Air flows through a pipe at a rate of 128 L/s. The pipe consists of two sections of diameters 0.21 m and 0.08 m with a smooth reducing section that connects them. A water manometer measures the pressure difference between the two pipe sections. Neglecting frictional effects, determine the differential height of water in meter unit between the two pipe sections. Take the air density to be 1.20 kg/m3.
Question
Air flows through a pipe at a rate of 128 L/s. The pipe consists of two sections of diameters 0.21 m and 0.08 m with a smooth reducing section that connects them. A water manometer measures the pressure difference between the two pipe sections. Neglecting frictional effects, determine the differential height of water in meter unit between the two pipe sections. Take the air density to be 1.20 kg/m3.
Solution
To solve this problem, we can use the principle of conservation of mass and Bernoulli's equation.
Step 1: Convert the flow rate from L/s to m^3/s 128 L/s = 0.128 m^3/s
Step 2: Calculate the velocity in the larger pipe section (V1) using the equation of continuity (A1V1 = A2V2). The area of a pipe can be calculated using the formula A = πd^2/4.
A1 = π(0.21 m)^2/4 = 0.0347 m^2 A2 = π(0.08 m)^2/4 = 0.00502 m^2
V1 = Q / A1 = 0.128 m^3/s / 0.0347 m^2 = 3.69 m/s
Step 3: Calculate the velocity in the smaller pipe section (V2) using the equation of continuity (A1V1 = A2V2).
V2 = Q / A2 = 0.128 m^3/s / 0.00502 m^2 = 25.5 m/s
Step 4: Use Bernoulli's equation to find the pressure difference between the two sections. Bernoulli's equation is P1 + 1/2ρV1^2 + ρgh1 = P2 + 1/2ρV2^2 + ρgh2. Since the pipe is horizontal, h1 = h2, and the equation simplifies to P1 + 1/2ρV1^2 = P2 + 1/2ρV2^2.
ΔP = P2 - P1 = 1/2ρ(V2^2 - V1^2) = 1/2 * 1.20 kg/m^3 * ((25.5 m/s)^2 - (3.69 m/s)^2) = 306.8 Pa
Step 5: Convert the pressure difference to a water height difference. The pressure due to a column of fluid is given by P = ρgh, so h = P / (ρg). The density of water is 1000 kg/m^3 and g is 9.81 m/s^2.
Δh = ΔP / (ρ_water * g) = 306.8 Pa / (1000 kg/m^3 * 9.81 m/s^2) = 0.0313 m
So, the differential height of water between the two pipe sections is 0.0313 meters.
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