Question

A sump pump is draining a flooded basement at the rate of 0.800 L/s, with an output pressure of 3.50 ? 105 N/m2. Neglect frictional losses in both parts of this problem.

(a) The water enters a hose with a 3.00 cm inside diameter and rises 2.80 m above the pump. What is its pressure at this point?

(b) The hose then loses 2.00 m in height from this point as it goes over the foundation wall, and widens to 4.00 cm diameter. What is the pressure now?

Answer 1

Solution:

a) Use the flow rate and the cross setional area to find the velocity,

Q = v*A

Since the area is circular, in terms of diameter, A = 1/4**d²
v = 4*Q / (*d²)

Remember to convert L/s to m³/s and the diameter from cm to m before plugging in numbers. The velocity will come out in terms of m/s, standard units.

Then use Bernoulli for the rest

P₁ + 1/2**v₁² + *g*h₁ = P₂ + 1/2**v₂² + *g*h₂

For the first problem, assume that the cross section is consistant, so the velocities are the same.

The velocity terms will cancel.

P₁ + *g*h₁ = P₂ + *g*h₂

Solve for P₂

P₂ = P₁ + *g*(h₁ - h₂) where = 1000kg/m³, g = 9.81 and h₁ - h₂ = -2.8m

= 3.23*10⁵ N/m²

b)For the next part, use continuuity to find the exit velocity

Q = constant ==> v₁*A₁ = v₂*A₂ = 0.8L/s = 8*10^-4 m³/s

Then use Benoulli including the velocity terms.

You can either start at the bottom again, P₁ = 3.23*10⁵ Pa, but the height drop of 2 m in height.

P₂ = P₁ + 1/2**(v₁² - v₂²) + *g*(h₁ - h₁)

Here v₁  = Q/A₁ = 4*8*10^-4 / (*0.03²) = 1.132 m/s and v₂ = 0.64m/s. and h₁ - h₂ = 2.8 - 2 = 0.8m

P₂ = 3.23*10⁵ + 0.5*1000*[1.132² - 0.64²] + 1000*9.81*0.8

= 3.313*10⁵ Pa.

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