CM 335 Assignment _3a Chapter 14.10 Problem Solution

CM 335 Assignment _3a Chapter 14.10 Problem Solution -...

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Unformatted text preview: Chapter 14 14.10 Solution Size of Crusher 36”x42” (given) Crusher Closed Setting‐4” (given) Weight of Stone‐115 lb/cf (given) Capacity of Crusher (Table 14.3), 200 tph for a unit weight of 100 lb/cf; must adjust for 115 lb/cf [(115 lb/cf)/(100 lb/cf)] X 200 tph = 230 tph Sizes of Screens –2 ½”, 1 ½”, ¾” (given) Efficiency of Screens‐90% (given) Determine the output of the crusher using Figure 14.5 on page 415: Size Range (in.) Percent Passing Screens Percent in Size Range Total Output Of Crusher (TPH) Over 2 ½” 2 ½”‐1 ½” 1 ½”‐3/4” ¾”‐0 100‐56 56‐37 37‐21 21‐0 TOTAL 44 19 16 21 100 230 230 230 230 Quantity Produced in Size Range (TPH) 101.2 43.7 36.8 48.3 230 Double check total of “Percent in Range Column” =100 and Total Output of Crusher = 230 Screen No.1— Using Figure 14.18, for a 2 ½” Screen it should produce 4.2 tph/sf for material with a unit weight of 100 lb/cf and with 100% efficiency Aggregate weight correction for 115 lb/cf: 1.15 X 4.2 tph = 4.83 tph/sf Screen Efficiency Factor from Table 14.7 = 1.25 for 90% efficiency Deck Factor from Table 14.8 = 1.0 for first deck Aggregate Size Factor from Table 14.9: 2 ½”/2 = 1 ¼” Now using Figure 14.5 you have to estimate what the % passing is for 1 ¼” aggregate on the 4” opening line of Figure 14.5. Looking at the 4” line I see that for 1 ½” it is 37% and for 1” it is 26%. Therefore, I take the difference between the two (11 %) and divide by two for 5.5% and then I round up to 6% and add it to 26% for a total of 32%. Now take our 32% and go to Table 14.9 and I am going to use 0.80 because we are approximately 30%. So our variables are as follows: Q=230 tph A=? C=4.83 tph/sf E=1.25 D=1.0 for first deck G=0.80 A= (230 tph)/((4.83 tph/sf)(1.25)(1.00)(.80) Multiply through on the bottom and eliminate the tph on both the numerator and the denominator; the sf moves to the numerator. A= (230sf)/(4.83) A=47.6 SF Now adjust for 10‐25% overcapacity; 47.6 sf X 1.1 = 52.4 sf We know that the screen deck is 4’ wide, therefore: 52.4 sf/ 4’ = 13.1’ You could list your answer as a 4’x13.1’ screen deck, but they would come in 6” increments, so I would say 4’x13.5’ 2 ½” screen Screen No.2— First determine what aggregate(tph) is falling through the first screen deck to the second screen deck: 43.7 + 36.8 + 48.3 = 128.8 128.8 tons X 90% = 115.9 tph Using Figure 14.18, for a 1 ½” Screen it should produce 3.3 tph/sf for material with a unit weight of 100 lb/cf and with 100% efficiency Aggregate weight correction for 115 lb/cf: 1.15 X 3.3 tph = 3.8 tph/sf Screen Efficiency Factor from Table 14.7 = 1.25 for 90% efficiency Deck Factor from Table 14.8 = 0.90 for second deck Aggregate Size Factor from Table 14.9: 1 ½”/2 = ¾” Now using Figure 14.5 you have to estimate what the % passing is for 3/4” aggregate on the 4” opening line of Figure 14.5. Looking at the 4” line I see that for 1” it is 26% and for 1/2” it is 16%. Therefore, I take the difference between the two (10 %) and divide by two for 5% and add it to 16% for a total of 21%. Now we need to take our 21% and multiply it by our total tph of 230 for 48.3 tph. We now need to adjust for the 90% efficiency of the 2 ½” screen. Therefore, 48.3 tph X .90 = 43.5 tph. Now calculate the percentage 43.5 tph /115.9 = 38% Now take our 38% and go to Table 14.9 and I am going to use 1.0 because we are approximately 40%. So our variables are as follows: Q=115.9 tph A=? C=3.80 tph/sf E=1.25 D=0.90 for second deck G=1.0 A= (115.9 tph)/((3.80 tph/sf)(1.25)(0.90)(1.0) Multiply through on the bottom and eliminate the tph on both the numerator and the denominator; the sf moves to the numerator. A= (115.9sf)/(4.28) A=27.1 SF Now adjust for 10‐25% overcapacity; 27.1 sf X 1.1 = 29.8 sf We know that the screen deck is 4’ wide, therefore: 29.8 sf/ 4’ = 7.45’ You could list your answer as a 4’x7.45’ screen deck, but they would come in 6” increments, so I would say 4’x8’ 1 ½” screen Screen No.3— First determine what aggregate(tph) is falling through the first screen deck to the second screen deck: 36.8 + 48.3 = 85.1 85.1 tons X 90% X 90%= 68.9 tph Using Figure 14.18, for a ¾” Screen it should produce 2.2 tph/sf for material with a unit weight of 100 lb/cf and with 100% efficiency Aggregate weight correction for 115 lb/cf: 1.15 X 2.2 tph = 2.53 tph/sf Screen Efficiency Factor from Table 14.7 = 1.25 for 90% efficiency Deck Factor from Table 14.8 = 0.75 for third deck Aggregate Size Factor from Table 14.9: ¾”/2 = 3/8” Now using Figure 14.5 you have to estimate what the % passing is for 3/8” aggregate on the 4” opening line of Figure 14.5. Looking at the 4” line I see that for ½” it is 16% and to estimate for ¼” I took a straight edge and put it on the dots for ¼” for the previous four opening sizes and it shows that it should be about 8% on the ¼”. Therefore, I take the difference between the two (8 %) and divide by two for 4% and add it to 8% for a total of 12%. Now we need to take our 12% and multiply it by our total tph of 230 for 27.6 tph. We now need to adjust for the 90% efficiency of the 2 ½” screen and the 1‐1/2” screen. Therefore, 27.6 tph X 0.90 X 0.90 = 22.4 tph less than one half of ¾”. Now calculate the percentage 22.4 tph /68.9 tph = 32.5% Now take our 32.5% and go to Table 14.9 and I am going to use 0.80 because we are approximately 30%. So our variables are as follows: Q=68.9 tph A=? C=2.53 tph/sf E=1.25 D=0.75 for third deck G=0.80 A= (68.9 tph)/((2.53 tph/sf)(1.25)(0.75)(.80) Multiply through on the bottom and eliminate the tph on both the numerator and the denominator; the sf moves to the numerator. A= (68.9 sf)/(1.90) A=36.3 SF Now adjust for 10‐25% overcapacity; 36.3 sf X 1.1 = 39.9 sf We know that the screen deck is 4’ wide, therefore: 39.9 sf/ 4’ = 10.0 sf List your answer as a 4’x10’ 3/4” screen. ...
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This note was uploaded on 01/07/2012 for the course CM 335 taught by Professor Staff during the Fall '08 term at BYU.

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