Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.6 m3s and exits through a 25-cm-diameter pipe. The pressure drop in the turbine is measured by a mercury manometer to be 1.2 m. For a combined turbine-generator efficiency of 83 percent, determine the net electric power output. Disregard the effect of the kinetic energy correction factors.

Net electric power output = 54.594KW

Explanation:

Enter dia = 0.3m, Exit dia = 0.25, Flow Rate (Vflow) = 0.6m³/s, Hhg = 1.2m, Efficiency = 83%, Net electric power output = ?

Vflow = A1V1, where A1 is the Area of flow and V1 is the Velocity

V1 (Velocity at entering) = Vflow/A1 = 0.6/{ (π/4) 0.3²} = 8.48m/s

V2 (Velocity at exiting) = Vflow/A1 = 0.6/{ (π/4) 0.25²} = 12.22m/s

ΔP = (Shg - 1) x (ρh2o) x g x Hhg where Shg is the specific gravity of Mercury, ρh2o is the density of water, g is the acceleration due to gravity and Hhg is the height drop of the manometer

ΔP = (13.6 - 1) x 1000 x 9.81 x 1.2 = 148327.2Pa

Applying Bernoulli's Equation between the entering and exit

(P1/ρg) + α1 ({V1^2) / 2g} + z1 = (P2/ρg) + α2 ({V2^2) / 2g} + z2 + Hturbine

where z1, z2 = 0 as there no is change in the datum head and α is the correction factor = 1

Hturbine = (P1/ρg) - (P2/ρg) + α[{ (Vq^2) / 2g} - { (V2^2) / 2g}]

Hturbine = (ΔP/ρg) + α[{ (V1^2) / 2g} - { (V2^2) / 2g}]

Hturbine = (148327.2/1000 x 9.81) + 1[{ (8.48^2) / 2 x 9.81 - { (12.22^2) / 2 x 9.81}] = 11.175m

The electrical Power Output is given by the equation Wturbine = η (turbine - generator) x ρ x Vflow x g x Hturbine

Wturbine = 0.83 x 1000 x 0.6 x 9.81 x 11.175 = 54.594 x 10^3 = 54.594 KW