Path of Least Resistance #1 - Questions and Answers

Question 1 An oil pump is drawing 44 kW of electric power while pumping oil with p= 860 kg/m^3 at a rate of 0.1 m^3/s. The inlet and outlet diameters of the piper are 8 cm and 12 cm, respectively. If the pressure rise of oil in the pump is measured to be 500 kPa and the motor efficiency is 90 percent, determine the mechanical efficiency of the pump. Answer 1 First we calculate the mechanical point power output from the efficienty relation: Wmech,out = ηmotorWel, E = 0.9 * 44 kW = 39.6 kW This serves as the power input for the mechanical efficiency. The output is obtained from the rates of the changes of the mechanical energy: = ρVgh + ( (V₂/2πr₂²)² - (V₁/2πr₁²)²) = VΔP + (ρV / 2π²) * (1/r₂⁴ - 1/r₁⁴) = 0.1 * 510⁵ Pa + (860 kg/m ³ * 0.1m³/s * (1/(610⁻²m)⁴ - 1/(4*10⁻²m)⁴) / 2π² Finally, the mechanical efficiency is: ηmech = Wmech / Wmech,out = 36.2kW / 39.6kW = 0.91 Answer 2 To determine the mechanical efficiency of the pump, we need to calculate the actual power output and compare it to the ideal power output. First, let's calculate the actual power output of the pump. The power output can be determined using the formula: Pactual = ρ * g * Q * H where: ρ is the density of the oil (860 kg/m ³), g is the acceleration due to gravity (9.8 m/s²), Q is the flow rate of the oil (0.1 m³/s), H is the pressure rise of the oil in the pump (500 kPa). Next, let's calculate the ideal power output of the pump. The ideal power output can be determined using the formula: Pideal = Pactual / Motor efficiency Given that the motor efficiency is 90 percent (or 0.9), we can calculate the ideal power output. Finally, the mechanical efficiency of the pump can be calculated by dividing the actual power output by the ideal power output: Mechanical efficiency = Pactual / Pideal Now, let's substitute the given values into the equations and calculate the mechanical efficiency. Given: ρ = 860 kg/m ³ Q = 0.1 m³/s H = 500 kPa Motor efficiency = 0.9 Calculating the actual power output: Pactual = 860 kg/m ³ * 9.8 m/s² * 0.1 m³ / s * 500 kPa Calculating the ideal power output: Pideal = Pactual / Motor efficiency Finally, calculating the mechanical efficiency: Mechanical efficiency = Pactual / Pideal Substituting the calculated values into the equation, we find that the mechanical efficiency of the pump is approximately 0.91. Answer 3 To determine the mechanical efficiency of the pump, we need to calculate the power output of the pump and compare it to the power input. The power output of the pump can be calculated using the equation: Poutput = ΔP * Q where ΔP is the pressure rise of the oil in the pump and q is the volume flow rate of the oil. Given that the pressure rise ΔP is 500 kPa and the volume flow rate Q is 0.1m³ / s, we can calculate the power output as follows: Poutput = ΔP * Q = 50010³ Pa * 0.1 m³ / s Next, we need to calculate the power input to the motor. The power input can be calculated using the equation: Pinput = Poutput / efficiency where the efficiency is given as 90 percent or 0.9. Pinput = Poutput / efficiency = (50010³ Pa * 0.1 m³/s) / 0.9 Now, we can determine the mechanical efficiency of the pump by comparing the power output to the power input: Mechanical efficiency = Poutput / Pinput Substituting the values, we get: Mechanical efficiency = (50010³ Pa * 0.1 m³/s) / [(50010³ Pa * 0.1 m³/s) / 0.9] Simplifying this expression, we find: Mechanical efficiency = 0.9 Therefore, the mechanical efficiency of the pump is 0.9 or 90 percent. Answer 4 Given: Electric power input to the pump, Pinput = 44 kW Density of the oil, ρ = 860 kg/m ³ Oil flow rate, Q = 0.1 m³/s Inlet diameter of the pipe, D₁ = 8 cm Outlet diameter of the pipe, D₂ = 12 cm Pressure rise of oil in the pump, ΔP = 500 kPa Motor efficiency, ηmotor = 90% First, let's calculate the cross-sectional areas of the inlet and outlet pipes using the given diameters: A₁ = πD₁²/4 and A₂ = πD₂²/4 Next, we can calculate the power input to the pump in terms of the flow rate and pressure rise: Pinput = ΔPQρg where g is the acceleration due to gravity. Rearranging this equation, we can solve for g: g = Pinput / (ΔPQρ) Now, we can calculate the power output from the pump: Poutput = ηmotor Pinput Finally, the mechanical efficiency of the pump, ηmechanical, can be calculated by dividing the power output by the power input: ηmechanical = Poutput / Pinput Now, let's substitute the given values into the equations and calculate the mechanical efficiency. Answer 5 Solution: Let's begin by calculating the hydraulic power delivered to the oil by the pump. The hydraulic power (Ph) can be calculated using the formula: Ph = ρ * g * Q * H where ρ is the density of the oil, g is the acceleration due to gravity, Q is the volume flow rate of the oil, and H is the pressure rise of the oil in the pump. Given values: ρ = 860kg/m³ Q = 0.1 m³/s H = 500 kPa First, we need to convert the pressure rise H from kilopascals (kPa) to pascals (Pa): H = 50010³ Pa Next, we can substitute the given values into the formula to calculate Ph: Ph = 860kg/m³ * 9.8 m/s² * 0.1 m³ / s * 50010³ Pa Simplifying the expression, we get: Ph = 4.24610⁶ W Now, let's calculate the power supplied to the pump motor. The electric power (Pelectric) supplied to the pump motor is given as 44 kW. Converting 44 kW to watts, we have: Pelectric = 4410³ W The motor efficiency (ηmotor) is given as 90 percent. We can convert this percentage to a decimal: ηmotor = 0.9 Finally, we can calculate the mechanical efficiency (ηmechanical) using the formula: ηmechanical = Ph / Pelectric * ηmotor Substituting the calculated values, we get: ηmechanical = 4.24610⁶ W / 4410³ W * 0.9 Simplifying the expression, we find: ηmechanical = 97.05% Therefore, the mechanical efficiency of the pump is 97.05%. Question 2 A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge destiny of +6.37 × 10⁻⁶ C/m². A charge of -0.500µC is now introduced at the center of the cavity inside the sphere. (a) What is the new carge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere? Answer 1 (a) Calculation: Solve for the new charge density on the outside of the sphere: As equation (22.10) mentions, the electric field at the surface of a conductor is given by E = σ / ε₀ We use the same technique as an example (22.5), the electric field at the surface of a charged conducting sphere is given by E= 1 / (4πε₀ * Q / r²) Substituting from the previous calculation and solve for the charge density on the outside of the sphere, then we get σout = Eε₀ = 1 / (4πε₀ * Q / r²) * ε₀ = Q / (4πr²) = -0.5 * 10⁻⁶ C / (4π(0.25m)²) = -6.366 × 10⁻⁷ C/m² In order to evaluate the new charge density on the outside of the sphere, we use the following relation: σnew = σout - σinduced = -6.37 × 10⁻⁶ C/m² - (-6.366 × 10⁻⁷ C/m²) Answer 2 (b) Calculation: Solve for the strength of the electric field just outside of the sphere: As equation (22.10) mentions, the electric field at the surface of a conductor is given by E = σ / ε₀ = 5.733 × 10⁻⁶ C/m² / 8.854 × 10⁻¹² N*m²/C² = 6.475 × 10⁵ N/C So, the strength of the electric field just outside of the sphere is 6.475 × 10⁵ N/C Answer 3 (c) Calculation: As equation (22.8) mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by ΦE = Qenclosed / ε₀ = -0.510⁻⁶C / 8.85410⁻¹² C²/Nm² = -5.647 × 10⁴ Nm²/C So, the electric flux through a spherical surface inside the inner surface of the sphere is -5.647 × 10⁴ N*m²/C Answer 4 Step 1: (a) To find the new charge density on the outside of the sphere, we need to consider the effect of the charge introduced inside the cavity. The charge inside the cavity induces an equal and opposite charge on the inner surface of the sphere. Since the sphere is a conductor, this induced charge redistributes on the outer surface. The new charge density on the outside of the sphere is given by the sum of the original charge density and the induced charge density. Let σ be the original charge density and σinduced be the induced charge density. Then the new charge density σnew is: σnew = σ + σinduced The induced charge density is equal in magnitude to the charge inside the cavity divided by the inner surface area of the sphere. Therefore: σinduced = Qinside / Ainner where Qinside is the charge inside the cavity and Ainner is the inner surface area of the sphere. Substituting the given values, we have: σinduced = -0.500µC / 4πrinner² Now we can substitute the values and calculate σnew: σnew = 6.37 * 10⁻⁶ C/m² + (-0.500 * 10⁻⁶ C / 4π(0.200 m)²) Step 2: (b) To calculate the strength of the electric field just outside the sphere, we can use Gauss's law. Gauss's law states that the electric field through a closed surface is proportional to the total charge enclosed by that surface. The electric field just outside the sphere can be found using the formula: E = σnew / ε₀ where ε₀ is the vacuum permittivity. Substituting the values, we have: E = σnew / ε₀ Step 3: (c) The electric flux through a spherical surface just inside the inner surface of the sphere can be calculated using the formula: Φ = Qenclosed / ε₀ where Qenclosed is the charge enclosed by the surface. In this case, the charge enclosed is the charge inside the cavity. Substituting the given values, we have: Φ = Qinside / ε₀ Answer 5 Outer radius of the conducting sphere, Router = 0.250 m Inner radius of the conducting sphere, Rinner = 0.200 m Surface charge density on the outer surface, σ = +6.37×10⁻⁶ C/m² Charge introduced inside the cavity, q = -0.500μC = -0.500×10⁻⁶ C (a) To find the new charge density on the outside of the sphere, we need to consider the charges on the inner and outer surfaces separately. The charge on the inner surface of the sphere is equal in magnitude but opposite in sign to the charge introduced inside the cavity. Therefore, the charge on the inner surface, Qinner = -q = -(-0.500×10⁻⁶C) = 0.500×10⁻⁶C. The total charge on the outer surface of the sphere, Qouter is the sum of the charge on the outer surface due to surface charge density and the charge on the inner surface due to the charge introduced inside the cavity. Since the inner and outer surfaces are concentric, the total charge on the outer surface is equal to the charge on the inner surface. Hence, Qouter = Qinner = 0.500×10⁻⁶C. The new charge density on the outside of the sphere, σnew is given by the ratio of the total charge on the outer surface to the area of the outer surface. σnew = Qouter / 4πRouter² Calculating the value: σnew ≈ 1.273×10⁻⁵ C/m² Therefore, the new charge density on the outside of the sphere is 1.273×10⁻⁵ C/m². (b) To calculate the strength of the electric field just outside the sphere, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε₀). The electric field just outside the sphere is radially outward and has the same magnitude at every point on the surface due to its symmetry. Therefore, we can use a Gaussian surface in the form of a sphere with a radius slightly greater than the outer radius of the conducting sphere. The electric flux through a closed surface is given by: ΦE = EA where E is the electric field and A is the area of the closed surface. The total charge enclosed by the Gaussian surface is Qtotal = Qouter = 0.500×10⁻⁶C. The electric flux through the Gaussian surface is also equal to Qtotal / ε₀ Therefore, we have: Qtotal / ε₀ = EA E = Qtotal / (ε₀A) E = Qouter / (ε₀A) Substituting the given values: E = 0.500 × 10⁻⁶ C / (ε₀ * 4πRouter²) The area of the closed surface is A = 4πRouter² Simplifying further: E = 0.500 × 10⁻⁶ C / (ε₀* 4π(0.250 m)²) Calculating the value: E ≈ 2.268×10⁵ N/C Therefore, the strength of the electric field just outside the sphere is 2.268×10⁵ N/C. (c) To find the electric flux through a spherical surface just inside the inner surface of the conducting sphere, we can use Gauss's law again. Consider a Gaussian surface in the form of a sphere with a radius slightly less than the inner radius of the conducting sphere. The electric field inside the conducting sphere is zero since the electric charges redistribute themselves on the inner surface to cancel out the electric field inside. Therefore, the electric flux through the Gaussian surface is zero, and the electric flux through a spherical surface just inside the inner surface of the sphere is also zero. In mathematical notation, Φε = 0. Hence, the electric flux through a spherical surface just inside the inner surface of the sphere is 0.