Fracturing treatment design process Problem well identification Fluid selection Treatment selection Treatment design Operational constraints Treatment result prediction Operational stimulation program Scheduling and logistics Site preparation Proppant selection Evaluation cycle Job execution 1 Real Time Pressure Analysis ¥ Nolte Fracture growth Analysis Ð Log‐log diagnos>c plot 2 Model comparison KGD model PKN model ¥ High and short fractures ¥ Rectangular ¥ Decreasing Net pressure ¥ Long and shallow fractures ¥ Ellip>cal ¥ Increasing Net Pressure 1/4 21μq E’3 ΔP = Pnet,w= 64πhfL2 1/4 16μq E’3 L ΔP = Pnet,w= πh f Low Fluid leakoff 1/6 E’q3 L(t)= 0.38 μhf 3 t2/3 1/6 μq3 ww= 1.48 E’hf3 t1/3 1/5 E’q3 L(t)= 0.39 μhf4 1/5 t4/5 ww= 2.18 μq2 E’hf High Fluid leakoff L(t)= q t1/2 2πCLhf t1/5 1/4 μq2 ww= 4 π E’CLhf t1/8 Model comparison KGD model PKN model ¥ High and short fractures ¥ Rectangular ¥ Decreasing Net pressure ¥ Long and shallow fractures ¥ Ellip>cal ¥ Increasing Net Pressure c 1/4 ΔP = Pnet,w= L 1/2 1/6 E’q3 L(t)= 0.38 μhf3 ΔP = t2/3 L ΔP = Pnet,w= c Low Fluid leakoff 1/5 E’q3 L(t)= 0.39 μhf4 c High Fluid leakoff t⅓ q t1/2 L(t)= 2πCLhf ΔP = c t⅕ t4/5 ΔP = c t⅛ Log‐Log fracture pressure diagnos>c plot Net Pressure, psi 1000 m=0 Stable height growth m= Restricted height growth m>1 Near well restriction m=1/8 to 1/5 contained height 100 mme, min 100 Methods to evaluate fracture geometry Source Cipolla and Wright 2000 6 What Happened? 7 What Happened (frac&pack 8 What Happened 9 What Happened KW‐7 main job 100 8000 90 80 6000 70 5000 60 50 4000 40 3000 30 Pump rate (bpm) & prop conc (ppg) Pressure (psi) 7000 2000 20 1000 10 0 0 0 500 1000 1500 2000 2500 3000 3500 4000 :me (s) 10 What Happened 11 ... 21μq E 3 ΔP = Pnet,w= 64πhfL2 1/4 16μq E 3 L ΔP = Pnet,w= πh f Low Fluid leakoff 1/6 E’q3 L(t)= 0 .38 μhf 3 t2 /3 1/6 μq3 ww= 1.48 E’hf3 t1 /3 1 /5 E’q3 L(t)= 0 .39 μhf4 1 /5 ... 6000 70 50 00 60 50 4000 40 30 00 30 Pump rate (bpm) & prop conc (ppg) Pressure (psi) 7000 2000 20 1000 10 0 0 0 50 0 1000 150 0 2000 250 0 30 00 35 00 4000 ... 1/6 E’q3 L(t)= 0 .38 μhf3 ΔP = t2 /3 L ΔP = Pnet,w= c Low Fluid leakoff 1 /5 E’q3 L(t)= 0 .39 μhf4 c High Fluid leakoff t⅓ q t1/2 L(t)= 2πCLhf ΔP = c t⅕ t4 /5 ΔP = c t⅛