Exercício Bombas

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Plan1
	PROJETO - DIMENSIONAMENTO DE BOMBAS HIDRÁULICAS
	FLUÍDO	ÁGUA	Sucção (Le):	Q(m³/s)	Vs (m/s)	Vd(m/s)	TABELA FINAL
	MATERIAL	AÇO CARBONO	3 Cotovelos 90°	0.0027777778	1.4147106053	2.2104853208	HbSistema (m)	Hbomba (m)	Q(m³/s)	ɳ (%)
	ρ (kg/m³)	1000	0.0055555556	2.8294212106	4.4209706417	14.05571386	4.572	0.0027777778	38
	µ (kg/m.s)	0.0009	Le/Dsint = 32	0.0083333333	4.2441318158	6.6314559625	34.79753409	4.4196	0.0055555556	52
	g (m/s²)	10	Le = 32*Dsint	0.0111111111	5.6588424211	8.8419412833	68.58747867	4.1148	0.0083333333	60
	Ds (m)	0.05	Le = 1,6 m	0.0138888889	7.0735530264	11.0524266042	116.488851	3.6576	0.0111111111	65
	Dd (m)	0.04	3 Cotovelos = 3*Le = 4,8 m	0.0166666667	8.4882636317	13.262911925	178.0763296	3.048	0.0138888889	62
	Ls (m)	20	0.0194444444	9.9029742369	15.4733972458	253.3499147	2.4384	0.0166666667	55
	Ld (m)	50	2 Válvulas Abertas	342.3096061	1.524	0.0194444444	50
	Ys (m)	15	Número de Reynolds (Nre)	44138971*Q
	Yd (m)	22	Le/Dsint = 7	CONCLUSÕES:
	Dsint=Ddint	3" sch. 40 = 3,068" = 0,078 m	Le = 7*Dsint	Q(m³/s)	Nre	fd	A BOMBA NÃO POSSUI POTÊNCIA SUFICIENTE PARA ATINGIR
	Le = 0,35 m	0.0027777778	122608.2527777778	0.0325	A ALTURA MANOMÉTRICA REQUERIDA NO PROJETO, POR ISSO
	SUCÇÃO	3 COTOVELOS 90°/ 2 VÁLVULAS ABERTAS	2 Válvulas = 2*Le = 0,7 m	0.0055555556	245216.5055555556	0.032	RECOMENDAMOS UMA BOMBA DE MAIOR EFICIÊNCIA E POTÊNCIA
	DESCARGA	4 COTOVELOS 90° / 3 VÁLVULAS ABERTAS	0.0083333333	367824.7583333333	0.0315
	∑Le = 0,7 + 4,8 = 5,5 m	0.0111111111	490433.0111111112	0.0315
	e/d	0.00058	0.0138888889	613041.2638888889	0.0315
	Descarga (Le):	0.0166666667	735649.5166666666	0.0315
	Q(m³/s)	H(m)	ɳ (%)	4 Cotovelos 90°	0.0194444444	858257.7694444444	0.0315
	0.0027777778	4.572	38
	0.0055555556	4.4196	52	Le/Ddint = 32	hLs = fd*(∑Les + Ls/ Dsint)*Vs²/2g
	0.0083333333	4.1148	60	Le = 32*Ddint	hLd = fd*(∑Led + Ld/ Ddint)*Vd²/2g
	0.0111111111	3.6576	65	Le = 1,6 m
	0.0138888889	3.048	62	4 Cotovelos = 4*Le = 6,4 m	fd	hLs (m)	hLd (m)
	0.0166666667	2.4384	55	0.0325	1.0632469889	5.8482249076
	0.0194444444	1.524	50	3 Válvulas Abertas	0.032	4.1875573715	23.0330088669
	0.0315	9.2747852721	51.0145157326
	Hb = (P2-P1/pg) + (Vd²-Vs²/2g) + (Yd-Ys) + hLs + hLd	Le/Ddint = 7	0.0315	16.4885071504	90.6924724135
	Le = 7*Ddint	0.0315	25.7632924225	141.706988146
	Como o tanque é aberto: P1 = P2 = Patm	Le = 0,35 m	0.0315	37.0991410884	204.0580629303
	Hb = (Vd²-Vs²/2g) + (Yd-Ys) + hLs + hLd	2 Válvulas = 2*Le = 1,05 m	0.0315	50.4960531481	277.7456967662
	Vs = 4Q/π(Ds²)	∑Led = 1,05 + 6,4 = 7,45 m	Hb = (Vd²-Vs²/2g) + (Yd-Ys) + hLs + hLd
	Vs = 509,2958179*Q
	Nre = Res = Red = ρ.V.Dint/µ	Hb1 (m)	14.0557138593
	Vd = 4Q/π(Dd²)	Hb2 (m)	34.7975340898
	Vd = 795,7747155*Q	Hb3 (m)	68.5874786703
	Hb4 (m)	116.4888509694
	Hb5 (m)	178.0763296397
	Hb6 (m)	253.3499146811
	Hb7 (m)	342.3096060937
Plan2
	PROJETO - DIMENSIONAMENTO DE BOMBAS HIDRÁULICAS
	FLUÍDO	ÁGUA	Sucção (Le):	Q(m³/s)	Vs (m/s)	Vd(m/s)	hLs = fds*(∑Les + Ls/ Ds)*Vs²/2g	Hb = (Vd²-Vs²/2g) + (Yd-Ys) + hLs + hLd
	MATERIAL	AÇO CARBONO	3 Cotovelos 90°	0.0027777778	1.4147106053	2.2104853208	hLd = fdd*(∑Led + Ld/ Dd)*Vd²/2g
	ρ (kg/m³)	1000	0.0055555556	2.8294212106	4.4209706417	Hb1 (m)	15.0741479195
	µ (kg/m.s)	0.0009	Le/Ds = 32	0.0083333333	4.2441318158	6.6314559625	fds	hLs (m)	Hb2 (m)	37.7425717691
	g (m/s²)	10	Le = 32*Ds	0.0111111111	5.6588424211	8.8419412833	0.0215	0.8724379351	Hb3 (m)	75.2692597886
	Ds (m)	0.05	Le = 1,6 m	0.0138888889	7.0735530264	11.0524266042	0.021	3.4085947233	Hb4 (m)	125.7829502443
	Dd (m)	0.04	3 Cotovelos = 3*Le = 4,8 m	0.0166666667	8.4882636317	13.262911925	0.0208	7.5962968118	Hb5 (m)	190.8612350351
	Ls (m)	20	0.0194444444	9.9029742369	15.4733972458	0.0206	13.374676438	Hb6 (m)	269.2587188513
	Ld (m)	50	2 Válvulas Abertas	0.0204	20.6950393913	Hb7 (m)	360.5584917599
	Ys (m)	15	Nres1	78595.0336328308	0.0202	29.5086914614
	Yd (m)	22	Le/Ds = 7	Nres2	157190.0672656616	0.02	39.7669384381
	Le = 7*Ds	Nres3	235785.1008984924
	SUCÇÃO	3 COTOVELOS 90°/ 2 VÁLVULAS ABERTAS	Le = 0,35 m	Nres4	314380.1345313233	fdd	hLd (m)
	DESCARGA	4 COTOVELOS 90° / 3 VÁLVULAS ABERTAS	2 Válvulas = 2*Le = 0,7 m	Nres5	392975.168164154	0.023	7.0574680215
	Nres6	471570.2017969848	0.0218	26.7570091945
	e/Ds	0.0009	∑Les = 0,7 + 4,8 = 5,5 m	Nres7	550165.2354298157	0.0215	59.3747853112
	e/Dd	0.00125	0.021	103.1004024008
	Descarga (Le):	Nred1	98243.7920272127	0.0208	159.5601465726
	Q(m³/s)	H(m)	ɳ (%)	4 Cotovelos 90°	Nred2	196487.5840544253	0.0206	227.5573167274
	0.0027777778	4.572	38	Nred3	294731.3760816379	0.0204	306.7236971423
	0.0055555556	4.4196	52	Le/Dd = 32	Nred4	392975.1681088507
	0.0083333333	4.1148	60	Le = 32*Dd	Nred5	491218.9601360632
	0.0111111111	3.6576	65	Le = 1,28 m	Nred6	589462.7521632758
	0.0138888889	3.048	62	4 Cotovelos = 4*Le = 5,12 m	Nred7	687706.5441904885
	0.0166666667	2.4384	55
	0.0194444444	1.524	50	3 Válvulas Abertas	Q(m³/s)	Nres	Nred	fds	fdd
	0.0027777778	78595.0336328308	98243.7920272127	0.0215	0.023
	Hb = (P2-P1/pg) + (Vd²-Vs²/2g) + (Yd-Ys) + hLs + hLd	Le/Dd = 7	0.0055555556	157190.0672656616	196487.5840544253	0.021	0.0218
	Le = 7*Dd	0.0083333333	235785.1008984924	294731.3760816379	0.0208	0.0215
	Como o tanque é aberto: P1 = P2 = Patm	Le = 0,28 m	0.0111111111	314380.1345313233	392975.1681088507	0.0206	0.021
	Hb = (Vd²-Vs²/2g) + (Yd-Ys) + hLs + hLd	3 Válvulas = 3*Le = 0,84 m	0.0138888889	392975.168164154	491218.9601360632	0.0204	0.0208
	0.0166666667	471570.2017969848	589462.7521632758	0.0202	0.0206
	Vs = 4Q/π(Ds²)	∑Led = 5,12 + 0,84 = 5,96 m	0.0194444444	550165.2354298157	687706.5441904885	0.02	0.0204
	Vs = 509,2958179*Q
	Nre =ρ.V.D/µ
	Vd = 4Q/π(Dd²)	Nres = 1000*Vs*0,05/0,0009
	Vd = 795,7747155*Q	Nres = 55555,55556*Vs
	Nred = 1000*Vd*0,04/0,0009
	Nred = 44444,44444*Vd
Plan3
	104
	2	52	156
	3	156	208
	4	260
	5	0