Answer:
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
