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.
ScanSolve AI Answer
🤍🦋Mashiyamahle🤍🦋, let's knock this out. Here's how to calculate the flow of water in the pipe using the Venturi meter: Given: Throat diameter, d_2 = 200 mm = 0.200 m Pipe diameter, d_1 = 300 mm = 0.300 m Manometer reading (mercury), h_m = 135 mm = 0.135 m Fluid in pipe: Water Manometer fluid: Mercury Assumptions: Density of water, _w = 1000 kg/m^3 Density of mercury, _m = 13600 kg/m^3 Acceleration due to gravity, g = 9.81 m/s^2 Coefficient of discharge for the Venturi meter, C_d = 0.98 (a typical value when not specified). Step 1: Calculate the cross-sectional areas of the pipe and the throat. The area of the pipe (A_1) and the throat (A_2) are calculated using the formula for the area of a circle, A = ()/(4)d^2. A_1 = ()/(4) (0.300 m)^2 = ()/(4) × 0.09 m^2 ≈ 0.070686 m^2 A_2 = ()/(4) (0.200 m)^2 = ()/(4) × 0.04 m^2 ≈ 0.031416 m^2 Step 2: Convert the manometer reading from mm of mercury to an equivalent head of water (h). The differential head h in terms of the flowing fluid (water) is given by: h = h_m ( (_m)/(_w) - 1 ) h = 0.135 m ( 13600 kg/m^31000 kg/m^3 - 1 ) h = 0.135 m (13.6 - 1) h = 0.135 m × 12.6 h = 1.701 m of water Step 3: Apply the Venturi meter flow rate formula. The formula for the volumetric flow rate (Q) through a Venturi meter is: Q = C_d A_1 A_2 sqrt(2gh)sqrt(A_1^2 - A_2^2) Step 4: Substitute the calculated values and solve for Q. First, calculate the term sqrt(2gh): sqrt(2gh) = sqrt(2 × 9.81 m/s)^2 × 1.701 m sqrt(2gh) = sqrt(33.37062 m)^2/s^2 ≈ 5.7767 m/s Next, calculate the denominator term sqrt(A_1^2 - A_2^2): A_1^2 = (0.070686 m^2)^2 ≈ 0.0049964 m^4 A_2^2 = (0.031416 m^2)^2 ≈ 0.00098696 m^4 sqrt(A_1^2 - A_2^2) = sqrt(0.0049964 m)^4 - 0.00098696 m^4 sqrt(A_1^2 - A_2^2) = sqrt(0.00400944 m)^4 ≈ 0.063320 m^2 Now, substitute all values into the flow rate formula: Q = 0.98 × (0.070686 m^2) × (0.031416 m^2) × (5.7767 m/s)0.063320 m^2 Q = 0.98 × 0.012829 m^3/s0.063320 Q = 0.98 × 0.20260 m^3/s Q ≈ 0.19855 m^3/s Rounding to three significant figures: Q ≈ 0.199 m^3/s What's next?