follow the steps to complete a mass balance and heat balance.

Lw starts at -26mm. its ends at -2mm with fluctuations ranging from 0mm to 3mm in between.

22 Exercise G - Performing a mass balance and a heat balance across the tower Objective To perform a mass balance and a heat balance on the system. Method To compare the mass of water lost from the sump tank with the mass of water lost through evaporation. To compare the heat gained by the air exiting the tower with the heat input to the system via the water heater. Equipment Required None Previous results obtained using UOP6-MKIl will be used to perform the exercise. Optional Equipment None Theory - Water evaporation - waker woif The mass of water lost from the sump tank in a given time t (secs) can be determined from Mass Balance the change in water level using the readings from the level sensor Lw (mm). Watur leval The coeresponding volume of water can be calculated using the cross sectional aroa of the sumptank: Plan dimensions of tank 0.285m by 0.22m Plan area of tank =0.2850.22=0.0927m2 Therefore volume of water lost in time =Lw0.06271000(m2) Rate of water loss by volume =(Lw0.06271000)/t(m2/5sec) Rate of water loss by mass (Make-up rate) a Pwuw "Lw*0.0627*1000)/t (kkg/s) The mass of water gained by the air can be determined from the chango in Specific Humidity (Humidity Ratio) of the air. Considering a fixed purcel of air, the Specific Humidity SH is the ratio of the mass of water vapour.(me) in the parcel to the lotal mass of air (m2+m, ) where ma is the mass of dry air present In practice the Specic Humidily is approximately equat to the ratio of the mass of water vapour (m.) in the parcel to the mass of dry air (ma) Therefore by using a Psychrometric App or Psychromatric chart to determine the Specific Humidity the change in the mass of water vapour between the iniet and outlet of the tower can be deterthined Mass fiow of air m2 at iniet and outlet can be calculated from =Q2Ou4 From the measurements of Dry bulb Temperature and Relative Humidity at inlet and outlet the increase in the mass of water vapour at the outlet can be caiculated: Increaseinmassofwaterwapout=m1(Outet)=m1(inlet)==(m3SH)atOutlet(m2SH)atInlet In practice a small amount of water may be lost from the system via the arifice at the top of the tower in the focm of small droplets. For this reason the loss of water from the sump tank wil usuaiy be larger than the calculated loss due to the increase in humidity of the ait. Heat Balance Heat input to the system consists of electrical power supplied to the water hoater (PWR) and work done by the water pump (P). However, the small stze of the DC motor driving the water pump on UOPG-MKII means that heat input to the water via the pump is negligible and can be ignored in a larger system heat in put via the pump might be significant and should be ithelided In the sy stem low hamidity air enters at the bose of the tower and high humidity air leaves at the top resulting in a change in enthaipy. The Liquid to Gas fatio L/G of a cooling tower is the ratio between the water and the air: mass fiow rates. Thermodynamics d ctate that the heat removed from the water must be equal to the heat absorbed by the surrounding air. Where L in the Mass flow of Liquid (Water) kg/sec G is the mass flow rate of Gas (Ain) kg/ sec Cp.w is the Specific heat of water k. i /kg/k Cpe is the Specific heat of air kl rkg/k Twe (T1) is the inlet Water Temperature (tep of packing) "C Tw (T2) is the Outlet Water. Temperature (bothom of packing) "C hei is the Enthalpy of the air at the outet (lop of packing) kikg Th is the Enthalpy of the air at the enity (bottom of packing) kuing Analysis of the results requires mass flowraten rather than volume flowrabes that are measured using the PC. To convert to mass Howrates. L=QwPbinG=QaPw In practice a small amount of heat will be transferred from the tower; sump tank elo, to the sumoundings: For this reason power suppied to the heater PWR may bo slightyy largor than the heat transterred in the system. 22 Exercise G - Performing a mass balance and a heat balance across the tower Objective To perform a mass balance and a heat balance on the system. Method To compare the mass of water lost from the sump tank with the mass of water lost through evaporation. To compare the heat gained by the air exiting the tower with the heat input to the system via the water heater. Equipment Required None Previous results obtained using UOP6-MKIl will be used to perform the exercise. Optional Equipment None Theory - Water evaporation - waker woif The mass of water lost from the sump tank in a given time t (secs) can be determined from Mass Balance the change in water level using the readings from the level sensor Lw (mm). Watur leval The coeresponding volume of water can be calculated using the cross sectional aroa of the sumptank: Plan dimensions of tank 0.285m by 0.22m Plan area of tank =0.2850.22=0.0927m2 Therefore volume of water lost in time =Lw0.06271000(m2) Rate of water loss by volume =(Lw0.06271000)/t(m2/5sec) Rate of water loss by mass (Make-up rate) a Pwuw "Lw*0.0627*1000)/t (kkg/s) The mass of water gained by the air can be determined from the chango in Specific Humidity (Humidity Ratio) of the air. Considering a fixed purcel of air, the Specific Humidity SH is the ratio of the mass of water vapour.(me) in the parcel to the lotal mass of air (m2+m, ) where ma is the mass of dry air present In practice the Specic Humidily is approximately equat to the ratio of the mass of water vapour (m.) in the parcel to the mass of dry air (ma) Therefore by using a Psychrometric App or Psychromatric chart to determine the Specific Humidity the change in the mass of water vapour between the iniet and outlet of the tower can be deterthined Mass fiow of air m2 at iniet and outlet can be calculated from =Q2Ou4 From the measurements of Dry bulb Temperature and Relative Humidity at inlet and outlet the increase in the mass of water vapour at the outlet can be caiculated: Increaseinmassofwaterwapout=m1(Outet)=m1(inlet)==(m3SH)atOutlet(m2SH)atInlet In practice a small amount of water may be lost from the system via the arifice at the top of the tower in the focm of small droplets. For this reason the loss of water from the sump tank wil usuaiy be larger than the calculated loss due to the increase in humidity of the ait. Heat Balance Heat input to the system consists of electrical power supplied to the water hoater (PWR) and work done by the water pump (P). However, the small stze of the DC motor driving the water pump on UOPG-MKII means that heat input to the water via the pump is negligible and can be ignored in a larger system heat in put via the pump might be significant and should be ithelided In the sy stem low hamidity air enters at the bose of the tower and high humidity air leaves at the top resulting in a change in enthaipy. The Liquid to Gas fatio L/G of a cooling tower is the ratio between the water and the air: mass fiow rates. Thermodynamics d ctate that the heat removed from the water must be equal to the heat absorbed by the surrounding air. Where L in the Mass flow of Liquid (Water) kg/sec G is the mass flow rate of Gas (Ain) kg/ sec Cp.w is the Specific heat of water k. i /kg/k Cpe is the Specific heat of air kl rkg/k Twe (T1) is the inlet Water Temperature (tep of packing) "C Tw (T2) is the Outlet Water. Temperature (bothom of packing) "C hei is the Enthalpy of the air at the outet (lop of packing) kikg Th is the Enthalpy of the air at the enity (bottom of packing) kuing Analysis of the results requires mass flowraten rather than volume flowrabes that are measured using the PC. To convert to mass Howrates. L=QwPbinG=QaPw In practice a small amount of heat will be transferred from the tower; sump tank elo, to the sumoundings: For this reason power suppied to the heater PWR may bo slightyy largor than the heat transterred in the system