Let's solve each problem step by step: --- ## **Q4 Solution** **Given:** - Engine speed (\( N \)) = 140 RPM - Power (\( P_{out} \)) = 800 HP - Efficiency (\( \eta \)) = 25% - Fuel = Liquid Butane - Time = 12 hours We are to find: 1. **Output torque** 2. **Fuel consumption in gallons for 12 hours** --- ### **Step 1: Output Torque** First, convert horsepower to watts: \[ 1\, \text{HP} = 745.7\, \text{W} \] \[ P_{out} = 800 \times 745.7 = 596,560\, \text{W} \] Recall the formula: \[ P = 2\pi N T \] Where: - \( P \) = Power in watts - \( N \) = Rotational speed in rev/sec - \( T \) = Torque in Nm Convert RPM to RPS: \[ N = 140\ \text{RPM} = \frac{140}{60} = 23.33\ \text{rev/sec} \] Rearrange for torque: \[ T = \frac{P}{2\pi N} \] Plug in values: \[ T = \frac{596,560}{2\pi \times 23.33} \] \[ T = \frac{596,560}{146.54} \approx 4,071\, \text{Nm} \] --- ### **Step 2: Fuel Consumption** #### **2.1 Find Total Input Power** \[ \text{Efficiency} = \frac{P_{out}}{P_{in}} \implies P_{in} = \frac{P_{out}}{\eta} \] \[ P_{in} = \frac{596,560}{.25} = 2,386,240\, \text{W} \] #### **2.2 Find Total Energy Consumed in 12 Hours** \[ \text{Energy} = P_{in} \times t \] \[ t = 12\, \text{hrs} = 12 \times 360 = 43,200\, \text{s} \] \[ \text{Total energy used} = 2,386,240 \times 43,200 = 103,077,568,000\, \text{J} \] #### **2.3 Energy Content of Butane** \[ \text{Heating value for liquid butane} \approx 45.7\, \text{MJ/kg} \] \[ = 45,700,000\, \text{J/kg} \] #### **2.4 Find Mass of Butane Consumed** \[ \text{Mass of butane} = \frac{\text{Total energy used}}{\text{Heating value}} \] \[ = \frac{103,077,568,000}{45,700,000} \approx 2,256\, \text{kg} \] #### **2.5 Convert Mass to Gallons** - **Density of liquid butane** ≈ .584 g/cm³ = 584 kg/m³ \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{2,256}{584} \approx 3.864\, \text{m}^3 \] \[ 1\, \text{m}^3 = 264.172\, \text{gal} \] \[ \text{Volume in gallons} = 3.864 \times 264.172 \approx 1,020\, \text{gal} \] --- ### **Final Answers for Q4** - **Output Torque:** \( \boxed{4,071\, \text{Nm}} \) - **Fuel Consumption in 12 hours:** \( \boxed{1,020\, \text{gallons}} \) --- ## **Q5 Solution** **Given:** - Formation (sandstone) density = \( 2.65\, \text{g/cm}^3 \) - Porosity (\( \phi \)) = 15% - Fluid density = \( .75\, \text{g/cm}^3 \) - Depth = \( 9,600\, \text{ft} \) - Mud weight = \( 12.5\, \text{ppg} \) Find: 1. Formation pressure 2. Overburden pressure 3. Fracture pressure --- ### **1. Formation Pressure** Formation pressure is usually **hydrostatic pressure** of the formation fluid: \[ P_{formation} = \rho_{fluid} \times g \times h \] - \( \rho_{fluid} = .75\, \text{g/cm}^3 = .75 \times 100 = 750\, \text{kg/m}^3 \) - \( h = 9,600\, \text{ft} \times .3048 = 2,926\, \text{m} \) - \( g = 9.81\, \text{m/s}^2 \) \[ P_{formation} = 750 \times 9.81 \times 2,926 = 21,536,805\, \text{Pa} \] \[ = 21.54\, \text{MPa} \] \[ 1\, \text{psi} = 6,894.76\, \text{Pa} \] \[ P_{formation} = \frac{21,536,805}{6,894.76} \approx 3,125\, \text{psi} \] --- ### **2. Overburden Pressure** Overburden pressure is the pressure exerted by the weight of the overlying rocks: \[ P_{overburden} = \rho_{rock} \times g \times h \] - \( \rho_{rock} = 2.65\, \text{g/cm}^3 = 2,650\, \text{kg/m}^3 \) \[ P_{overburden} = 2,650 \times 9.81 \times 2,926 = 76,024,629\, \text{Pa} \] \[ = 76.02\, \text{MPa} \] \[ P_{overburden} = \frac{76,024,629}{6,894.76} \approx 11,022\, \text{psi} \] --- ### **3. Fracture Pressure** A typical estimate for fracture pressure is: \[ P_{fracture} \approx .7 \times P_{overburden} \] \[ P_{fracture} = .7 \times 11,022 = 7,715\, \text{psi} \] --- ### **Final Answers for Q5** - **Formation Pressure:** \( \boxed{3,125\, \text{psi}} \) - **Overburden Pressure:** \( \boxed{11,022\, \text{psi}} \) - **Fracture Pressure:** \( \boxed{7,715\, \text{psi}} \) --- If you need more details or clarification on any step, just ask!