately 15%. Trial 2, which had restricted air flow, had the second highest temperature change with the water cooling 2.5⁰C. This trial also had the second best closure of the energy balance. Trial 3, which had heated air, had the worst closure of all of the trials, and the second worst temperature change. Thus, a correlation can be seen regarding the inlet temperatures of the air and water and the overall cooling of the water in the tower. This data illustrates that the most efficient cooling occurs when there is a larger difference between the water and air inlet temperatures.
In addition the energy balance, the mass balance of the water for the overall system was performed. First, the velocity of the water leaving the make-up tank needs to be calculated. The cross-sectional area of the make-up tank is 0.01 m^2. Using the cross-sectional area of the tank, change in height of the tank, and the time measurements taken, the velocity of water leaving the make-up tank was calculated, as shown below for Trial 1. v_(water leaving make-up)=(0.47 cm)/(5 mins)*(1 m)/(100 cm)*(1 min)/(60 sec)=0.0000157 m/s
Then, using the calculated velocity, the mass flow rate can be calculated by multiplying the velocity by the cross-sectional area of the tank and the density of water. m ̇_( water leaving make-up)=0.0000157 m/s*0.01 m^2*997 kg/m^3 =0.000156 (kg H2O)/sec
Table 8 summarizes the velocity and mass flow rate of the water leaving the make-up tank.
Table 8. Data for Water Leaving the Make-Up Tank
Trial Velocity (m/s) Mass Flow Rate (kg H2O/sec)
1 0.0000157 0.000157
2 0.0000213 0.000212
3 0.0000203 0.000202
4 0.0000164 0.000164
5 0.0000188 0.000187
Since the mass flow rate of the moist air at the top and bottom of the tower are known, the amount of water gained by air during the cooling process can be calculated by subtracting the two mass flow rates. An example is shown below for Trial 1. m ̇_( water gained by air)=0.000219 kg/s-0.000112 kg/s=0.000107 kg/s
Table 9.
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
78.0
Analyzing Table 10, the best closure for the mass balance is Trial 3, which was the worst of the trials regarding the cooling efficiency. However, Trial 1 has the next best mass balance closure, and it is only 0.6% higher than Trial 3. Trial 5, which had heated and restricted air flow, has the lowest closure Thus, this data does not show any relationship between the inlet temperatures of the water and air and the closure of the mass balance. Trials 2 and 5 do indicate that with restricted air flow, the closure of the water mass balance decreases. Although, Trial 4 does not conform to this pattern. It may be an outlier or indicate that there is no such pattern.
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…
moist air,bottom )=0.0095 kg/(kg dry air)*0.0118 (kg dry air)/sec=0.000112 kg/s
After the mass flow rate of dry air was found, the enthalpy value for the moist air was calculated by multiplying the enthalpy value for dry air by the mass flow rate of dry air. An example is shown below for Trial 1. h_( moist air,bottom)=47 kJ/(kg dry air)*0.0118 (kg dry air)/sec=0.556 kJ/s=0.556 kW
All of the values found from the psychrometric chart, calculated mass flow rates, and enthalpies of the moist air for each trial are presented in Table 3.
Table 3. Psychrometric Chart Readings for Air at the Base of the Tower
Trial Humidity
(kg/kg of dry air) Humid Volume (m^3/kg of dry air) Enthalpy of Dry Air
(kJ/kg of dry air) Mass Flow Rate Dry Air
(kg dry air/s) Mass Flow Rate of Moist Air
(kg H20/s) Enthalpy of Moist Air
(kW)
1 0.0095 0.85 47 0.0118 0.000112 0.556
2 0.0095 0.85 47 0.0106 0.000101 0.498
3 0.0098 0.86 54 0.0119 0.000117 0.644
4 0.0100 0.85 46 0.0118 0.000118 0.544
5 0.0110 0.86 57 0.0118 0.000118 0.613
The same calculations for the air at the base of the tower were performed for the air leaving the tower at the top. Thus, Table 4 summarizes the psychrometric chart values, the mass flow rates, and the enthalpy values of the moist air for each trial.
Table 4. Psychrometric Chart Readings for Air at the Top of the Tower
Trial Humidity
(kg/kg of dry air) Humid Volume (m^3/kg of dry air) Enthalpy of Dry Air
(kJ/kg of dry air) Mass Flow Rate Dry Air
(kg dry air/s) Mass Flow Rate of Moist Air
(kg H20/s) Enthalpy of Moist Air
(kW)
1 0.019 0.88 75 0.0114 0.000219 0.857
2 0.021 0.88 80 0.0090 0.000210 0.818
3 0.022 0.89 90 0.0103 0.000256 1.037
4 0.016 0.86 65 0.0101 0.000188 0.764
5 0.015 0.87 68 0.0093 0.000159 0.723
With the enthalpy values calculated for air at the top and bottom of the tower, the change in enthalpy of the air was calculated by subtracting the enthalpy values, as shown below for Trial 1.
〖∆h〗_(moist air)=0.857 kW-0.556 kW=0.301 kW
The change in enthalpy of the moist air for each trial is present in Table 5.
Table 5. Change in Enthalpy of Air over the Tower
Trial Δ Enthalpy of Air (kW)
1 0.301
2 0.321
3 0.393
4 0.220
5 0.110
For the water being cooled in the tower, the mass flow rate was determined to be a constant 21 grams per minute for all of the trials. Thus, the change in enthalpy of water was easily calculated using the specific heat capacity of water, 4.186 J/(g*⁰C) , as shown below for Trial 1.
〖∆h〗_water=4.186 J/(g*⁰C)*(〖27.3〗^0 C-〖24.5〗^0 C)*21 g/s*(1 kJ)/(1000 J)=0.246 kJ/s=0.246 kW
The change in enthalpy of the water for each trial is presented below in Table 6.
Table 6. Change in Enthalpy of Water over the Tower
Trial Mass Flow Rate (g/sec) Change in Temperature (⁰C) Δ Enthalpy of Water (kW)
1 21 2.8 0.246
2 2.5 0.219
3 1.7 0.149
4 1.6 0.140
5 2.0 0.176
After the enthalpy changes were calculated for the air and water in the system, the energy balance of the system was analyzed. The percent difference of the calculated enthalpy values was determined, as shown below for Trial 1.
% difference=|0.246 kW-0.301 kW|/[(0.246 kW+0.301 kW)/2] *100%=18.3%
Table 7 summarizes the change in enthalpy values for the water and air and the percent difference of the values for each trial.
Table 7. Percent Difference Calculations for Overall Energy Balance
Trial Δ Enthalpy of Air (kW) Δ Enthalpy of Water (kW) % Difference
1 0.301 0.246 18.3
2 0.321 0.219 31.5
3 0.393 0.149 62.0
4 0.220 0.140 36.2
5 0.110 0.176 59.7
From Table 7, the effect of each of the variables being tested on the enthalpy values, and in turn the temperature changes, can be observed. Trial 1, the base trial, has the largest change in temperature of the water at 2.8⁰C. Additionally, this trial has the best closure of the energy balance by approximately 15%. Trial 2, which had restricted air flow, had the second highest temperature change with the water cooling 2.5⁰C. This trial also had the second best closure of the energy balance. Trial 3, which had heated air, had the worst closure of all of the trials, and the second worst temperature change. Thus, a correlation can be seen regarding the inlet temperatures of the air and water and the overall cooling of the water in the tower. This data illustrates that the most efficient cooling occurs when there is a larger difference between the water and air inlet temperatures.
In addition the energy balance, the mass balance of the water for the overall system was performed. First, the velocity of the water leaving the make-up tank needs to be calculated. The cross-sectional area of the make-up tank is 0.01 m^2. Using the cross-sectional area of the tank, change in height of the tank, and the time measurements taken, the velocity of water leaving the make-up tank was calculated, as shown below for Trial 1. v_(water leaving make-up)=(0.47 cm)/(5 mins)*(1 m)/(100 cm)*(1 min)/(60 sec)=0.0000157 m/s
Then, using the calculated velocity, the mass flow rate can be calculated by multiplying the velocity by the cross-sectional area of the tank and the density of water. m ̇_( water leaving make-up)=0.0000157 m/s*0.01 m^2*997 kg/m^3 =0.000156 (kg H2O)/sec
Table 8 summarizes the velocity and mass flow rate of the water leaving the make-up tank.
Table 8. Data for Water Leaving the Make-Up Tank
Trial Velocity (m/s) Mass Flow Rate (kg H2O/sec)
1 0.0000157 0.000157
2 0.0000213 0.000212
3 0.0000203 0.000202
4 0.0000164 0.000164
5 0.0000188 0.000187
Since the mass flow rate of the moist air at the top and bottom of the tower are known, the amount of water gained by air during the cooling process can be calculated by subtracting the two mass flow rates. An example is shown below for Trial 1. m ̇_( water gained by air)=0.000219 kg/s-0.000112 kg/s=0.000107 kg/s
Table 9.
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
78.0
Analyzing Table 10, the best closure for the mass balance is Trial 3, which was the worst of the trials regarding the cooling efficiency. However, Trial 1 has the next best mass balance closure, and it is only 0.6% higher than Trial 3. Trial 5, which had heated and restricted air flow, has the lowest closure Thus, this data does not show any relationship between the inlet temperatures of the water and air and the closure of the mass balance. Trials 2 and 5 do indicate that with restricted air flow, the closure of the water mass balance decreases. Although, Trial 4 does not conform to this pattern. It may be an outlier or indicate that there is no such pattern.