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A barometric condenser is connected to a single-effect evaporator. The flow rate of cooling water used in the barometric condenser is 5000 kg/h. The inlet and outlet temperatures of cooling water are 20°C and 50°C, respectively. The enthalpy of the vapor leaving from evaporator is 2000 kJ/kg. Find the amount of the vapor leaving from the evaporator. Cpwater=4.18 kJ/kg°C. 

          A barometric condenser is connected to a single-effect evaporator. The flow rate of cooling water used in the barometric condenser is 5000 kg/h. The inlet and outlet temperatures of cooling water are 20°C and 50°C, respectively. The enthalpy of the vapor leaving from evaporator is 2000 kJ/kg. Find the amount of the vapor leaving from the evaporator. Cpwater=4.18 kJ/kg°C.
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A barometric condenser is connected to a single-effect evaporator. The flow rate of cooling water used in the barometric condenser is 5000 kg/h. The inlet and outlet temperatures of cooling water are 20°C and 50°C, respectively. The enthalpy of the vapor leaving from evaporator is 2000 kJ/kg. Find the amount of the vapor leaving from the evaporator. Cpwater=4.18 kJ/kg°C.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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A barometric condenser is connected to a single-effect evaporator. The flow rate of cooling water used in the barometric condenser is 5000 kg/h. The inlet and outlet temperatures of cooling water are 20°C and 50°C, respectively. The enthalpy of the vapor leaving from evaporator is 2000 kJ/kg. Find the amount of the vapor leaving from the evaporator. Cpwater-4.18 kJ/kg°C. A barometrie condenser is connccted to a single-effect evaporator.The flow rate of cooling water used in the barometric condenser is 5000 kg/h.The inlet and outlet temperatures of cooling water are 20C and 50C.respcctively.The enthalpy of the vapor leaving from evaporator is 2000 kJ/kg.Find the amount of the vapor leaving from the evaporator.Cpwater-4.18 kJ/kgC
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Transcript

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00:01 Hello students, in this question as there is a constant pressure so we have pv is equal to rt so now substituting the given values we have 6 .02 into 3 .8 will be equal to m into 0 .287 into 309 .2 now we get m that is equal to 6 .02 into 3 .8 upon 0 .287 into 309 .2 after solving this we get 0 .257 kg per minute or 15 .42 kg per now we have mass of air removed divided by mass flow rate multiplied by 10 ,000 so now substituting values 15 .42 upon 20000 into 10 000 after solving this we get mass flow rate mass flow mass of air will be 7 .71 kg now for part b mass of stream condensed in air per minute pressure value of 29 degree in steam that is equal to 4 .0306 kilopascal now we have value g that is 34 .984 meter cube per kg so we can find p that will be equal to 6 .02 minus 4 .0306 so this will be equal to 1 .9894 kilopascal now using p v is equal to m r t so we have v that is equal to 15 .42 into 0 .287 into 302 divided by 1 .9894 now after solving we get v is 671 .8171 meter cube per minute…
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