Revised Conveyor Belt Cal

7/31/2019 Revised Conveyor Belt Cal

1/66

andys BELT CONVEYOR

DESIGN COMPUTATION:

TRANSPORT CAPACITY

QG =

Where:

QG = weight of material per linear meter.

Lv = belt load (TPH)

V = velocity (m/s)

QG =

= 44.44 Kg/m


2/66

I.HORIZONTAL CONVEYOR

#1 , 2 & 5 SEGMENT

Belt type ST400

Belt width 1000mm

Conveyor length 38 m

Carrying idler dia. 108mm

Space between carrying idler 0.3m

Return idler dia (plain) 108mm

Space between idler 1m

Weight of belt per linear meter (QB) = 13.5Kg/m

Weight of carrying idler per linear meter (QC) = 53.3 Kg/m

Weight of return idler per linear meter (QR) = 12.8 Kg/m

Total weight of moving parts per meter = 79.6 Kg/m

CORRECTED LENGTH (LC)

LC = C * L

Where:

C = correction factor = 1.95

L = conveyor length = 38m

LC = 1.95(38) = 74.1m

MOTIONAL RESISTANCE


3/66

a) Empty belt tension (Tx)

Tx = g * (Qb + QRC + QRr )* fx * LC

Where:

g = local gravity = 9.81m/s

fx = coefficient of friction = 0.02

Tx = 9.81(79.6)(0.02)(74.1)

= 1,157.7 Newton

b) Tension to move the load horizontally (TY)

Ty = g * (Qb + QG )* Fy * LC

Where:

QG = convey material = 44.44 Kg/m

Qb = belt weight = 13.5 kg/m

fY = coefficient of friction = 0.03

Ty = 9.81(13.5 + 44.44)(0.03)(74.1)

= 1,262.6 Newton

c) Special resistance (TS )

TS = 0.25(TX + TY)

= 0.25(1,157.7 + 1,262.6)

= 605 Newton

d) Effective tension (Te)


4/66

Te = TX + TYn + Ts

= 1,157.7 + 1,262.6 + 605

= 3,025.51 Newton

e) Slacked side tension to prevent slip (T2)

T2 = K*Te

Where;K = 0.50 @ 180 wrap angle

T2 = 0.50(3,025.51)

= 1,512.75 Newton

f) Return side tension (T3)

T3 = T2 + TX

= 1,512.75 + 1,157.7

= 2,670.5 Newton

g) Maximum tension (T1)

T1 = Te + T2

= 3,025.51 + 1,512.75

= 4,538.26 Newton

h) Tension to limit sag @ 2% (Tsag)

Tsag = g*sf*( Qb + QG )*s


5/66

Where:

Sf = sag factor = 6.3

= idler space = 0.3m

Tsag = 9.81(6.3)(13.5 + 44.44)(0.3)

= 1,074.5 Newton

i) Take up (TG)

TG = 2*T3

= 2(2,670.5)

= 5,341 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:

Ks = factor of safety = 8

B = belt width = 1m

T = (4,538.26 * 8)/1

= 36.31 KN/m]

ABSORBED POWER (PA)

PA = Te * V

Where:


6/66

V = belt speed = 2.5 m/s

PA = (3,025.51)(2.5)

= 7,563.7 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (7,563.7)/0.90

= 8.4 KW

RECOMMENDED POWER: 10KW for # 1,2 & 5 segment

# 3 & 15 SEGMENT

Belt type ST400

Belt width 1000mm


7/66











LC = C * L

Where:



LC = 1.95(38) = 64.35m

MOTIONAL RESISTANCE



8/66


Where:



Tx = 9.81(79.6)(0.02)(64.35)

= 1,005.4 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(13.5 + 44.44)(0.03)(64.35)

= 1,096.4 Newton


TS = 0.25(TX + TY)

= 0.25(1,005.4 + 1,096.4)

= 525.48 Newton


Te = TX + TYn + Ts


9/66

= 1,005.4 + 1,096.4 + 525.48

= 2,627.4 Newton


T2 = K*Te


T2 = 0.50(2,627.4)

= 1,313.7 Newton


T3 = T2 + TX

= 1,313.7 + 1,005.4

= 2,319Newton


T1 = Te + T2

= 2,627.4 + 1,313.7

= 3,941 Newton



Where:



10/66


Tsag = 9.81(6.3)(13.5 + 44.44)(0.3)

= 1,074.5 Newton

i) Take up (TG)

TG = 2*T3

= 2(2319)

= 4,638 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (3,941 * 8)/1

= 31.5 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (2,627)(2.5)


11/66

= 6,568 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (6,568)/0.90

= 7.3 KW

RECOOMENDED POWER: 8.5 KW for # 3 segment

# 4 SEGMENT

Belt type ST630Belt width 1000mm




12/66



Space between idler 1.5m


Weight of carrying idler per linear meter (QC) = 29 Kg/m


Total weight of moving parts per meter = 55 Kg/m


LC = C * L

Where:



LC = 1.65(78) = 128.7m

MOTIONAL RESISTANCE



Where:


13/66



Tx = 9.81(55)(0.02)(128.7)

= 1,392 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(17.5 + 44.44)(0.03)(128.7)

= 2344 Newton


TS = 0.25(TX + TY)

= 0.25(1,392 + 2,344)

= 934Newton


Te = TX + TYn + Ts

= 1,392 + 2,344.5 + 934

= 4,670.6 Newton


14/66


T2 = K*Te


T2 = 0.50(4,670.6)

= 2335.3 Newton


T3 = T2 + TX

= 2,335.3 + 1,392

= 3,727 Newton


T1 = Te + T2

= 4,670 + 2,335.3

= 7,005 Newton



Where:



Tsag = 9.81(6.3)(17.5 + 44.44)(0.55)

= 2,104Newton


15/66

i) Take up (TG)

TG = 2*T3

= 2(3,727)

= 7,454.5 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (7,005 * 8)/1

= 56 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (4,670.6)(2.5)

= 11,676 watts

MOTOR POWER


16/66

PM = PA/

Where:

= efficiency = 0.90

PM = (11,676)/0.90

= 12,973 KW

RECOOMENDED POWER: 15 KW for # 4 segment

# 6 , 16 & 11 SEGMENT

Belt type EP200

Belt width 1000mm







17/66



Weight of return idler per linear meter (QR) = 16 Kg/m



LC = C * L

Where:



LC = 1.95(15) = 29.25m

MOTIONAL RESISTANCE



Where:



Tx = 9.81(105.5)(0.02)(29.25)

= 605.45 Newton


18/66


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(9.5 + 44.44)(0.03)(29.25)

= 464 Newton


TS = 0.25(TX + TY)

= 0.25(605.45 + 464)

= 267.36Newton


Te = TX + TYn + Ts

= 605.45 + 464 + 267.36

= 1,336.7 Newton


T2 = K*Te


T2 = 0.50(1,336.7)


19/66

= 668.4 Newton


T3 = T2 + TX

= 668.4 + 605.45

= 1,273.8 Newton


T1 = Te + T2

= 1336.7 + 668.4

= 2005 Newton



Where:



Tsag = 9.81(6.3)(9.5 + 44.44)(0.2)

= 666.2Newton

i) Take up (TG)

TG = 2*T3

= 2(1,274)


20/66

= 2,547 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (2005 * 10)/1

= 20.05 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (1,336.7)(2.5)

= 3,341 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (3,341)/0.90


21/66

= 3.7 KW

RECOMMENDED POWER: 5 KW for # 6 & 11 segment

# 7 SEGMENT

Belt type ST1800

Belt width 1000mm






Weight of belt per linear meter (QB) = 28Kg/m





22/66


LC = C * L

Where:



LC = 1.45(202) = 292.9m

MOTIONAL RESISTANCE



Where:



Tx = 9.81(54.3)(0.02)(292.9)

= 3,120.46 Newton



23/66

Ty = g * (Qb + QG )* Fy * LC

Where:


Qb = belt weight = 28 kg/m


Ty = 9.81(28 + 44.44)(0.03)(292.9)

= 6,241 Newton


TS = 0.25(TX + TY)

= 0.25(3,120.5 + 6241)

= 2,340 Newton


Te = TX + TYn + Ts

= 3,120 + 6,241 + 2,340

= 11,702 Newton


T2 = K*Te


T2 = 0.50(11,702)

= 5851 Newton


T3 = T2 + TX


24/66

= 5851 + 3120

= 8971 Newton


T1 = Te + T2

= 11,701 + 5851

= 17,552 Newton



Where:



Tsag = 9.81(6.3)(28 + 44.44)(1.2)

= 5,369 Newton

i) Take up (TG)

TG = 2*T3

= 2(8,971)

= 17,942 Newton


25/66

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (17,552 * 8)/1

= 175 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (11,702)(2.5)

= 29,254 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (29,254)/0.90

= 32.5 KW

RECOOMENDED POWER: 39 KW for # 7 segment


26/66

# 8 SEGMENT

Belt type ST2000

Belt width 1000mm






Weight of belt per linear meter (QB) = 29 Kg/m






27/66

LC = C * L

Where:


L = conveyor length = 210 m

LC = 1.45(210) = 304.5m

MOTIONAL RESISTANCE



Where:



Tx = 9.81(52.3)(0.02)(304.5)

= 3,123 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:



28/66



Ty = 9.81(29 + 44.44)(0.03)(304.5)

= 6577.7 Newton


TS = 0.25(TX + TY)

= 0.25(3,274 + 6534)

= 2425 Newton


Te = TX + TYn + Ts

= 3,123 + 6,577.7 + 2425

= 12,125.8 Newton


T2 = K*Te


T2 = 0.50(12,125.8)

= 6,062 Newton


T3 = T2 + TX

= 6,062 + 3,123

= 9,185 Newton


29/66


T1 = Te + T2

= 12,125.8 + 6,062.9

= 18,188 Newton



Where:



Tsag = 9.81(6.3)(29 + 44.44)(1.3)

= 5,897 Newton

i) Take up (TG)

TG = 2*T3

= 2(9,185)

= 18,371.6 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


30/66


B = belt width = 1m

T = (18,188 * 8)/1

= 181.8 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (12,125)(2.5)

= 30,314 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (30,314)/0.90

= 33.6 KW

RECOMMENDED POWER: 40 KW for # 8 segment


31/66

# 9 SEGMENT

Belt type EP 315

Belt width 1000mm




Return idler dia (plain) 108 mm


Weight of belt per linear meter (QB) = 13.5 Kg/m


Weight of return idler per linear meter (QR) = 16 Kg/m



LC = C * L

Where:


32/66



LC = 1.95(20) = 39m

MOTIONAL RESISTANCE



Where:



Tx = 9.81(109.5)(0.02)(39)

= 838 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(13.5 + 44.44)(0.03)(39)


33/66

= 665 Newton


TS = 0.25(TX + TY)

= 0.25(838 + 665)

= 375.6 Newton


Te = TX + TYn + Ts

= 838 + 664.5 + 375.6

= 1,878 Newton


T2 = K*Te

Where;

K = 0.50 @ 180 wrap angleT2 = 0.50(1,878)

= 939 Newton


T3 = T2 + TX

= 939 + 838

= 1,777 Newton


T1 = Te + T2

= 1,878 + 939


34/66

= 2,817 Newton



Where:



Tsag = 9.81(6.3)(13.5 + 44.44)(0.2)

= 715.6 Newton

i) Take up (TG)

TG = 2*T3

= 2(1,777)

= 3,554 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (2,817* 10)/1


35/66

= 28.17 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (1,878)(2.5)

= 4,695 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (4,695)/0.90

= 5.2 KW

RECOMMENDED POWER: 6.2 KW for # 9 segment


36/66

# 10 SEGMENT

Belt type ST 1000

Belt width 1000mm





Space between idler 1.5 m






LC = C * L

Where:



LC = 1.65(115) = 189.75 m


37/66

MOTIONAL RESISTANCE



Where:



Tx = 9.81(48)(0.02)(189.75)

= 1788 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(19.5 + 44.44)(0.03)(189.7)

= 3,568 Newton



38/66

TS = 0.25(TX + TY)

= 0.25(1788 + 3,568)

= 1,339 Newton


Te = TX + TYn + Ts

= 1,788 + 3,568 + 1339

= 6,695Newton


T2 = K*Te


T2 = 0.50(6,695)

= 3,348 Newton


T3 = T2 + TX

= 3,348+ 1788

= 5,136Newton


T1 = Te + T2

= 6696.7 + 3348

= 10043 Newton


39/66



Where:



Tsag = 9.81(6.3)(19.5 + 44.44)(0.8)

= 3159 Newton

i) Take up (TG)

TG = 2*T3

= 2(5136)

= 10,272 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (10,043* 8)/1

= 80 KN/m


40/66

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (6696)(2.5)

= 16,739 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (16,739)/0.90

= 18.5 KW



41/66

# 12 SEGMENT

Belt type ST 4000

Belt width 1000mm





Space between idler 5 m

Weight of belt per linear meter (QB) = 48 Kg/m





LC = C * L

Where:



LC = 1.25(467) = 583.7 m


42/66

MOTIONAL RESISTANCE



Where:



Tx = 9.81(62.18)(0.02)(583.7)

= 7,121 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(48 + 44.44)(0.03)(583.7)

= 15,874 Newton


TS = 0.25(TX + TY)

= 0.25(7,121+ 15,874)

= 5,748.8 Newton


43/66


Te = TX + TYn + Ts

= 7,121 + 15,874 + 5,748.8

= 28,744 Newton


T2 = K*Te

Where;

K = 0.50 @ 180 wrap angleT2 = 0.50(28,744)

= 14,372Newton


T3 = T2 + TX

= 14,372+ 7,121

= 21,493 Newton


T1 = Te + T2

= 28,743 + 14,372

= 43,115 Newton



44/66


Where:



Tsag = 9.81(6.3)(48 + 44.44)(2.3)

= 13,134 Newton

i) Take up (TG)

TG = 2*T3

= 2(21,493)

= 42, 986 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (43,115* 8)/1

= 345 KN/m

ABSORBED POWER (PA)

PA = Te * V


45/66

Where:


PA = (28,743)(2.5)

= 71,859 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (71,859)/0.90

= 79.8 KW


# 13 SEGMENT

Belt type ST 630

Belt width 1000mm


46/66


Carrying idler dia. 108 mm









LC = C * L

Where:



LC = 1.65(73) = 120.45 m

MOTIONAL RESISTANCE



47/66


Where:



Tx = 9.81(55.12)(0.02)(120.45)

= 1,302.7 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(17.5 + 44.44)(0.03)(120.45)

= 2,194 Newton


TS = 0.25(TX + TY)

= 0.25(1302.7+ 2194)

= 874 Newton


Te = TX + TYn + Ts


48/66

= 1302.7 + 2194 + 874

= 4371 Newton


T2 = K*Te


T2 = 0.50(4371)

= 2185.6 Newton


T3 = T2 + TX

= 2,185.6 + 1302.7

= 3488Newton


T1 = Te + T2

= 4371 + 2185.6

= 6556.8 Newton



Where:



49/66


Tsag = 9.81(6.3)(17.5 + 44.44)(0.55)

= 2104 Newton

i) Take up (TG)

TG = 2*T3

= 2(3448)

= 6976 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (6556* 8)/1

= 52 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (4371)(2.5)


50/66

= 10,928 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (10,928)/0.90

= 12.1 KW


# 14 SEGMENT

Belt type ST 1000

Belt width 1000mm




51/66









LC = C * L

Where:



LC = 1.65(134) = 221 m

MOTIONAL RESISTANCE



Where:


52/66



Tx = 9.81(48)(0.02)(221)

= 2083.7 Newton


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(19.5 + 44.44)(0.03)(221)

= 4158 Newton


TS = 0.25(TX + TY)

= 0.25(2084+ 4158)

= 1560 Newton


Te = TX + TYn + Ts

= 2084 + 4158 + 1560

= 7802 Newton


53/66


54/66

i) Take up (TG)

TG = 2*T3

= 2(59850

= 11,969 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (11,703* 8)/1

= 94 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (7802)(2.5)

= 19505 watts

MOTOR POWER


55/66

PM = PA/

Where:

= efficiency = 0.90

PM = (19,505)/0.90

= 21.6 KW


# 14 SEGMENT

Belt type ST 1000

Belt width 1000mm







56/66






LC = C * L

Where:



LC = 1.65(134) = 221 m

MOTIONAL RESISTANCE



Where:



Tx = 9.81(48)(0.02)(221)

= 2083.7 Newton


57/66


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(19.5 + 44.44)(0.03)(221)

= 4158 Newton


TS = 0.25(TX + TY)

= 0.25(2084+ 4158)

= 1560 Newton


Te = TX + TYn + Ts

= 2084 + 4158 + 1560

= 7802 Newton


T2 = K*Te


T2 = 0.50(7802)


58/66

= 3901 Newton


T3 = T2 + TX

= 3901 + 2084

= 5985 Newton


T1 = Te + T2

= 7802 + 3901

= 11,703 Newton



Where:



Tsag = 9.81(6.3)(19.5 + 44.44)(0.8)

= 3159 Newton

i) Take up (TG)

TG = 2*T3

= 2(59850


59/66

= 11,969 Newton

j) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (11,703* 8)/1

= 94 KN/m

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (7802)(2.5)

= 19505 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (19,505)/0.90


60/66

= 21.6 KW


II.TRANSFER POINT

Belt type EP630

Belt width 1000mm


Elevating height 5m




Space between idler 1 m




61/66




LC = C * L

Where:



LC = 1.95(12) = 23.4 m

MOTIONAL RESISTANCE



Where:



Tx = 9.81(58.3)(0.02)(23.4)

= 267.6 Newton


62/66


Ty = g * (Qb + QG )* Fy * LC

Where:




Ty = 9.81(13.5 + 44.44)(0.03)(23.4)

= 399 Newton

c) Tension to lift the load (TH)

Tz = g * QG *H

Where:

H = height = 5m

Tz = (9.81 )(44.44)(5)

= 2179 Newton

d) Special resistance (TS )

TS = 0.25(TX + TY + TH)

= 0.25(267.6 + 399 + 2179)

= 711 Newton

e) Effective tension (Te)

Te = TX + TYn + TH + Ts


63/66

= 267.6 + 399 +2179 +711

= 3557 Newton

f) Slope tension (TH)

TH = g * QB *H

= (9.81)(13.5)(5)

= 662 Newton

g) Slacked side tension to prevent slip (T2)

T2 = K*Te


T2 = 0.50(3,578.7)

= 1789 Newton

h) Return side tension (T3)

T3 = T2 + TX TH

= 1779 + 267.6 662

= 1,411 Newton

i) Maximum tension (T1)

T1 = Te + T2

= 3578 + 1789


64/66

= 5367 Newton

j) Tension to limit sag @ 2% (Tsag)


Where:



Tsag = 9.81(6.3)(13.5 + 44.44)(0.45)

= 1611 Newton

k) Take up (TG)

TG = 2*T3

= 2(1411)

= 2822 Newton

l) Unit tension (T)

T = (T1*Ks)/B

Where:


B = belt width = 1m

T = (5367* 10)/1

= 53.68KN/m


65/66

ABSORBED POWER (PA)

PA = Te * V

Where:


PA = (3578)(2.5)

= 8946 watts

MOTOR POWER

PM = PA/

Where:

= efficiency = 0.90

PM = (8946)/0.90

= 9.9 KW

RECOMMENDED POWER: 12 KW for TRANSFER POINT


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Revised Conveyor Belt Cal

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Transcript of Revised Conveyor Belt Cal