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1 ton = 3.51 kJ/sec 1278 ton = 1278 x 3.51 kJ/sec = 4485.78 kJ/sec from property table for saturated ammonia @ 48 celcius H3 = specific enthalpy vapor= 1470.96 kJ /kg + Cp(5) {because of the 5 degree overheat}=1470.96 kJ /kg + 2.175 (5) = 1481.835 kJ/kg K H4= specific enthalpy vapor= 396.13 kJ /kg (h2) let the flow rate required be m kg/sec m(1481.835-396.13) = 4485.78 m = 4.13167 kg/sec From property table of ammonia @ -25 degrees h2 = 1411.4 kj/KgK h3 = 1481.835 ideal work of compressor = m ( h3 -h2) actual power consumed = m( h3-h2)/0,85 =4.13167 ( 1481.835-396.13)/0.85 =343.68 kJ {due to the isoentropic inefficiency} performance coefficient = cooling effect / compressor work = 4485.78/343.68= 13.0521

1 ton = 3.51 kJ/sec 1278 ton = 1278 x 3.51 kJ/sec = 4485.78 kJ/sec from property table for saturated ammonia @ 48 celcius H3 = specific enthalpy vapor= 1470.96 kJ /kg + Cp(5) {because of the 5 degree overheat}=1470.96 kJ /kg + 2.175 (5) = 1481.835 kJ/kg K H4= specific enthalpy vapor= 396.13 kJ /kg (h2) let the flow rate required be m kg/sec m(1481.835-396.13) = 4485.78 m = 4.13167 kg/sec From property table of ammonia @ -25 degrees h2 = 1411.4 kj/KgK h3 = 1481.835 ideal work of compressor = m ( h3 -h2) actual power consumed = m( h3-h2)/0,85 =4.13167 ( 1481.835-396.13)/0.85 =343.68 kJ {due to the isoentropic inefficiency} performance coefficient = cooling effect / compressor work = 4485.78/343.68= 13.0521

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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this is an asnwer but i want you write it with handwritten .

1 ton = 3.51 kJ/sec 1278 ton = 1278 x 3.51 kJ/sec = 4485.78 kJ/sec from property table for saturated ammonia @ 48 celcius H3 = specific enthalpy vapor= 1470.96 kJ /kg + Cp(5) {because of the 5 degree overheat}=1470.96 kJ /kg + 2.175 (5)= 1481.835 kJ/kg K H4= specific enthalpy vapor= 396.13 kJ /kg (h2) let the flow rate required be m kg/sec m(1481.835-396.13) = 4485.78 m = 4.13167 kg/sec From property table of ammonia @ -25 degrees h2 = 1411.4 kj/KgK h3 = 1481.835 ideal work of compressor = m ( h3 -h2) actual power consumed = m( h3-h2)/0,85 =4.13167 ( 1481.835-396.13)/0.85 =343.68 kJ {due to the isoentropic inefficiency} performance coefficient = cooling effect / compressor work = 4485.78/343.68= 13.0521
Transcribed Image Text:1 ton = 3.51 kJ/sec 1278 ton = 1278 x 3.51 kJ/sec = 4485.78 kJ/sec from property table for saturated ammonia @ 48 celcius H3 = specific enthalpy vapor= 1470.96 kJ /kg + Cp(5) {because of the 5 degree overheat}=1470.96 kJ /kg + 2.175 (5)= 1481.835 kJ/kg K H4= specific enthalpy vapor= 396.13 kJ /kg (h2) let the flow rate required be m kg/sec m(1481.835-396.13) = 4485.78 m = 4.13167 kg/sec From property table of ammonia @ -25 degrees h2 = 1411.4 kj/KgK h3 = 1481.835 ideal work of compressor = m ( h3 -h2) actual power consumed = m( h3-h2)/0,85 =4.13167 ( 1481.835-396.13)/0.85 =343.68 kJ {due to the isoentropic inefficiency} performance coefficient = cooling effect / compressor work = 4485.78/343.68= 13.0521
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