Perfo - Fernando.docx

DATOS DE CAMPO: 1. 2. 3. 4. 5. 6. 7.

Sección: 4.5 × 4 Taladro de alivio: 1 Diámetro del taladro de alivio: 102mm = 0.102m Diámetro de producción: 0.045m Altura de arco: 0.5m 𝛾 = 0.05 𝑟𝑎𝑑3° 𝑚𝑚 𝛼 = 10 = 0.01𝑚/𝑚

8. 𝛽 = 20

𝑚 𝑚𝑚 𝑚

= 0.02𝑚/𝑚

9. Explosivo: Emulnor



EMULNOR 1000:

Calor de Explosión= 785

𝐾𝑐𝑎𝑙 𝑘𝑔

= 3.2866𝑀𝐽 𝑚3

𝑙

Volumen de Gas = 920𝑘𝑔 = 0.92 𝑘𝑔 

EMULNOR 3000:

Calor de Explosión= 920

𝐾𝑐𝑎𝑙 𝑘𝑔

𝑙 𝑘𝑔

Volumen de Gas = 880

= 3.8519𝑀𝐽

= 0.88

𝑚3 𝑘𝑔

𝜌𝑒𝑥𝑝𝑙𝑜𝑠𝑖𝑣𝑜=1.14𝑔/𝑐𝑚3 

EMULNOR 5000:

Calor de Explosión= 4.2287𝑀𝐽 𝑙

𝑚3

Volumen de Gas = 870𝑘𝑔 = 0.87 𝑘𝑔

𝜌𝑒𝑥𝑝𝑙𝑜𝑠𝑖𝑣𝑜=1.169𝑔/𝑐𝑚3 SOLUCION:  𝑆=

EMULNOR 1000:

5 3.2866 1 0.92 × + × = 0.7282 6 5.0 6 0.85

𝑆𝑎𝑛𝑓𝑜 =  𝑆=

0.7282 = 0.8669 0.84

EMULNOR 2000:

5 3.8519 1 0.88 × + × = 0.8145 6 5.0 6 0.85

𝑆𝑎𝑛𝑓𝑜 = 

𝑆=

0.8145 = 0.9696 0.84

EMULNOR 5000:

5 4.2287 1 0.87 × + × = 0.8754 6 5.0 6 0.85

𝑆𝑎𝑛𝑓𝑜 =

0.8754 = 1.0421 0.84

∅𝐸𝑄𝑈𝐼𝑉𝐴𝐿𝐸𝑁𝑇𝐸: ∅𝑒𝑞𝑢 = ∅𝑣𝑎𝑐𝑖𝑜 × √𝑛 = 0.102 × √1 = 0.102𝑚 ∅𝑒𝑞𝑢 = 10.2𝑐𝑚 AVANCE: H = 0.15 + 34.1(∅𝑒 ) − 39.4(∅𝑒 )2 = 3.2182𝑚 I = 95%(H) = 95%(3.2182) = 3.0573𝑚 <> 10𝐹𝑇 = 10 𝑝𝑖𝑒𝑠 𝐵𝑚𝑎𝑥 = 1.7∅𝑒 = 0.1734 𝑓 = 𝐻𝛼 + 𝛽 F=0.0522 𝐵𝑝 = 𝐵𝑚𝑎𝑥 − 𝐹 = 0.1754 − (3.218 × 0.01 + 0.02) 𝐵𝑝 = 0.1212 𝑚

𝑎 = 𝑚 + 2𝑦 ∅𝑝 ∅ 𝑎 = (𝐵𝑝 + ) √2 + 2( 𝑐𝑜𝑠45°) 2 2 𝑎 = 0.2753𝑚

CALCULO DEL FACTOR DE ROCA: C = 0.4 CALCULO PARA LA DENSIDAD DE CARGA (dq) PARA EL ARRANQUE (EMULNOR 5000): 3

2 ∅ 𝐶 𝐵 [55∅𝑝 ( 𝑚𝑎𝑥 ) (𝐵𝑚𝑎𝑥 − 2𝑒 )(0.4)] ∅𝑒

𝑑𝑞 =

𝑆𝑒𝑚𝑢𝑙𝑛𝑜𝑟5000

3

0.1734 2 0.102 0.4 [55(0.045) ( 0.102 ) (0.1734 − 2 )(0.4)] 𝑑𝑞 =

1.0421

𝑑𝑞 = 0.6443 kg/ml 3

CC = 4 𝐻 × 𝑑𝑞 =

3 4

(3.2182) × (0.7323) = 1.5551

𝑘𝑔 𝑡𝑎𝑙

(EMULNOR 5000)

CALCULO PARA LA PRIMERA AYUDA (I CUADRANTE): α = 0.2753m 𝑑𝑞 = 6443 𝑘𝑔/𝑚𝑙 a) Calculo del burden (I cuadrante):

√𝛼 × 𝑑𝑞 × 𝑆𝑒𝑚𝑢𝑙𝑛𝑜𝑟 5000 𝐼 𝐵𝑚𝑎𝑥 = 8.8 × 10−2 ( ) ∅𝑃 × 𝐶 √0.2753 × 0.6443 × 1.0421 𝐼 𝐵𝑚𝑎𝑥 = 8.8 × 10−2 ( ) 0.045 × 0.4 𝐼 𝐵𝑚𝑎𝑥 = 0.2820 𝑚 𝐼 𝐵𝑝𝐼 = 𝐵𝑚𝑎𝑥 − 𝐼 = 0.2820 − 0.0522 = 0.2298𝑚

L=ℎ +

∅𝑝 2

+ 𝐵𝑝𝐼

𝐿 = 0.1812 +

0.045 + 0.2298 2

𝐿 = 0.3741

𝑋 = 𝐿√2

𝑎𝐼 = 𝑥 + 2𝑦 𝑎𝐼 = 0.5290 + 2(

∅𝑝 𝐶𝑜𝑠45°) 2

𝑎𝐼 = 0.5290 + 2(0.0159) = 0.5608

DISPARO:

𝑎𝐼 = 0.5608

CALCULO DE LA DENSIDAD DE CARGA: 𝐼

𝑑𝑞 =

𝐼 32.3 × ∅𝑝 × 𝑐 × 𝐵𝑚𝑎𝑥 1.5

𝑎 𝑆5000 × [𝑠𝑒𝑛(𝑎𝑟𝑡𝑔( 𝐼 ))] 2𝐵𝑚𝑎𝑥 𝑑𝑞 𝐼 =

32.3 × 0.045 × 0.4 × 0.2820 0.2753 1.5 1.0421 × [𝑠𝑒𝑛(𝑎𝑟𝑡𝑔(0.2820)]

𝑑𝑞 𝐼 = 0.5415

3

CC = 4 𝐻 × 𝑑𝑞 =

3 4

(3.2182) × (0.5415) = 1.307

𝑘𝑔 𝑡𝑎𝑙

CALCULO DE LA SEGUNDA AYUDA: 𝑎𝐼 = 0.5608 𝐼𝐼 𝐵𝑚𝑎𝑥 = 8.8 × 10−2 𝐼𝐼 𝐵𝑚𝑎𝑥 = 8.8 × 10−2

√𝑑𝑞×𝑎𝐼 ×𝑆5000 ∅𝑝 ×𝐶 √0.6443×0.56508×1.0421 0.045×𝐶

𝐼𝐼 𝐵𝑚𝑎𝑥 = 0.4033𝑚 𝐼𝐼 𝐵𝑝𝐼𝐼 = 𝐵𝑚𝑎𝑥 − 𝐹 = 0.4033 − 0.0522 = 0.354m

𝑥 = (𝐵𝑝𝐼𝐼 +

∅𝑝 2

+ 𝐵𝑝 +

∅𝑒 2

)√2

∅𝑝 𝑎 = 𝑥 + 2𝑦 = 0.5488 + 2( 𝑐𝑜𝑠45°) 2

𝑎 = 0.7719 + 2(0.0159) = 0.8037𝑚

𝐼𝐼

𝑑𝑞 =

𝐼𝐼 32.3 × ∅𝑝 × 𝐶 × 𝐵𝑚𝑎𝑥 1.5

𝑆5000

𝑑𝑞 𝐼𝐼 =

𝑎𝐼 × [𝑠𝑒𝑛(𝑡𝑔 ( 𝐼𝐼 ))] 2𝐵𝑚𝑎𝑥 −1

32.3 × 0.045 × 0.4 × 0.4033 1.5 0.5608 1.0421 × [𝑠𝑒𝑛(𝑡𝑔−1 ( ))] 2 × 0.4033

𝑑𝑞 𝐼𝐼 = 0.5217𝐾𝐺/𝑚𝑙 3

𝐾𝐺

4

𝑇𝐴𝐿

CC = 𝐻 × 𝑑𝑞 = 1.2592