6.5 in., 100 lb/ft; length $=15\%$ of total depth $D$
Design safety factor, $SF$
1.5
Engine efficiency
85%
Transmission (engine-to-drawworks) efficiency
90%
Fuel consumption rate
20 gal/hr, density 6.6 lb/gal
Heat of combustion
20,000 BTU/lb
Unit conversions
1 BTU = 779 ft-lbf; 1 HP = 33,000 ft-lbf/min
Buoyancy and travelling-block/hook weight are explicitly ignored throughout.
Find. (a) The maximum depth the drilling-line/block-and-tackle capacity allows. (b) The maximum hoisting (block) speed available when the hook carries its highest possible load.
Approach. Size the allowable hook load from the drilling line's rated strength (frictionless sheaves, since no reeving-efficiency table is given), express the string's own weight as a function of depth from the DC/DP split, solve for the governing depth, then find the surface power actually delivered to the drawworks from the engine's fuel-energy balance and convert that power to a hoisting speed at the depth-(a) hook load.
Part (a) — allowable hook load from the drilling line. With $n=8$ lines and no reeving-efficiency table given, assume frictionless sheaves, so each of the 8 lines shares the hook load equally. Allowable line pull $=90{,}000/SF=90{,}000/1.5=60{,}000$ lbf, so $$W_{max}=n\times60{,}000=8\times60{,}000\Rightarrow\boxed{W_{max}=480{,}000\ \text{lbf}}$$
Part (a) — string weight vs. depth. Drill collars occupy $0.15D$ at 100 lb/ft and drill pipe the remaining $0.85D$ at 20 lb/ft: $$W(D)=0.85D(20)+0.15D(100)=17D+15D=32D\ \text{lb}$$
Part (a) — solve for maximum depth. Setting $W(D)=W_{max}$: $32D=480{,}000\Rightarrow D=480{,}000/32$, so $\boxed{D_{max}=15{,}000\ \text{ft}}$. The highest possible hook load therefore occurs at 15,000 ft, where $W=480{,}000$ lbf — this is the load used in Part (b).
Part (b) — available fuel energy rate. Mass burn rate $=20\ \text{gal/hr}\times6.6\ \text{lb/gal}=132$ lb/hr, so the heat input rate is $$\dot{Q}=132\times20{,}000=2{,}640{,}000\ \text{BTU/hr}$$ Converting to mechanical-equivalent power: $2{,}640{,}000\times779=2{,}056{,}560{,}000$ ft-lbf/hr $=34{,}276{,}000$ ft-lbf/min, so $$HP_{fuel}=34{,}276{,}000/33{,}000\Rightarrow\boxed{HP_{fuel}=1{,}038.7\ \text{HP}}$$
Part (b) — power delivered to the drawworks. Apply the two efficiencies in series (engine, then transmission to drawworks): $$HP_{drawworks}=1{,}038.7\times0.85\times0.90\Rightarrow\boxed{HP_{drawworks}=794.6\ \text{HP}}$$
Part (b) — hoisting speed at maximum hook load. With no further mechanical losses assumed between drawworks output and the travelling block (frictionless sheaves, as in Part (a)), overall power balance gives $HP_{drawworks}\times33{,}000=W_{max}\times v$. Using the Part-(a) maximum hook load $W_{max}=480{,}000$ lbf: $$v=\frac{794.6\times33{,}000}{480{,}000}\Rightarrow\boxed{v_{max}=54.6\ \text{ft/min}}$$
Check: frictionless sheaves ($E_n=1$) are assumed for both the fast-line tension in Part (a) and the block-speed power balance in Part (b), since no reeving-efficiency table is supplied — the established convention when such a table is absent. Part (b) assumes the maximum hoisting speed is evaluated at the same maximum hook load found in Part (a) ("when the hook load is at its highest possible level"), i.e. at 15,000 ft.
