Page 120 - Demo

%u062c%u0645%u064a%u0639 %u0627%u0644%u062d%u0642%u0648%u0642 %u0645%u062d%u0641%u0648%u0638%u0629 %u0640 %u0627%u0625%u0644%u0639%u062a%u062f%u0627%u0621 %u0639%u0649%u0644 %u062d%u0642 %u0627%u0645%u0644%u0624%u0644%u0641 %u0628%u0627%u0644%u0646%u0633%u062e %u0623%u0648 %u0627%u0644%u0637%u0628%u0627%u0639%u0629 %u064a%u0639%u0631%u0636 %u0641%u0627%u0639%u0644%u0647 %u0644%u0644%u0645%u0633%u0627%u0626%u0644%u0629 %u0627%u0644%u0642%u0627%u0646%u0648%u0646%u064a%u0629120- 118-cargo unloading efficiency = 1/ (85/100 + 15/25) %u00d7 100% = 69%production capacity of hoods = number of hoods x theoretical capacity of each hood x maximum capacity that can be achieved x cargo unloading efficiency.Example: If you know that the bulk wheat imported into the country is mechanically unloaded from vessel to grain silos by means of three air suction or hoods, each theoretical capacity of 200 tons / hour, and a conveyor belt with a theoretical capacity of 400 tons / hour, and the largest capacity that can be achieved does not exceed 90% of designed theoretical capacity, and unloading efficiency decreases as the grain level decreases inside storage holds of the ship. It is 100% when discharging or unloading 80% of volume of the ship%u201fs cargo, while it decreases to 30% for unloading the rest of the cargo, which is 20%. Required:1 -Calculate cargo unloading efficiency?2 -Calculate possible production capacity for hoods?3 -If you know that actual (production) capacity of hoods is equivalent to 195 tons/hour, what do you think the reasons for this lower capacity than possible production capacity of hoods?4 -If you know that daily operating time at grain silo berth is 18 hours per day, and reduction factor in operating time is equivalent to 10%. What is the possible productivity of grain silo berth / per day.5 -If you know that berth occupancy factor is 80% after deducting holidays and vacations equivalent to 45 days. What is the possible productivity of grain berth / annually?6- If we assume that daily quantity that grain silos administration undertakes to unload by the General Authority for Ports and by shipping companies and terms of the import tender is 2000 tons / day, which is less than the amount of daily productivity for actual performance, what is your justification for this situation if you know that the delay fine is 5000 dollars For each day of delay after the expiry of specified period for unloading the ship, while you will receive $2,500 for each day submitted on the set date for unloading.1- cargo unloading efficiency = 1 / (80/100+20/30) x 100% = appro. 68%2 -production capacity of the hoods = number of hoods x theoretical power of each hood x the maximum capacity that can be achieved x cargo unloading efficiency.= 3 x200 x90% x68% = 367.23 -By comparing possible productivity (367.2tons / hour) and actual productivity of grain silos (195 tons / hour), we find that actual productivity is equivalent to only 53% of the possible productivity after taking into account the coefficients of decrease from theoretical capacity with which machines are designed, and this percentage reflects the extent of significant decrease .The actual performance is less than the achievable performance, which may be attributed to the presence of some technical and administrative deficiencies related to installation, operation, maintenance and management of machines.