Geothermal drilling in the Godavari rift: heat, permeability and a small tremor
A renewable-energy company is testing a geothermal prospect near Chintoor in the Godavari valley of Andhra Pradesh. A 3,000 m borehole records a mean annual surface temperature of 30°C, 78°C at 1,000 m, 126°C at 2,000 m and 174°C at 3,000 m. The company needs at least 150°C at the wellhead for a small binary-cycle plant. A pump test at 2,500 m yields only 2 litres per second from fractured sandstone, whereas the plant design needs 30–50 litres per second. The site lies on a fault bounding a Gondwana rift basin. A nearby seismic station records a small tremor with an S–P interval of 2.4 s; the epicentre is about 18 km from the borehole. Assume Vp = 6.0 km/s and Vs = 3.5 km/s. District officials ask whether the company should drill deeper, stimulate the existing well, or move to another fault block.
1. Calculate the geothermal gradient and the depth at which 150°C should be reached. Explain why this gradient is higher than the 25–30°C/km typical of a stable shield.
Read the worked answer
Gradient = (174 − 30)°C ÷ 3 km = 48°C/km. The 1–3 km interval gives (174 − 78) ÷ 2 = 48°C/km, so the gradient is consistent. Depth to 150°C = (150 − 30) ÷ 48 = 2.5 km. At 2.5 km the predicted temperature is 30 + (48 × 2.5) = 150°C, exactly the plant target.
The gradient exceeds the shield average because this is a rift basin, not stable craton. Crustal thinning raises heat flow; faults allow convective circulation of hot fluids; and the low thermal conductivity of Gondwana shale and sandstone acts as a thermal blanket, steepening the gradient in the sedimentary pile. The high gradient therefore does not prove that fluids can be produced. Temperature and permeability are separate tests: the borehole passes the thermal test but fails the flow test.
2. Estimate the hypocentral depth of the tremor and judge whether it shows magma at 2.5 km.
Read the worked answer
Use d = Δt × (VpVs)/(Vp − Vs). VpVs/(Vp − Vs) = (6.0 × 3.5)/(6.0 − 3.5) = 21/2.5 = 8.4 km/s. So d = 2.4 × 8.4 = 20.2 km.
If the epicentre is 18 km away, depth = √(20.2² − 18²) = √(408 − 324) = √84 ≈ 9.2 km. The tremor is therefore a crustal fault event at about 9 km depth, not a magma body at 2.5 km. It is consistent with brittle failure in the upper crust and with an active rift fault.
This changes the interpretation of the geothermal system. The heat is more likely conducted through rift sediments and circulated along faults than supplied by shallow magma. The 2.5 km fracture may still be a useful fluid pathway, but the tremor gives no evidence that it is magmatically heated or unusually permeable.
3. Recommend the next operational step: deepen the well, stimulate the existing well, or move to another fault block. Justify the decision with the temperature and flow data.
Read the worked answer
The temperature target is already met at 2.5 km, so the failure is hydraulic, not thermal. Deepening to 3,000 m adds only about 24°C while increasing cost and risk; it will not help if the fault is clay-sealed. Blind deepening is therefore the weakest option.
The next step should be a diagnostic injection or packer test and a fault-image survey. If the fracture is open but poorly connected, a limited hydraulic stimulation of the damage zone may raise flow. If it is clay-filled or healed, stimulation is unlikely to succeed and a deviated sidetrack into a nearby fault damage zone is technically better than a vertical deepening.
Because the site is on a fault near villages, stimulation needs a microseismic monitoring array and a traffic-light protocol: stop injection above a small magnitude threshold. Set an abandonment rule: if sustainable flow remains below about 10 L/s, a 30–50 L/s plant cannot be supplied, so the project should move to another fault block rather than spend more on the same hole.