Water flows at a rate of 0.035 m3/s in a horizontal pipe whose diameter is reduced from x cm to y cm by a reducer. If the pressure at the centerline is measured to be P1 kPa and P2 kPa before and after the reducer, respectively, determine the irreversible head loss in "meter unit" in the reducer. Take the kinetic energy correction factors to be 1.05. Take the density of water to be ρ = 1000 kg/m3. (Round the final answer to two decimal places.)Here,x=13 cmy=8 cmP1=492 kPaP2=432kPa
Question
Water flows at a rate of 0.035 m3/s in a horizontal pipe whose diameter is reduced from x cm to y cm by a reducer. If the pressure at the centerline is measured to be P1 kPa and P2 kPa before and after the reducer, respectively, determine the irreversible head loss in "meter unit" in the reducer. Take the kinetic energy correction factors to be 1.05. Take the density of water to be ρ = 1000 kg/m3. (Round the final answer to two decimal places.)Here,x=13 cmy=8 cmP1=492 kPaP2=432kPa
Solution
To solve this problem, we need to use the Bernoulli's equation which is a statement of the conservation of energy principle for flowing fluids. It accounts for gravitational potential energy, kinetic energy, and fluid pressure.
The Bernoulli's equation is given by:
P1/ρ + V1²/2 + gz1 = P2/ρ + V2²/2 + gz2 + hL
where: P1 and P2 are the pressures at the two points, ρ is the density of the fluid, V1 and V2 are the velocities of the fluid at the two points, g is the acceleration due to gravity, z1 and z2 are the heights of the two points above some reference level, hL is the head loss between the two points.
Given that the pipe is horizontal, z1 = z2, so the height term can be ignored. The equation then simplifies to:
P1/ρ + V1²/2 = P2/ρ + V2²/2 + hL
We can rearrange this to find the head loss:
hL = (P1 - P2)/ρ + (V1² - V2²)/2
We know the pressures P1 and P2, and the density ρ. We need to find the velocities V1 and V2.
The velocity of a fluid in a pipe is given by the equation V = Q/A, where Q is the flow rate and A is the cross-sectional area of the pipe.
The cross-sectional area of a pipe with diameter d is given by A = πd²/4.
So, V1 = Q/A1 = Q/(π*(x/100)²/4) and V2 = Q/A2 = Q/(π*(y/100)²/4).
Substituting these into the equation for hL gives:
hL = (P1 - P2)/ρ + (Q²/(π*(x/100)²/4)² - Q²/(π*(y/100)²/4)²)/2
Substituting the given values:
hL = (492000 Pa - 432000 Pa)/1000 kg/m³ + ((0.035 m³/s)²/(π*(13/100 m)²/4)² - (0.035 m³/s)²/(π*(8/100 m)²/4)²)/2
Solving this gives the head loss in meters. Remember to multiply the final answer by the kinetic energy correction factor of 1.05 and round to two decimal places.
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