This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the calculations for the given hydraulics problems: QUESTION 5: HYDRAULICS 5.1.1. Force that must be applied to the lever to lift the load of 2 Mg if the efficiency of the lever has a slip of 20%. Step 1: Convert given values to SI units and calculate the load force. Ram piston diameter D = 220 mm = 0.22 m Plunger diameter d = 30 mm = 0.03 m Mechanical advantage of the lever MA_lever = 15 Load mass m = 2 Mg = 2000 kg Acceleration due to gravity g = 9.81 m/s^2 Load force F_ram = m × g = 2000 kg × 9.81 m/s^2 = 19620 N Step 2: Calculate the ideal mechanical advantage of the hydraulic jack. The ideal mechanical advantage of the jack is the ratio of the ram area to the plunger area. IMA_jack = A_ramA_plunger = ( D^2)/(4)( d^2)/(4) = (D^2)/(d^2) IMA_jack = (0.22 m)^2(0.03 m)^2 = (0.0484)/(0.0009) = 53.777... Step 3: Calculate the total ideal mechanical advantage of the system (lever and jack). IMA_total = MA_lever × IMA_jack = 15 × 53.777... = 806.666... Step 4: Account for the efficiency due to slip. A slip of 20% means the efficiency = 100\% - 20\% = 80\% = 0.80. The actual mechanical advantage is MA_actual = × IMA_total. The force applied to the lever (effort) is F_effort = F_ramMA_actual. F_effort = 19620 N0.80 × 806.666... = 19620 N645.333... = 30.402 N The force that must be applied to the lever is approximately 30.40 N. 30.40 N 5.1.2. Number of strokes needed to lift the load by 180 mm if there is an efficiency of 95%. Step 1: Convert given values to SI units. Load lift height H = 180 mm = 0.18 m Plunger stroke length L_s = 60 mm = 0.06 m Efficiency = 95\% = 0.95 Step 2: Calculate the volume of fluid displaced by the ram and the plunger. Volume lifted by ram V_ram = A_ram × H = ( D^2)/(4) × H = (0.22 m)^24 × 0.18 m = 0.006842 m^3 Volume displaced per plunger stroke V_plunger = A_plunger × L_s = ( d^2)/(4) × L_s = (0.03 m)^24 × 0.06 m = 0.00004241 m^3 Step 3: Calculate the ideal number of strokes. N_ideal = V_ramV_plunger = 0.006842 m^30.00004241 m^3 = 161.333... strokes Step 4: Account for the efficiency to find the actual number of strokes. Since there is an efficiency of 95%, more strokes are needed. N_actual = N_ideal = (161.333...)/(0.95) = 169.824... strokes Since the number of strokes must be a whole number, we round up. The number of strokes needed is 170 strokes. 170 strokes 5.2. A two-cylinder water pump is running at 120 r/min. The diameter of the plunger is 150 mm and the stroke length is 250 mm. 5.2.1. The volume of water delivered in l/s. Step 1: Convert given values to SI units. Number of cylinders n_c = 2 (assuming single-acting cylinders) Running speed N = 120 r/min = (120)/(60) r/s = 2 r/s Plunger diameter d = 150 mm = 0.15 m Stroke length L_s = 250 mm = 0.25 m Step 2: Calculate the area of the plunger. A_plunger = ( d^2)/(4) = (0.15 m)^24 = 0.017671 m^2 Step 3: Calculate the volume delivered per second. For a two-cylinder single-acting pump, there are two delivery strokes per revolution. Volume delivered per stroke V_stroke = A_plunger × L_s = 0.017671 m^2 × 0.25 m = 0.0044178 m^3 Volume delivered per second Q = n_c × V_stroke × N = 2 × 0.0044178 m^3 × 2 r/s = 0.017671 m^3/s Step 4: Convert the volume to litres per second. 1 m^3 = 1000 l Q = 0.017671 m^3/s × 1000 l/m^3 = 17.671 l/s The volume of water delivered is approximately 17.67 l/s. 17.67 l/s 5.2.2. The pressure needed to pump the water through a height of 30 m. Step 1: Identify the given values. Height h = 30 m Density of water = 1000 kg/m^3 Acceleration due to gravity g = 9.81 m/s^2 Step 2: Calculate the pressure using the hydrostatic pressure formula. P = g h P = 1000 kg/m^3 × 9.81 m/s^2 × 30 m = 294300 Pa Step 3: Convert the pressure to kilopascals. P = 294300 Pa = 294.3 kPa The pressure needed is 294.3 kPa. 294.3 kPa 5.3. An irrigation system fills an empty circular dam with a diameter of 50 m and a height of 20 m with a pressure of 250 MPa. Determine the amount of work needed to fill the dam to capacity. Step 1: Identify the given values and calculate the volume of the dam. Dam diameter D = 50 m Dam height H = 20 m Density of water = 1000 kg/m^3 Acceleration due to gravity g = 9.81 m/s^2 The pressure of 250 MPa is the pump's capability and is not directly used to calculate the work done to fill the dam (potential energy). Volume of the dam V = ( D^2)/(4) × H V = (50 m)^24 × 20 m = ( × 2500)/(4) × 20 m^3 = 12500 m^3 V ≈ 39269.9 m^3 Step 2: Calculate the mass of the water in the dam. m = × V = 1000 kg/m^3 × 12500 m^3 = 12500000 kg m ≈ 39269908 kg Step 3: Determine the average height the water is lifted. When filling an empty dam, the water is lifted from the bottom to an average height of half the dam's height. h_avg = (H)/(2) = 20 m2 = 10 m Step 4: Calculate the work needed to fill the dam (potential energy gained by the water). W = m g h_avg W = (12500000 kg) × 9.81 m/s^2 × 10 m W = 125000000 × 9.81 J W ≈ 3851000000 J W ≈ 3.851 × 10^9 J The amount of work needed to fill the dam to capacity is approximately 3.85 × 10^9 J (or 3.85 GJ). 3.85 × 10^9 J What's next? Send 'em! 📸