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A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)
Correct answer is between '30,32'. Can you explain this answer?
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Most Upvoted Answer
A pump with an efficiency of 80% is used to draw groundwater from a we...
Q = discharge required by crop
A = 108 hac
Δ = 1.44 m
B = 120 days
∴
Water required to be pumped = 0.15/0.8 = 0.1875m3/s
Power required by pump = γQH/0.8
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A pump with an efficiency of 80% is used to draw groundwater from a we...
To calculate the minimum required horsepower (h.p.) of the pump, we need to consider several factors such as the efficiency of the pump, the water application efficiency in the field, the area of the field, the base period and delta for the crop, and the lowest level of water in the well.
Given information:
Efficiency of the pump = 80%
Area of the field = 108 hectares
Base period for the crop = 120 days
Delta for the crop = 144 cm
Water application efficiency in the field = 80%
Lowest level of water in the well = 10 m
Let's calculate the total water requirement for the field:
Water requirement = Area of the field x Delta
Water requirement = 108 hectares x 144 cm
Since 1 hectare = 10,000 m² and 1 cm = 0.01 m, we can convert the units as follows:
Water requirement = 108 x 10,000 m² x 144 x 0.01 m
Now, let's calculate the total volume of water required:
Total volume of water = Water requirement x Water application efficiency
Total volume of water = (108 x 10,000 m² x 144 x 0.01 m) x 80%
Since the unit weight of water is given as 9810 N/m³, we can calculate the weight of water:
Weight of water = Total volume of water x Unit weight of water
Weight of water = (108 x 10,000 m² x 144 x 0.01 m) x 80% x 9810 N/m³
To convert the weight of water into work done per unit time (power), we need to divide it by the time taken (base period) in seconds:
Power = (Weight of water / Base period) x g
Power = [(108 x 10,000 m² x 144 x 0.01 m) x 80% x 9810 N/m³] / (120 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute)
To convert power into horsepower (h.p.), we need to divide it by 746 W/h.p.:
Minimum required horsepower (h.p.) = Power / 746 W/h.p.
Now, let's calculate the minimum required horsepower:
Minimum required horsepower = [(108 x 10,000 m² x 144 x 0.01 m) x 80% x 9810 N/m³] / (120 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute x 746 W/h.p.)
Calculating the above expression, we find that the minimum required horsepower is between 30 and 32.
Given information:
Efficiency of the pump = 80%
Area of the field = 108 hectares
Base period for the crop = 120 days
Delta for the crop = 144 cm
Water application efficiency in the field = 80%
Lowest level of water in the well = 10 m
Let's calculate the total water requirement for the field:
Water requirement = Area of the field x Delta
Water requirement = 108 hectares x 144 cm
Since 1 hectare = 10,000 m² and 1 cm = 0.01 m, we can convert the units as follows:
Water requirement = 108 x 10,000 m² x 144 x 0.01 m
Now, let's calculate the total volume of water required:
Total volume of water = Water requirement x Water application efficiency
Total volume of water = (108 x 10,000 m² x 144 x 0.01 m) x 80%
Since the unit weight of water is given as 9810 N/m³, we can calculate the weight of water:
Weight of water = Total volume of water x Unit weight of water
Weight of water = (108 x 10,000 m² x 144 x 0.01 m) x 80% x 9810 N/m³
To convert the weight of water into work done per unit time (power), we need to divide it by the time taken (base period) in seconds:
Power = (Weight of water / Base period) x g
Power = [(108 x 10,000 m² x 144 x 0.01 m) x 80% x 9810 N/m³] / (120 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute)
To convert power into horsepower (h.p.), we need to divide it by 746 W/h.p.:
Minimum required horsepower (h.p.) = Power / 746 W/h.p.
Now, let's calculate the minimum required horsepower:
Minimum required horsepower = [(108 x 10,000 m² x 144 x 0.01 m) x 80% x 9810 N/m³] / (120 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute x 746 W/h.p.)
Calculating the above expression, we find that the minimum required horsepower is between 30 and 32.
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A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer?
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A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer?.
A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is ________ . (round off to two decimal places) (Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m3)Correct answer is between '30,32'. Can you explain this answer?.
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