**Example of Ring hammer balancing process –**

**A granulator has 4 rows of hammer, suppose 2 rows have 18 hammers per row and 2 rows have 20 hammers per rows for a total of 76 hammers per set.**

**Note – Quantity of hammers and their weight will vary depending upon size of granulator**

**1. Weight each hammer from a new set individually and accurately. Mark the weight on each hammer. Also list the weights on a sheet of paper.**

Suppose the weights of hammers in kg are –

22.50, 22.30, 22.00, 22.10, 22.80, 22.40, 23.00, 22.10, 22.40,23.10, 23.20, 22.50, 22.80, 23.90,23.40, 22.70, 22.40, 22.80, 23.50, 23.60, 22.70, 23.30, 22.60, 22.60, 23.20, 21.20, 22.00, 21.50, 22.40, 22.10, 21.90, 21.60,22.60, 22.00, 23.50, 22.40, 21.80, 21.60, 22.90, 22.50, 23.50, 23.60, 22.50,22.40, 23.00, 22.00, 23.50, 22.60, 22.80, 22.40, 22.40, 22.20, 23.50, 21.90, 22.60, 23.50, 22.70, 23.00, 21.80, 21.20, 21.60, 22.70, 22.70, 22.60, 22.90, 22.80, 22.90, 22.80, 23.00, 22.80, 22.50, 23.20, 23.00, 21.60, 23.5, 23.00

**2. Arrange the ring hammer in descending order **

23.90, 23.60, 23.60, 23.50, 23.50, 23.50, 23.50, 23.50, 23.50, 23.50, 23.40, 23.30, 23.20, 23.20, 23.20, 23.10, 23.00, 23.00, 23.00, 23.00, 23.00, 23.00, 22.90, 22.90, 22.90, 22.80, 22.80, 22.80, 22.80, 22.80, 22.80, 22.80, 22.70, 22.70, 22.70, 22.70, 22.70, 22.60, 22.60, 22.60, 22.60, 22.60, 22.60, 22.50, 22.50, 22.50, 22.50, 22.50, 22.40, 22.40, 22.40, 22.40, 22.40, 22.40, 22.40, 22.40, 22.30, 22.20, 22.10, 22.10, 22.10, 22.00, 22.00, 22.00, 22.00, 21.90, 21.90, 21.80, 21.80, 21.60, 21.60, 21.60, 21.60, 21.50, 21.20, 21.20

**3. On Sheet of paper Make four column A,B,C &D and arrange the ring hammers from maximum weight to minimum weight as mention below and add the weight of each hammer column wise.**

**4. Opposing rows of hammers ( A and B, C and D must be balanced so that variance in the total weight of each opposing rows of hammers does not exceed the limit value.**

**5. If the sum of opposite rows of hammers are not equal then interchange the ring hammer of opposite rows in such a way that the the sum of opposite rows should be equal or within limit valu**

*e.*

In fig. 2 we can see that the opposite rows C and D ring hammer weight sum are equal so these opposite rows are balanced. But the total weight sum of opposite rows A and B are not equal so we need to interchange the ring hammer between these opposite rows. The process of interchanging the ring hammer is mention below –

**I.** *Find the average of two opposite unbalance rows.*

From fig.2, average weight = (A+B)/2 = (451.5+450.9)/2 = 451.2

**II.** *Find the balancing weight of each unbalance rows*

Balancing Weight of row A = row A total weight – average = 451.5 – 451.2 = 0.3 kg

Balancing Weight Of row B = row B total weight – average = 450.9 – 451.2 = – 0.3 kg

**III.** *Interchange the ring hammer of unbalance opposite rows*

We have to interchange one or more ring hammer from each unbalance rows in such a way that the difference between two unbalance rows hammer weight is equal or approx to balancing weight and choose heavy weight from the row which have positive balanced weight.

**6. After interchanging the weight of unbalanced opposite rows again find the sum of total weight of rows. Finally we get balanced opposite rows.**

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