COMMON FISH FARMING CALCULATIONS

Authors:
Koushik Roy

M.F.Sc. Aquaculture (1st class 1st),
B.Sc. Industrial Fish and Fisheries (1st class 1st),
ICAR-JRF, NET and ARS (Mains) Examination Qualified, Senior Research Fellow, ICAR – Central Inland Fisheries Research Institute, (Govt. Of India), Barrackpore

Dr. M.S. Chari

Ph.D.,
Professor, Dept. of Fisheries, College of Agriculture, I.G.K.V. Raipur, Chhattisgarh

Dr. S.R. Gaur

Ph.D.,
Professor and Head, Dept. of Fisheries, College of Agriculture, I.G.K.V. Raipur, Chhattisgarh

First Edition 2015
FOREWORD

The present technical manual is divided into nine chapters, encompassing a wide range of managerial aspects commonly encountered in a fish farm. It is not directly subjective; rather the main strategy is to provide knowledge through problem-solution oriented approach and practical field experiences. The methodologies adopted and discussed in this technical manual are strictly field oriented and region specific. This manual will prove to be quite handy as a quick reference guide for fishery professionals, progressive aqua-famers, fish culture enthusiasts and amateur researchers who are working at the field level.

CONTENTS

CHAPTER CHAPTER NAME PAGE NO.
1. Determining pond areas 1
2. Determining average depth of ponds 6
3. Estimating volume (water capacity) of ponds 7
4. Calculating treatments 12
5. Calculating water requirements, water filling time and flow 15 rates
6. Calculating fish stocks in a pond 19
7. Determining feed conversion ratio (FCR) 21
8. Calculation of feed requirement 23
9. Special treatments 30
10. Appendix – Conversion charts 32
11. Suggested Readings 34
CHAPTER-1: DETERMINING POND AREAS Very few ponds are perfectly shaped squares, rectangles, triangles or circles. Common area formulas may need a little correction.
1. For Square/ Rectangular Shaped Ponds
If a pond is a perfect square or rectangle, the following formula applies:
A = l x w Where A = Area (in square meter)
l = length in meter
w = width in meter
Example #1.1. Measurements show a regular rectangular pond to have dimensions of 120 m length and 80 m breadth. What is the surface area?
120 m
80 m
A = l x w A = 120 x 80
A = 9600square meter (m²).
Convert it into hectares/acre by the following formula:
Area (in hectares) = Area (in square meter) /10000
= 9600/ 10000 = 0.96 ha (hectares).
OR, Area (in acres) = Area (in square meter) /4048 = 9600/ 4048 = 2.37 acres.
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
Example #1.2. An irregular rectangular pond has the following dimensions. What is the area?
120 m
60 m 80 m
150 m
Area = [(l₁ + l₂)/ 2] x [(w₁ + w₂)/ 2]
= [(120+150)/ 2] x [(60 + 80)/ 2]
= (270/ 2) x (140/ 2)
= 135 x 70
= 9450 square meter (m²)
Convert it into hectares/acre by the following formula: Area (in hectares) = Area (in square meter) / 10000
= 9450/ 10000 = 0.945 ha (hectares).
OR, Area (in acres) = Area (in square meter) / 4048
= 9450/ 4048 = 2.33 acres.
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
2. For Triangular Shaped Ponds
Calculating pond areas for triangular-shaped ponds is easy. Just use the following formula:
Area = (a x b)/ 2
Where, a = short side (in meter)
b = long side (in meter)
The above formula used for calculating area works well for any type of triangular pond.
Example #1.3. Calculate the area of a triangular pond with the dimension given below.
a = 80 m
b = 120 m
Area = (a x b)/ 2 = (80 x 120)/ 2
= 9600/ 2
= 4800 square meter (m²)
Convert it into hectares/acre by the following formula:
Area (in hectares) = Area (in square meter) / 10000
= 4800/ 10000 = 0.48 ha (hectares).
OR, Area (in acres) = Area (in square meter) / 4048 = 4800/ 4048 = 1.19 acres.
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
3. For Trapezoidal Shaped Ponds To calculate areas for trapezoid-shaped ponds having four sides and a nearly or exactly 90 degree angle, use the following formula:
Area = [(a + b) x h]/ 2
Where,
h a

b
Example #1.4. Calculate the area of a trapezoid- shaped pond having the dimensions given in the figure below.

h = 40 m a = 60 m

b = 90 m
Area = [(a + b) x h]/ 2
= [(60 + 90) x 40]/ 2
= [150 x 40]/ 2 = 6000/ 2
= 3000 square meter (m²)
Convert it into hectares/acre by the following formula:
Area (in hectares) = Area (in square meter) / 10000
= 3000/ 10000 = 0.3 ha (hectares).
OR, Area (in acres) = Area (in square meter) / 4048
= 3000/ 4048 = 0.74 acres.
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
4. For Circular Shaped Ponds
If a pond is circular or near circular, the following formula applies:
Area = 3.14x (d/ 2)² Where, d = Diameter of the pond (in meters)
Example #1.5. Calculate the area of a circular shaped pond having the diameter of 80 meter.
80 m
Area = 3.14 x (d/ 2)²
= 3.14 x (80/ 2)²
= 3.14 x 40² = 3.14 x 1600
= 5024 square meter (m²)
Convert it into hectares/acre by the following formula:
Area (in hectares) = Area (in square meter) / 10000
= 5024/ 10000 = 0.5 ha (hectares).
OR, Area (in acres) = Area (in square meter) / 4048
= 5024/ 4048 = 1.24 acres.
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
5. Irregularly-Shaped Ponds

For calculating areas for irregularly-shaped or mixed- shaped ponds, divide the pond into regularshaped sections and calculate those areas. Then add the areas of each section to determine the total area for the pond.

CHAPTER-2: DETERMINING AVERAGE DEPTH OF PONDS

Calculating the volume of ponds requires an accurate estimate of the pond’s average depth. Two people are needed to calculate the depth of a pond – one to measure depth and one to record the measurements.

For ponds of 5 acres or less, a minimum of ten measurements is needed. For ponds over 5 acres, a minimum of 20 measurements must be taken. Take the measurements along an Sshaped figure across the pond. Refer to the figure below:

To calculate average depth, total the measurements, and divide by the number of measurements.

A3.5-4.0 meter long 3/4-inchPVC pipe with a 30-45 cm diameter 1-2 cm thick wooden disc/ metal plate at the bottom makes a good depth measuring device. The pipe’s measurements should be graduated in 1 meter increments.

CHAPTER-3: ESTIMATING VOLUME (WATER CAPACITY) OF PONDS
For accurate estimation of pond volume in terms of their water holding capacity, following data will be required.

1. Area of the pond (see Chapter-1).
2. Average depth of the pond (see Chapter-2).
3. Pond Volume = Pond Area x Avg. depth.
4. Conversion of volume (in cubic meter) into Liter.

*NOTE: Only measure the inside length and inside width of the pond for area or volume estimation (i.e.- the watery margin areas of the pond) but NOT the bundh or dyke length and width.

6. For Square/ Rectangular Shaped Ponds
If a pond is a perfect square or rectangle, the following formula applies:
V = l x w x h
Where V = Volume (in cubic meter)
l = length in meter
w = width in meter
h = average depth in meter
Example #3.1. Measurements show a regular rectangular pond to have dimensions of 120 m length and 80 m breadth and 1.5 m depth. What is the volume?
120 m
80 m
Avg. depth = 1.5 m
V = l x w x h
= 120 x 80 x 1.5
= 14400cubic meter (m³). Convert it into litres by the following formula:
Volume (in litres) = Volume (in m³) x 1000
= 14400 x 1000 = 14400000 litres
*NOTE: A 1 m³unit holds 1000 litres of water.₇
Example #3.2. An irregular rectangular pond has the following dimensions. What is the volume?
120 m
60 m 80 m
Avg. depth = 1.5 m
150 m
Volume = [(l₁ + l₂)/ 2] x [(w₁ + w₂)/ 2] x h
= [(120 + 150)/ 2] x [(60 + 80)/ 2] x 1.5
= (270/ 2) x (140/ 2) x 1.5
= 135 x 70 x 1.5
= 14175 cubic meter (m³) Convert it into litres by the following formula:
Volume (in litres) = Volume (in m³) x 1000
= 14175 x 1000 = 14175000 litres
*NOTE: A 1 m³unit holds 1000 litres of water.
7. For Triangular Shaped Ponds
Calculating pond volume for triangular-shaped ponds is easy. Just use the following formula:
Area = [(a x b)/ 2] x h
Where, a = short side (in meter) b = long side (in meter)
h = average depth (in meter)
Example #3.3. Calculate the area of a triangular pond with the dimension given below.
a = 80 m
Average depth = 1.5 m
b = 120 m
Area = [(a x b)/ 2] x h
= [(80 x 120)/ 2] x 1.5
= (9600/ 2) x 1.5
= 7200 cubic meter (m³)
Convert it into litres by the following formula: Volume (in litres) = Volume (in m³) x 1000
= 7200 x 1000 = 7200000 litres
*NOTE: A 1 m³unit holds 1000 litres of water.
8. For Trapezoidal Shaped Ponds
To calculate volume for trapezoid-shaped ponds having four sides and a nearly or exactly 90 degree angle, use the following formula:
Volume = [ (a + b) x h / 2] x d Where, a
h avg. depth = d

b
Example #3.4. Calculate the area of a trapezoid- shaped pond having the dimensions given in the figure below.
h = 40 m a = 60 m
Avg. depth, d = 1.5 m

b = 90 m
Area = [{(a + b) x h}/ 2] x d
= [{(60 + 90) x 40}/ 2] x 1.5
= [(150 x 40)/ 2] x 1.5 = (6000/ 2) x 1.5
= 3000 x 1.5
= 4500 cubic meter (m³)
Convert it into litres by the following formula:
Volume (in litres) = Volume (in m³) x 1000 = 4500 x 1000 = 4500000 litres
*NOTE: A 1 m³unit holds 1000 litres of water.
9. For Circular Shaped Ponds
If a pond is circular or near circular, the following formula applies:
Volume = [3.14 x (d/ 2)²] x h
Where, d = Diameter of the pond (in meters)
h = Average depth
Example #3.5. Calculate the volume of a circular shaped pond having the diameter of 80 meter and an average depth of 1.5 m.
d = 80 m Avg. depth, h = 1.5 m
Volume = [3.14 x (d/ 2)²] x h
= [3.14 x (80/ 2)²] x 1.5
= 3.14 x 40² x 1.5
= 3.14 x 1600 x 1.5 = 7536 cubic meter (m³)
Convert it into litres by the following formula:
Volume (in litres) = Volume (in m³) x 1000
= 7536 x 1000 = 7536000 litres
*NOTE: A 1 m³unit holds 1000 litres of water.
10. Irregularly-Shaped Ponds

For calculating volume for irregularly-shaped or mixed- shaped ponds, divide the pond into regular shaped sections and calculate those volumes. Then add the volume of each section to determine the total volume for the pond.

CHAPTER 4: CALCULATING TREATMENTS 1. Basic Treatment Formula
Most chemical treatments can be calculated by using the following formula:
Amount of chemical needed (in grams) = V (in m³) x CF (=1) x ppm desired x [100/ A.I. (in %)]
Where, V= Volume of pond needing treatment (in m³)
CF= Conversion factor (i.e. 1 gm/m³)
ppm = Desired concentration of chemical needed for water volume being treated
A.I. = Active ingredient of the chemical to be used (in %)
*NOTE: Most chemicals are considered 100% AI. The percent AI is the percentage of the active ingredient of the product used and is found on the product’s label.

Example #4.1. A pond needs to be treated with an herbicide. The pond is 1.2 hectares or 2.965 acres and has an average depth of 1.5 m. The recommended treatment is 2 ppm. The active ingredient is 100%. How much amount of the herbicide is needed for the treatment?

Given,
Pond area = 1.2 ha or 2.965 acres
Average depth of the pond = 1.5 m
Recommended treatment for herbicide = 2 ppm
A.I. of the herbicide = 100% I. Convert the area (in hectares/ acres) into area (in square meter) by the following formula:
Area (in square meter) = Area (in hectares) x 10000
= 1.2 x 10000 = 12000 m² OR, Area (in square meter) = Area (in acres) x 4048
= 2.965 x 4048 = 12002.32 m² 12000 m² II. Calculate the volume of the pond by the following formula:
Volume (in cubic meter) = Area (in square meter) x Average depth (in meter) = 12000 x 1.5 = 18000 m³
III. Apply the treatment formula:
Amount of chemical needed (in grams) = V (in m³)xCF (=1) x ppm desired x [100/ A.I. (in %)]
= 18000 x 1 x 2 x (100/100)
= 18000 x 1 x 2 x 1
= 36000 grams
IV. Convert chemical needed (in grams) into chemical needed (in kilograms) by the following formula:
Chemical needed (in kilograms) = Chemical needed (in grams)/ 1000
= 36000/ 1000 = 36 kg.

CONCLUSION: Hence, 36 kg of herbicide (with 100% A.I.) will be required to treat the pond having 1.2 ha or 2.965 acres area and 1.5 m average depth, in order to obtain 2 ppm concentration.

Example #4.2. A pond needs to be treated with Deltamethrin. The pond is 1.2 hectares or 2.965 acres and has an average depth of 1.5 m. The recommended treatment is 1.5 ppm. The active ingredient is 72%. How much amount of the pesticide is needed for the treatment?

Given,
Pond area = 1.2 ha or 2.965 acres Average depth of the pond = 1.5 m
Recommended treatment for herbicide = 1.5 ppm
A.I. of Deltamethrin (pesticide) = 72%
I. Convert the area (in hectares/ acres) into area (in square meter) by the following formula:
Area (in square meter) = Area (in hectares) x 10000
= 1.2 x 10000 = 12000 m²
OR, Area (in square meter) = Area (in acres) x 4048
= 2.965 x 4048 = 12002.32 m² 12000 m² II. Calculate the volume of the pond by the following formula:
Volume (in cubic meter) = Area (in square meter) x Average depth (in meter)
= 12000 x 1.5 = 18000 m³
III. Apply the treatment formula: Amount of chemical needed (in grams) = V (in m³)xCF (=1) x ppm desired x [100/ A.I. (in %)]
= 18000 x 1 x 1.5 x (100/72)
= 18000 x 1 x 1.5 x 1.39 = 37530 grams
IV. Convert chemical needed (in grams) into chemical needed (in kilograms) by the following formula:
Chemical needed (in kilograms) = Chemical needed (in grams)/ 1000
= 37530/ 1000 = 37.53 kg.

CONCLUSION: Hence, 37.53 kg of pesticide (with 72% A.I.) will be required to treat the pond having 1.2 ha or 2.965 acres area and 1.5 m average depth, in order to obtain 1.5 ppm concentration.

*NOTE: Formalin, Salt, Potassium permanganate, Copper sulphate, Methylene blue used for prophylaxis/ therapy in fish farming is CONSIDERED to have 100% A.I.
CHAPTER 5: CALCULATING WATER REQUIREMENTS, WATER FILLING TIME AND FLOW RATES
POINT TO REMEMBER: A 1 m³ unit holds 1000 litres of water. A 1 hectare-meter pond holds 10000 litres of water. A 1 acre-meter pond holds 4000 litres of water
For determining minimum water flow rate to fill the pond within a fixed time (t, in days) use the following formula:
Flow rate (in litres per minute) = Pond volume (in litres)/ [t (in days) x 24 x 60] Now, convert minutes into days by the following formula:
Filling time (in days) = Filling time (in minutes)/ (60 x 24)
The following examples will illustrate the steps and calculations required to estimate the water requirements for various production units.

Example #5.1. A farmer wants to construct four ponds in the same area and fill all the ponds from one well. The ponds vary in size and the owner wants to be able to fill any pond within7 days. The sizes of the ponds are 6 acres (2.43 ha), 4 acres (1.62 ha), 5.5 (2.23 ha) acres and 2.5acres (1.01 ha). The average water depth in each pond is 5 feet. What flow rate in litres per minute (lpm) is required from the service well to fill any pond in at least 7 days?

I. First determine the volume of water in the largest pond. If the largest pond can be filled in 7 days, then any smaller ponds will fill in 7 days or less. Area of largest pond = 6 acres or 2.43 ha.
a) Convert the area (in hectares/ acres) into area (in square meter) by the following formula:
Area (in square meter) = Area (in hectares) x 10000
= 2.43 x 10000 = 24300 m²
OR, Area (in square meter) = Area (in acres) x 4048
= 6 x 4048 = 24288 m² 24300 m²
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
b) Calculate the volume of the pond by the following formula: Given, average depth required = 5 feet Converting feet into metre:
Depth (in metre) = Depth (in feet) x 0.3
= 5 x 0.3 = 1.5 m
*NOTE: 1 feet is equal to 0.30 metre. Now, Volume (in cubic meter) = Area (in square meter) x Average depth (in meter)
= 24300 x 1.5 = 36450 m³
II. Convert volume (in cubic metre) into litres:
Volume (in litres) = Volume (in m³) x 1000
= 36450 x 1000 = 36450000 litres
*NOTE: A 1 m³ unit holds 1000 litres of water. 1 hectare-meter pond holds 10000 litres of water. 1 acre-meter pond holds 4000 litres of water.
III. Determine the minimum flow rate needed to fill the pond in 7days.
For determining minimum water flow rate to fill the pond(s) within a fixed time (t, in days) use the following formula:
Flow rate (in litres per minute) = Pond volume (in litres)/ [t (in days) x 24 x 60]
= 36450000/ (7 x 24 x 60) = 36450000/ 10080
= 3616.07 3700 litres per minute (lpm)
CONCLUSION: A flow rate of 3700 lpm should be adequate to fill all the ponds within 7 days.

Example #5.2. With the 3700 lpm flow rate from Example #5.1, what would be the filling time in days for the smallest pond of 2.5acres (1.01 ha) and 5 feet average water depth?

I. First determine the volume of water in the pond.
Given, area of the pond = 2.5 acres or 1.01 ha.
a) Convert the area (in hectares/ acres) into area (in square meter) by the following formula:
Area (in square meter) = Area (in hectares) x 10000 = 1.01 x 10000 = 10100 m²
OR, Area (in square meter) = Area (in acres) x 4048
= 2.5 x 4048 = 10120 m² 10100 m²
*NOTE: 1 hectare (ha) land is equal to 10000 m² i.e. - a pond which is 100 m in length and 100 m in breadth. 1 ha area is equivalent to 2.47 acres. 1 acre land is equal to 4048 m².
b) Calculate the volume of the pond by the following formula:
Given, average depth required = 5 feet
Converting feet into metre:
Depth (in metre) = Depth (in feet) x 0.3 = 5 x 0.3 = 1.5 m
*NOTE: 1 foot is equal to 0.30 metre.
Now, Volume (in cubic meter) = Area (in square meter) x Average depth (in meter)
= 10100 x 1.5 = 15150 m³
II. Convert volume (in cubic metre) into litres:
Volume (in litres) = Volume (in m³) x 1000 = 15150 x 1000 = 15150000 litres
*NOTE: A 1 m³unit holds 1000 litres of water. 1 hectare-meter pond holds 10000 litres of water. 1 acre-meter pond holds 4000 litres of water
III. Estimating the filling time of the pond at a given flow rate: Given, flow rate = 3700 lpm (litres per minute)
With a flow rate of Flpm, the filling time (in minutes) can be calculated by the following formula:
Filling time (in minutes) = Pond volume (in litres)/ Flow rate (in lpm)
= 15150000/ 3700
= 4094.594 4095 minutes
IV. Convert minutes to days:
Convert minutes into days by the following formula:
Filling time (in days) = Filling time (in minutes)/ (60 x 24)
= 4095/ (60 x 24)
= 4095/ 1440
= 2.84 days 3 days
CONCLUSION: Almost 3 days will be required to fill the 2.5 acres (or 1.01 ha) pond upto a depth of 5 feet at a flow rate of 3700 lpm.
HOW TO DETERMINE FLOW RATES IN PIPES AT FISH FARM?
Fish culturists often need to adjust the supply of water in a pipe to obtain a desired water exchange or flow rate.
The flow rate through a pipe is easiest to determine by using a container of a known volume. Following steps needs to be followed:

I. A 1 litre container can be used or a larger container of known volume. II. Also needed is a stop watch or watch with a second hand.
III. Turn on the water, then place container under pipe to collect water.
IV. With your watch, determine the time that it takes for the container to fill completely.

V. Then make the following calculation:
Flow rate (in lpm) = [Volume of container (in litres) x 60]/ Total seconds to fill container This procedure should be repeated several times to confirm consistent results and determine an average value.
Example #5.3.A 10 litres container filled up in 1 minute and 20 seconds. What is the flow rate of the supply pipe?
Given, Volume of the container = 10 litres
Time taken to fill container = 1 min 20 secs = 80 seconds
Flow rate (in lpm) = [Volume of container (in litres) x 60]/ Total seconds to fill container
= (10 x 60)/ 80 = 600/ 80
= 7.5 lpm
CONCLUSION: Flow rate of the supply pipe is 7.5 lpm.
CHAPTER 6: CALCULATING FISH STOCKS IN A POND The following situations are related to stocking fish in ponds. Fish stocking rates are usually based on the surface area of water
Example #6.1. A pond is 0.8 hectare in size and 4,500 fish per ha is the desired stocking rate. How many fish are needed?
Given, Pond area = 0.8 ha.
Stocking rate = 4500 fish/ha. Therefore, applying the following formula:
No. fish seed required = Area of pond (in hectares) x Stocking rate (in fish/ hectare) 0.8hax 4,500 fish/ ha
= 3600 fish
CONCLUSION: 3600 fish seeds will be required to stock the pond at required stocking density.
Example #6.2. A fish farmer wants to stock 3600 fish in a pond. The fish he wants weigh 120 kg per 1,000 fish. How many kg of fish should be stocked?
Given, No. of fish seed required = 3600
Total sample weight of 1000 fish = 120 kg
No. of fish in the sample, N = 1000
Therefore, applying the following formula:
Fish seed required (in kg) = No. of fish seed required x [Total weight of N no. of fish (in kg)/ N]
= 3600 x (120/ 1000)
= 3600 x 0.12
= 432 kg
CONCLUSION: 432 kg fish seed will be required to stock the pond at required stocking density.

Example #6.3. A fish buyer wants to sample some fish from his pond to check the number of fish that he ordered. He samples 150 fish that weigh 18 kg. The total kg of fish stocked into his pond was 432 kg. How many fishes were stocked?

Given,
Total kg of fish stocked in pond = 432 kg
Total sample weight of 150 fish = 18 kg
No. of fishes in the sample, N = 150
Therefore, applying the following formula:
Number of fish seed stocked = N x [Total weight of fish seed stocked(in kg)/ Total weight of N no. of fish (in kg)]
= 150 x (432/ 18)
= 150 x 24
= 3600
CONCLUSION: 3600 fishes were stocked in the pond.
CHAPTER-7: DETERMINING FEED CONVERSION RATIO (FCR)

Feed conversion ratios are calculated to determine the cost and efficiency of feeding. It is affected by the quality of feed, size and condition of fish, number of good feeding days related to temperature and water quality, and feeding practices adopted. Ideally feed conversion ratio should be from 1 to 2.

Use the following procedure to determine the feed conversion ratio for your fish. Keep the following records:
1. Amount of feed fed daily. 2. Initial weight (in kg) and number of fish stocked.
3. Estimated or final weight (in kg) of standing crop (fish), which is determined by pond sampling.
Now for determining FCR during a particular period, use the following formula:
FCR = Feed given to fish during the period (in kg)/ [Final fish weight (in kg) – Initial fish weight (in kg)]

POINT TO REMEMBER: Generally under field conditions, 1 million (10 lakh) carp spawn weigh 1.4 kg. 1 lakh carp fry weigh 13 kg. 1000 carp fingerling weigh 15 kg. As a thumb rule, these figures can also be applied for other fishes like catfishes, murrels, perches, etc.

Example #7.1: 8000 fingerlings were stocked in a pond. Initially, 1000 fingerlings weighed 15 kg. Later, the fish were sampled and the average weight of fish was found to be 45 kg per 1000 fingerling. During this time 300 kg of feed were fed. What is the feed conversion ratio?

I. Calculating initial weight of fish stocked Given,
Stocking rate = 8000 fingerlings
Fish seed weight = 15 kg/ 1000 fingerlings
So, total weight of fish seed stocked (initial weight) = 8000 x 15/1000
= 8 x 15 = 120 kg
II. Calculating final weight of fish stocked
Given, Stocking rate = 8000 fingerlings
After sampling, average weight of the sample = 45 kg/ 1000 fingerlings
So, total weight of fish seed after sampling (final weight) = 8000 x 45/1000
= 8 x 45 = 360 kg
III. Calculating FCR
Given,
Amount of feed fed (from stocking upto the day before sampling) = 300 kg Therefore, applying the following formula:
FCR = Feed given to fish during the period (in kg)/ [Final fish weight (in kg) – Initial fish weight (in kg)]
= 300/ (360-120)
= 300/240 = 1.25 CONCLUSION: FCR is 1.25, i.e. - during the period fish consumed an average of 1.25 kg of feed to gain 1 kg in weight.
*NOTE: Ideally feed conversion ratio should be from 1 to 1.5 and must not exceed 2.
CHAPTER-8: CALCULATION OF FEED REQUIREMENT Recommended feeding rates for different culture practices:
1. Nursery culture (Spawn to Fry; 15 days culture): In first week, give feed @4 times (400%) of the initial body weight/day. In second week, feed @8 times (800%) of the initial body weight/day.
2. Rearing culture (Fry to Fingerling; 3 months culture): In first month, give feed @4% of the initial body weight/day. In second month, feed @6% of the initial body weight/day. In third (last) month, feed @8% of the initial body weight/day.
3. Grow-out culture (Fingerling to Table sized; 8-11 months):
Give feed @2-3% of the initial body weight/day.

POINT TO REMEMBER: Generally under field conditions, 1 million (10 lakh) carp spawn weigh 1.4 kg. 1 lakh carp fry weigh 13 kg. 1000 carp fingerling weigh 15 kg. As a thumb rule, these figures can also be applied for other fishes like catfishes, murrels, perches, etc.

Now for calculating feed requirement per day, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
Where, N = Recommended feeding rate (in %)
Now for calculating total feed requirement, use the following formula:
Total feed requirement (in kg) = Food requirement (in kg) x Number of days
8.1. FEED REQUIREMENT FOR NURSERY CULTURE:

Example #8.1: A farmer stocked 8 million spawn in his nursery pond. His target is to produce fry in 2 weeks. Calculate the daily ration and the amount of feed required during this culture. Also calculate the amount of feed ingredients required, if the feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.

I. Calculating initial weight of fish seed stocked
Given, Stocking rate = 8 million
Fish seed weight = 1.4 kg/ 1 million spawn So, total weight of fish seed stocked (initial weight) = 8 x 1.4/1
= 8 x 1.4 = 11.2 kg
*NOTE - 1 million (10 lakh) carp spawn weigh 1.4 kg.
II. Calculating feed requirement (FIRST WEEK)
Recommended feeding rate = 4 times or 400% per day
*NOTE: During nursery culture, in first week, give feed @4 times (400%) of the initial body weight per day. In second week, feed @8 times (800%) of the initial body weight per day.
Now for calculating feed requirement per day, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
= [400/100] x 11.2
= 4 x 11.2 = 44.8 kg per day
Now for calculating total feed requirement, use the following formula:
Total feed requirement (in kg) = Food requirement (in kg) x Number of days
= 44.8 x 7
= 313.6 kg in first week

CONCLUSION: In first week, 44.8 kg feed (22.4 kg rice bran + 22.4 kg oil cake) will be required per day. Total feed requirement is 313.6 kg feed in first week (156.8 kg rice bran + 156.8 kg oil cake).

*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
III. Calculating feed requirement (SECOND WEEK)
Recommended feeding rate = 8 times or 800% per day
*NOTE: During nursery culture, in first week, give feed @4 times (400%) of the initial body weight per day. In second week, feed @8 times (800%) of the initial body weight per day.
Now for calculating daily feed requirement, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
= [800/100] x 11.2
= 8 x 11.2 = 89.6 kg per day
Now for calculating total feed requirement, use the following formula:
Daily ration (in kg) = Food requirement (in kg)/ Number of days
= 89.6 x 7 = 627.2 kg
*NOTE: 1 week has 7 days

CONCLUSION : In second week, 89.6 kg feed (44.8 kg rice bran + 44.8 kg oil cake) will be required per day. Total feed requirement is 627.2 kg feed in second week (313.6 kg rice bran + 313.6 kg oil cake).

*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
IV. Calculating total feed requirement (during the culture period) Total feed required = Total feed requirement (first week) + Total feed requirement (second week)
= 313.6 kg + 627.2 kg = 940.8kg CONCLUSION: The farmer would require a total of 940.8 kg feed (470.4 kg rice bran + 470.4kg oil cake) for growing 8 million spawn in his entire culture period.
*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
8.2. FEED REQUIREMENT FOR REARING CULTURE:

Example #8.2: A farmer stocked 2 lakh fry in his rearing pond. His target is to produce fingerling in 3 months. Calculate the daily ration and the amount of feed required during this culture. Also calculate the amount of feed ingredients required, if the feed is to be made of rice bran and oil cake @1:1 by weight.

I. Calculating initial weight of fish seed stocked Given,
Stocking rate = 2 lakh
Fish seed weight = 13 kg / 1 lakh fry
So, total weight of fish seed stocked (initial weight) = 2 x 13/1
= 2 x 13 = 26 kg
*NOTE - 1 lakh carp fry weigh 13 kg.
II. Calculating feed requirement (FIRST MONTH)

Recommended feeding rate = 4% per day₂₅ *NOTE: In first month, give feed @4% of the initial body weight/day. In second month, feed @6% of the initial body weight/day. In third (last) month, feed @8% of the initial body weight/day.

Now for calculating feed requirement per day, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
= [4/100] x 26
= 0.04 x 26 = 1.04 kg per day
Now for calculating total feed requirement, use the following formula:
Total feed requirement (in kg) = Food requirement (in kg) x Number of days
= 1.04 x 30
= 31.2 kg
*NOTE: 1 month has 30 days

CONCLUSION: In first month, 1.04 kg feed (0.52 kg rice bran + 0.52 kg oil cake) will be required per day. Total feed requirement is 31.2 kg feed in first month (15.6 kg rice bran + 15.6 kg oil cake).

*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
III. Calculating feed requirement (SECOND MONTH)
Recommended feeding rate = 6% per day

*NOTE: During rearing culture, in first month, give feed @4% of the initial body weight/day. In second month, feed @6% of the initial body weight/day. In third (last) month, feed @8% of the initial body weight/day.

Now for calculating feed requirement per day, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
= [6/100] x 26
= 0.06 x 26 = 1.56 kg per day
Now for calculating total feed requirement, use the following formula:
Total feed requirement (in kg) = Food requirement (in kg) x Number of days
= 1.56 x 30 = 46.8 kg
*NOTE: 1 month has 30 days

CONCLUSION : In second month, 1.56 kg feed (0.78 kg rice bran + 0.78 kg oil cake) will be required per day. Total feed requirement is 46.8 kg feed (23.4 kg rice bran + 23.4 kg oil cake).

*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
IV. Calculating feed requirement (THIRD MONTH)
Recommended feeding rate = 8% per day

*NOTE: During rearing culture, in first month, give feed @4% of the initial body weight/day. In second month, feed @6% of the initial body weight/day. In third (last) month, feed @8% of the initial body weight/day.

Now for calculating feed requirement per day, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
= [8/100] x 26 = 0.08 x 26 = 2.08 kg per day
Now for calculating total feed requirement, use the following formula:
Total feed requirement (in kg) = Food requirement (in kg) x Number of days
= 2.08 x 30 = 62.4 kg
*NOTE: 1 month has 30 days.

CONCLUSION : In third (last) month, 2.08 kg feed (1.04 kg rice bran + 1.04 kg oil cake) will be required per day. Total feed requirement is 62.4 kg (31.2 kg rice bran + 31.2 kg oil cake).

*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
V. Calculating total feed requirement (during the culture period) Total feed required = Total feed requirement (first month) + Total feed requirement (second month) + Total feed requirement (third month)
= 31.2 kg + 46.8 kg + 62.4 kg = 140.4 kg
CONCLUSION: The farmer would require a total of 140.4 kg feed (70.2 kg rice bran + 70.2 kg oil cake) for growing 2 lakh fry in his entire culture period.
*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
8.3. FEED REQUIREMENT FOR GROW-OUT CULTURE

Example #8.3: A farmer stocked 8000 fingerling in his rearing pond. His target is to produce table sized fish in 10 months. Calculate the daily ration and the amount of feed required during this culture. Also calculate the amount of feed ingredients required, if the feed is to be made of rice bran and oil cake @1:1 by weight.

I. Calculating initial weight of fish seed stocked Given,
Stocking rate = 8 thousand fingerling
Fish seed weight = 15 kg / 1 thousand fingerling
So, total weight of fish seed stocked (initial weight) = 8 x 15/1
= 8 x 15 = 120 kg *NOTE - 1000 carp fingerling weigh 15 kg.
II. Calculating feed requirement (MONTHLY) Recommended feeding rate = 3% per day
*NOTE: During grow-out culture, give feed @2-3% of the initial body weight/day.
Now for calculating feed requirement per day, use the following formula:
Feed requirement per day (in kg) = [Recommended feeding rate (N) /100] x Total weight of seed stocked in the pond (in kg)
= [3/100] x 26
= 0.03 x 120 = 3.6 kg per day
Now for calculating total feed requirement, use the following formula:
Total feed requirement (in kg) = Food requirement (in kg) x Number of days = 3.6 x 30
= 108 kg per month
*NOTE: 1 month has 30 days
CONCLUSION: In each month, 3.6 kg feed (1.8 kg rice bran + 1.8 kg oil cake) will be required daily. Total monthly feed requirement is 108 kg (54 kg rice bran + 54 kg oil cake).
*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
III. Calculating total feed requirement (during the culture period)
Given,
Culture period = 10 months
Total feed required during the culture period = Monthly feed requirement (in kg) x Culture period (in months)
= 108 kg x 10 = 1080 kg
CONCLUSION: The farmer would require a total of 1080 kg feed (540 kg rice bran + 540 kg oil cake) for growing 8 thousand fingerlings in his entire culture period.
*NOTE: Feed is to be made of finely powdered rice bran and oil cake @1:1 by weight.
CHAPTER-9: SPECIAL TREATMENTS

In this chapter we will only learn about the application of some special chemicals used in fish culture operation viz.- Copper sulphate (CuSO₄), Formaldehyde or Formalin (HCHO) and Potassium permanganate (KMnO₄). Special care is required during their application or treatment calculation.

9.1. COPPER SULPHATE TREATMENT Find out the total alkalinity value (in ppm) of pond water. Divide the total alkalinity (in ppm) of pond water by 100 to get the required concentration of copper sulphate (in ppm).
Now for calculating CuSO₄ requirement (in ppm), use the following formula:
Copper sulphate required (in ppm) = Total Alkalinity (in ppm)/ 100 POINT TO REMEMBER: However in any case the treatment of Copper sulphate should not exceed 2 ppm in ponds. Copper sulphate has 100% A.I.

Example #9.1.A pond needs to be treated with Copper sulphate. The pond is 1.2hectaresor 2.965 acres and has an average depth of 1.5 m. The total alkalinity of pond water is 200 ppm. How much amount of copper sulphate is needed for the treatment?

Given, Total alkalinity of pond water = 200 ppm
Now for calculating CuSO₄ requirement (in ppm), use the following formula:
Copper sulphate required (in ppm) = Total Alkalinity (in ppm)/ 100
= 200/ 100
= 2 ppm
So, a treatment of 2 ppm Copper sulphate will be required in the pond. Follow Example #4.1 of Chapter – 4 for determining the amount of Copper sulphate needed (in grams/ kg) to treat this particular pond in order to obtain 2 ppm concentration.
*NOTE: Copper sulphate has 100% A.I.
9.2. FORMALDEHYDE (FORMALIN) TREATMENT

Treatment of formalin is temperature dependent. Low concentration of formalin is required at higher temperatures. High concentration of formalin is required at lower temperatures. Aeration should always be given in pond or tanks when formalin is applied. POINT TO REMEMBER: Formalin has 100% A.I. As a thumb rule, 1 ml formalin is equivalent to 1 mg of formalin.

1. For pond treatment, use formalin @15-25 ppm depending on temperature.
2. For bath treatment (less than 60 minutes), use formalin @150-250 ppm depending upon temperature.

Follow Example #4.1 of Chapter – 4 for determining the amount of Formalin needed (in gm/kg) to treat a pond in order to obtain a desired concentration.
*NOTE: Generally 1 ml formalin is equivalent to 1 mg in weight. Formalin has 100% A.I.
9.3. POTASSIUM PERMANGANATE TREATMENT Treatment of Potassium permanganate is dependent on organic load present in water. As a thumb rule, add 2-4 ppm (mg per litre) Potassium permanganate in a pond.
To determine the concentration (in ppm) of KMnO₄ needed in a particular pond, a “Demand test” is required. It is given as follows:

1. Take 10 litres of pond water in a container.
2. Add 2 ppm (i.e. – 2 mg per litre) Potassium permanganate into it and wait.
3. If the purple colour persists for 4 hours, then the pond needs 2 ppm KMnO₄. Stop testing.
4. However if the purple colour changes to brownish in less than 4 hours, then further add 2 ppm (i.e. – 2 mg per litre) Potassium permanganate into it and again wait.
5. If the purple colour persists for 4 hours, then the pond needs 4 ppm (2 ppm + 2 ppm) KMnO₄. Stop testing.
6. Again, if the purple colour changes to brownish in less than 4 hours, further add 2 ppm (i.e. – 2 mg per litre) Potassium permanganate into it and again wait.
7. If the purple colour persists for 4 hours, then the pond needs 6 ppm (2 ppm + 2 ppm + 2 ppm) KMnO₄. Stop testing.
8. Again, if the purple colour changes to brownish in less than 4 hours. Stop testing. The pond is unsuitable for Potassium permanganate application. Reduce the organic load of pond in such case by dewatering, desilting and liming the pond.

POINT TO REMEMBER: Do not add Potassium permanganate in a pond greater than 6 ppm. KMnO₄ has 100% A.I.
*NOTE: As a thumb rule, add 2-4 ppm (mg per litre) Potassium permanganate in a pond.
Follow Example #4.1 of Chapter – 4 for determining the amount of Potassium permanganate needed (in gm/kg) to treat a pond in order to obtain a desired concentration. *NOTE: KMnO₄ has 100% A.I.

APPENDIX

Table 1 – Conversion in Volume

From To
cubic cubic feet fluid gallon cubic liter cubic inches (ft³) ounce (gal) centimeter (l) meter

(in³) (fl oz) (cm³) (m³) cubic 1 0.000579 0.5541 0.00433 16.39 0.0164 0.00001 inches

(in³)
cubic feet 1728 1 9575 7.481 0.000283 28.32 0.0283
(ft³)
fluid 1.805 0.00104 1 0.0078 29.57 0.0296 0.00002
ounce
(fl oz)
gallon 231 0.1337 128 1 3785 3.785 0.0038
(gal)
cubic 0.061 0.0000353 0.0338 0.000264 1 0.001 0.000001 centimeter
(cm³)
liter 60.98 0.0353 33.81 0.2642 1000 1 0.001 (l)
cubic 610000 5.31 33800 264.2 1000000 1000 1
meter
(m³)
*NOTE: cubic centimeter (cm³) = milliliter (ml).

Table 2 – Conversion in Weight

From To
ounce (oz) pound (lb) gram (g) kilogram (kg) ounce (oz) 1 0.0625 28.35 0.0284 pound (lb) 16 1 453.6 0.4536

gram (g) 0.0353 0.0022 1 0.001 kilogram (kg) 35.27 2.205 1000 1
Table 3 – Conversion in Length
From To
inches feet (ft) yard (yd) centimeter (cm) meter (m)

(in)
inches (in) 1 0.0833 0.0278 2.540 0.0254
feet (ft) 12 1 0.3333 30.48 0.3048
yard (yd) 36 3 1 91.44 0.9144
centimeter (cm) 0.3937 0.0328 0.0109 1 100 meter (m) 39.37 3.281 1.0936 100 1

Table 4 – Conversion of various volumes to attain one part per million (ppm)

Amount Active Ingredient
2.71 pounds
1.235 grams
1.24 kilograms
0.0283 grams
1 milligram
8.34 pounds
1 gram
0.0038 grams
3.8 grams

Unit of Volume Parts per million acre-feet 1 ppm acre-feet 1 ppm acre-feet 1 ppm cubic feet 1 ppm

liter 1 ppm million gallons 1 ppm cubic meter 1 ppm gallon 1 ppm thousand gallons 1 ppm

Box 1 – Relationship between percent (gm/100 ml), parts per thousand (gm/l) and parts per million (mg/l) concentration solution

1 percent solution = 10 ppt solution
1 percent solution = 10000 ppm solution
1 ppt solution = 1000 ppm solution

Box 2 – Miscellaneous conversion factors

1 acre-foot= 43,560 cubic feet
1 acre-foot= 325,850 gallons
1 acre-foot of water= 2,718,144 pounds
1 cubic foot of water= 62.4pounds
1 gallon of water= 8.34 pounds
1 gallon of water= 3,785 grams
1 liter of water= 1,000 grams
1 fluid ounce= 29.57 grams
1 fluid ounce= 1.043 ounces
Centigrade to Fahrenheit = (C° x 9/5) + 32° Fahrenheit to Centigrade = (F° - 32°) x 5/9

SUGGESTED READINGS

Dorman, L. 1977. Aquaculture producer’s quick reference handbook. Cooperative Extension Program, University of Arkansas. Pub No.MP435-1M-1-07RV.

Jensen, G. 1998. Handbook for common calculations in fin fish aquaculture. Louisiana Agricultural Experiment Station, Louisiana Cooperative Extension Service. Pub No. 8903 (500). http://www.agctr.lsu.edu/wwwac

Piper, R. G., McElwain, I. B., Orme, L. E., McCraren, J. P., Fowler, L. G. and Leonard, J. R. 1982. Fish Hatchery Management. U.S. Department of Interior, Fish and Wildlife Service, Washington, D.C.

United States Fish and Wildlife Service. 1981. Diseases of Hatchery Fish. Superintendent of Documents, U.S. Government Printing Office.
USDA/Soil Conservation Service. 1982. Aquaculture pumping plants for water management. Engineering Technical Note.
Wellborn, T. L. 1987. Catfish Farmer's Handbook. Extension Wildlife and Fisheries Department, Mississippi Cooperative Extension Service, Mississippi State University.