A small water slide is to be installed inside a swimming poo

A small water slide is to be installed inside a swimming pool. See Fig. C6.3. The slide manufacturer recommends a continuous water flow rate Q of 1.39 × 10³ m³/s (about 22 gal/min) down the slide, to ensure that the customers do not burn their bottoms. A pump is to be installed under the slide, with a 5.00-m-long, 4.00-cm-diameter hose supplying swimming pool water for the slide. The pump is 80 percent efficient and will rest fully submerged 1.00 m below the water surface. The roughness inside the hose is about 0.0080 cm. The hose discharges the water at the top of the slide as a free jet open to the atmosphere. The hose outlet is 4.00 m above the water surface. For fully developed turbulent pipe flow, the kinetic energy flux correction factor is about 1.06. Ignore any minor losses here. Assume that = 998 kg/m³ and = 1.00 × 10⁶ m²/s for this water. Find the brake horsepower (that is, the actual shaft power in watts) required to drive the pump.

4.00 m Ladder Tube Na Weee sliding board Pump Water 1.00 m

Solution

The shaft power of the pump is found using the relation,

SHP = W / t where W = F x d (or) SHP = (p2 - p1) * A * V / eta

Bernoulli\'s eq. for pipe flow including the head is given by,

P / r*g + Z + alpha * V^2 / 2*g = constant

Also, using the mass conservation relation, Q = rho * A * V we find the velocity V2 at outlet as,

V2 = Q / rho * A = 1.39*10^-3 / 998 * 0.0004 * 3.14 = 1.1089 * 10^-3 m/s

The above relation considers the kinetic flux correction factor alpha.

Applying the Bernoulli\'s principle across the pump flow,

P1 / rho * g + Z1 + alpha * V1^2 / 2*g = P2 / rho * g + Z2 + alpha * V2^2 / 2*g + H_friction   ---------- eq 1

H friction = f * L/ D * V^2 / 2g

f = 64 / Re = 64 * nu / V * D

f = 64 * 1 * 10 ^-6 / 1.1089 * 10^-3 * 0.04 = 1.44

So, H friction = 1.44 * 4 / 0.04 * 1.1089^2 / 2 * 9.8 = 8.85 m/s2

The actual pressure head in the pipe is given by,

(P2 - P1) = rho * g [ (Z2 - Z1) + 1/2g * (V2^2 - V1^2) ]

P2 - P1 = 998 * 9.8 [ 4 + 0.05 * 1.22 * 10^-6 ] = 39121.6 or 39.12 kPa

The mechanical efficiency of a pump is given by,

eta = P - Pl / P (or) available power / output power

Availabe power = (P2 - P1) * A * V = 39.12 * 0.0004 * 1.1089* 10^-3 = 0.017 W

SHP = Available power / eta = 0.017 / 0.80 = 0.021 W