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PERHITUNGAN POMPA Diketahui v= 2 m/s g= ρ= 0.682 kg/m3 L= d= 0.0762 m P1= 0.08 kg/m2 vd= P2= 0.08 kg/m2 vs= Ditanya: DAYA POMPA? Wp= ϒ x Q x Htotal PENYELESAIAN: (berat jenis AMONIA) ϒ= ρ x g [kg/m2.s2] ϒ= 6.6904 (luas permukaan pipa) A= [m2] A= 0.0046 (DEBIT AMONIA) Q= v x A [m3/s] Q= 0.0091 (Head Total) Htot= (Head statis Total) ha= 2.6142 [m] (perbedaan head tekanan) ∆hp= hp1-hp2 hp1= 10 x P1/ ∆hp= 0.0000 hP1= 0.1196 (mayor dan minor losses) hl Re= v x d/ѵ Re= 62975.2066 (aliran Turbulen) (mayor Losses) hf= α x( L/d) x(v2/2g) [m] hf= 17.9086 (minor losses belokan pi hf= f x (v2/2g) [m] f(90)= hf= 5.0638 , hf= f x (v2/2g) [m] f(45)= hf= 0.0962 (ujung masuk pipa) f= 0.5600 (vd2-vs2 hf= 0.1142 (ujung keluar pipa) f= 1 hf= 0.2039 (jumlah Head Lose Minor) Hlm= 5.4781 Hlt 23.3867 (Head Total) Ht= 26.2685 ѵ(viscos 1/4 x π x d2 ha + ∆hp + hl + (vd2-vs2)/2g

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Sheet1PERHITUNGAN POMPADiketahuiv=2m/sg=9.81m/s2=0.682kg/m3L=252md=0.0762m(viscos)=0.00000242[m2/s]P1=0.08kg/m2vd=2.5m/s2P2=0.08kg/m2vs=1m/s2Ditanya: DAYA POMPA?Wp= x Q x HtotalPENYELESAIAN:(daya amoniak)(berat jenis AMONIA)= x g[kg/m2.s2]Whp= x Q x Htotal=6.6904Whp=1.6021[kW](luas permukaan pipa)A=1/4 x x d2[m2](daya pompa)A=0.0046(DEBIT AMONIA)Q=v x A[m3/s]Q=0.0091(Head Total)Htot=ha + hp + hl + (vd2-vs2)/2g(Head statis Total)ha=2.6142[m](perbedaan head tekanan)hp=hp1-hp2hp1=10 x P1/[m]hP2=0.1196hp=0.0000hP1=0.1196(mayor dan minor losses)hl=0.020 + 0.005/dRe=v x d/=0.0266Re=62975.2066(aliran Turbulen)(mayor Losses)hf= x( L/d) x(v2/2g)[m]hf=17.9086(minor losses belokan pipa)hf=f x (v2/2g)[m]f(90)=1.12924.838hf=5.0638,hf=f x (v2/2g)[m]f(45)=0.2360.472hf=0.0962(ujung masuk pipa)f=0.5600(vd2-vs2/2g)0.2676hf=0.1142(ujung keluar pipa)f=1hf=0.2039(jumlah Head Lose Minor)Hlm=5.4781Hlt23.3867(Head Total)Ht=26.2685

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