Powder factor for Opencast Mines and production calculation ,Blasting numerical covered

This is the second part of powder factor series in this lecture we will calculate powder factor determination for opencast mines with very basic concept.

[@alok8781]

Sub grade drilling means bench height agar 10 metre hai hum drill hole 11 metre( B. H+ 10%)karenge lekin powder factor calculation ke time 1 metre minus karke hole depth 10 metre rakhenge jo ki bench height ke saath same hoga. 9m agar lenge to proper volume nahin niklega

[@MiningClasses]

see basically two cases are there 1) if in question Bench height is given then subgrade drilling is done beyond the bench height i.e. if bench height is 10 meter its 10% will be the subgrade drilling now most important thing to note length of drilling is 10 (B.H) + 1(SUBGRADE DRILLIGN) = 11 M 2) if in question only length of hole is given (if bench height is not mentioned) and subgrade drilling is given as discussed in this video suppose total length or height of hole is 10m and 1 meter is subgrade drilling then it means bench height is 10-1 = 9m for production calculation bench height is taken not the hole length but for explosive consumption total length of hole is considered @ALOK

[@kamalnarayan269]

Great sir. Thank to appse the mistake

[@yogeshbhure7322]

Hi sir..sir blasting,breaking,loading and transportation lead in km ka rate kaise nikalte he sir?? Per tonnage ke hisab se?? Koi formula he uska to plz rpy kro sir

[@noursane7445]

great great bro, I share this nice video it in my group Mining engineering / Génie minier . Good continuation.

[@MiningClasses]

Thanks for sharing

[@shahbaazkhan2208]

In this problem we should not minus the subgrade drilling depth from bench height in terms of production calculation....because we are calculating production from original bench height....10m-1m=9m....this 9m is not the bench height....original height is 10m

[@kamalnarayan269]

Great buddy, Pure comment section me koi to samjhdar mila.....👍🏻

[@prashantkumar9056]

excellent sir ji.. 👍

[@YogeshKumar-hb2ow]

Sir please more videos on mining

[@amaron769]

Very very thanks sahab

[@abhi20118]

please put new videos on underground blasting

[@SunilKumar-hn3fe]

Volume of cylinder me 1000 se multiply kyo karte hai

[@deepakjangid3085]

Find out production in tonne given blast design No. Of holes= 60 Burden=3m Spacing=4m Bench height=10m Bank density of material= 2.5te/m3

[@deepakjangid3085]

Can anyone solve this question?

[@kishanmahant3109]

An SDL of 1.0 tonne capacity operates with a cycle time of 6 minutes. The dimension of the face is 4 m x 3 m. Five blasts are conducted per shift with an average pull of 1.2 m. If the density of blasted coal is 0.8 tonne/m3, the time required by the SDL in the shift to lift all the prepared coal in hours is______.

[@MiningClasses]

see first you have to solve how much coal is produced in blasting the face so as i have discussed in my blasting you can figure it out ,,, it will form a shape , determine which shape our blast design will make,,then with the help of shape determined calculate volume, after calculating volume to get in tonne multiply it with specific gravity you will get production in tonne hence cycle time is given calculate how it lift at once ,how much time it take for one pass divide it by total production see shape is cuboid show use formula = L×B×H = 4×3×1.2 = 14.4 cubic meter (pull is 1.2m) in tonne = volume × spe.gravity = 14.4×0.8 = 11.52tonne in a single blast in shift 5 blast is carried on so 11.52× 5 = 57.6 tonne in a shift (5 round of blasting) now as given in question it take 6 minute for a cycle time means it take 6 minute to load one bucket in belt and again comes to load and its capacity is one 1 tonne so in a pass it can load 1 tonne only no. of passes required to load 57.6 tonne of coal = total coal / coal loaded in one pass = 57.6/1 =57.6 passes = say 58 passes hence for one pass it takes 6 minute so for 58 passes time will be required = 58 × 6 = 348 minute convert it in hours 348/60 = 5.8 hours (note - method will be same however you can check the calculation from you end ) hope you got it watch this video of mine kzbin.info/www/bejne/eJPVqGawh5WYn6c

[@kishanmahant3109]

@@MiningClasses very thankfully 🙏🏻🙏🏻 Sir , you are my good guidence sir always .

[@aayushtailang7684]

5.76 hr

[@amitpandey483]

🙏🙏🙏 very nice 👌 sir

[@ganeshparida8536]

In calculating the volume of explosive don't we have to substracting the subgrade drilling length

[@MiningClasses]

yes we have to subtract subgrade drilling length

[@gamingwithsubhamnonia9745]

Sir you didn't subtracted

[@MiningClasses]

please watch video in 3.58 in which i have subtracted (10-1) in subgrade drilling from bench height

[@amlanjyotimajumdar3746]

Sir question me bench height 10m hai aur subdrilling 1m but aapne volume of cuboid nikalte wakt 10m ko total drilling height le liya aur (10-1) kiya, jo galat hai,sirf 10* 5*4 hoga for the volume of cuboid.Agar total drilling length diya hota hai toh hum sub drilling minus karte hai.

[@MiningClasses]

please note that concept is to minus subgrade drilling from total height or length of drill hole (as sub grade does not encounter in production) in above question hole is 10m of length including 1meter of sub grade drilling so for production 10-1 =9m hence to calculate volume of cuboid will be 9*5*4

[@alok8781]

@@MiningClasses Lekin volume calculation ke time Bench height ko pura lena padega kyun ki sub grade Bench height se 10 % jyada rehta hai

[@MiningClasses]

@@alok8781 YES YOU ARE RIGHT

[@sksaida8842]

Thanks sir this is useful to me

[@MiningClasses]

Always welcome

[@yogenshres]

dont we have to take the variable constraint like discontinuties for volume production??

[@MiningClasses]

generally while solving numerical this factor are not considered and if it is given then we have to deduct it from production what we have calculated ,in case of actual mining activities it is measured by survey

[@kamalnarayan269]

U r right bro

[@Upscaspirants18r]

Minral production 360 kaise nikale vo bhiya

[@Cricket_feverx]

Sir,, major project k liye guide kijiye.. Please

[@MiningClasses]

blasting pe kr skte ho electronic detonator pe bna lo

[@amitgorai7839]

Bhai apnae explosive ka volume nikalana kae samay subgrade koa deduct nahi kiya.Only apa total production mae subgrade koa deducted kiya hae.Explosive ka volume nikalna kae time subgrade koa deduct kare ki nahi direct (10-3 kiya = 7 huya par subgrade koa deducted karte toa 6m Height hota explosive colume ka) 6:58 mae dekhiya

[@MiningClasses]

voulme of explosive calculate krne k liye sub grade drilling ko mines nhi krna h , suppose 10 m hole k length h aur 1 meter sub grade drilling to total length of hole ho gya 11 m. ab agar stemming 5 meter rkha jata to total charging hogi 11-5 =6 m explosvie charging krte smay pura hole k deapth ko consider kiya jata h isliye sub grade drilling ko minus nhi kiya jata h in case of production sub grade drilling jo 1 meter bottom of hole m subgrade drilling ki gyi h vo hissa vlast nhi hota ,agr blast nhi hoga to production kaise hoga isliye sub grade drilling ko production m minus kr dete h suppose agr hole k length 11 meter h to exact 11 meter ki blasting nhi ho pata 11 se kuch km hi hota h isliye 11 se thoda jyada (sub grade) drilling krte h so 11 meter tk ek bench height bne jisse machine ko operate krne m dikkat na ho. hope u get it @amit gorai

[@kidambisaikarthik9934]

Sir how to find out burden?

[@kishanmahant3109]

Please solve this question sir

[@kishanmahant3109]

Ventilation term m numericals sir please 🙏🏻 🙏🏻🙏🏻

[@MiningClasses]

i will cover all the numerical for this topic very soon

[@kranthikumar2648]

Please make videos on numerical problems..

[@devendrasinghsolanki3601]

Volume of cuboid ka q liya hai sir, bcoz blast hole to cylinder shape me hi hota hai

[@MiningClasses]

cylinder shape borehole k blast hota

[@RamSingh-gc2um]

Burden se hamesh spacing jada hoti hai mere bhae

[@MiningClasses]

yes you are right

[@shivamgupta-sf4fr]

👌👌👍

[@kranthikumar2648]

Thanks bro👍👍🙏🙏

[@neeraj1204]

Nice sir

[@MiningClasses]

Thanks and welcome

[@suyashsharma33]

Nice

[@superhero2767]

Sir..... Aap ka instagram ka link bhej do use pe msg kar duga

[@MiningClasses]

u can ask your doubt in comment or mail me [e-mail] for any query

[@anniekabaghe5742]

Please from now on use English it not only India is watch but the world of Mining is watching

[@MiningClasses]

@annie kabaghe thank you for compliment i will upload in english also with subtitle

[@kishanmahant3109]

Data related to explosive and blasthole are given below. Assuming 1 kcal to be 4.2 kJ, the Power of explosive in GW in the blasthole (round offto one decimal place) is Diameter of the borehole : 200 mm Charge length : 8 m Density of ANFO : 0.8 g/co Heat of explosion : 912 cal/g VOD : 4500 m/s Initiation : Bottom

[@MiningClasses]

@Kishan Mahant first calculate how much explosive present in the hole use volume formula as hole is in cylindrical form so = πr^2h = 3.14× (10)^2 × 80 (convert all in cm because spe. gravity given in cc i.e cubic centimetres) = 25120 cc now we have = density of anfo = 0.8g/cc so weight of anfo in the hole = 25120 × 0.8 = 20096 g now calculate energy given by explosive as heat of explosion = 912cal/g nd we have 20096 g so = 20096× 912 = 18327552 cal convert it in kcal by dividing it 1000 = 18327.55 kcal now 1 kcal = 4.2kJ so 18327.55 × 4.2 = 76975.7184 KJ (joule is s.i. unit of energy means on blasting this much energy will released) we have to calculate power in GW POWER = energy/ time so we have determined energy id 76975.7184 KJ SO VOD = 4500M/S in explosivr coloum length charing is only upto 8 m hence time taken to travel for 8m length will be = 1/4500 * 80 = 0.0017 sec so power = energy /time = 76975718.4(joule)/ 0.00178 sec = 432.4 GW DIVIDE IT BY 10^9 to grt answer in GW hopefully you got it.

[@amitgorai7839]

Bro 8m joa charge lengh hae uya 800cm hoga .Jab volume nikal raha hoa explosive ka 80, nahi hoga

[@kishanmahant3109]

Sir please 🙏🏻🙏🏻 solve 🙏🏻🙏🏻

[@vijaykumar-9]

English class