8. [15 points] Energy Equation Water at the Lake William Dam enters a hydraulic turbine through a 30 cm diameter pipe at a rate of 0.6 m3/s and exits through a 25-cm diameter pipe. For the system, elevation effects are negligible and the pressure drop across the turbine is measured to be 148 kPa. For a combined turbine-generator efficiency of 83 percent, calculate (a) hturbine and (b) Wtubrine.

a) h_turbine is 19.033 m

b) W_turbine is 92981.48 J

Explanation:

Given that;

inlet diameter of pipe d1 = 30cm = 0.3 m

outlet diameter of pipe d2 = 25cm = 0.25 m

flow rate through pipe Q = 0.6 m³/s

pressure drop Δp = 148 kPa = 148 × 10³ Pa

turbine generator efficiency y = 83%

a)

find h_turbine

now applying energy equation at inlet and outlet of pipe

p1/pg + V1²/2g + z1 = p2/pg + v2²/2g + z2 let this be equ 1

now Q = A1V1  

0.6 = π/4(0.3)²V1

V1 = 8.48 m/s                            

Q = A2V2

0.6 = π/4(0.25)²V2

v2 = 12.22 m/s  

from the diagram consider outlet of pipe asdatum so; z1 = h, z2 = 0o

so we substitute our values into equation 1

p1/pg + v1²/2g + z1 = p2/pg + v2²/2g + z2  

p1/pg + (8.48)²/(2×9.81) + h = p2/pg + (12.22)²/(2×9.81) + 0

h + 3.665 = p2/pg - p1/pg + 7.611

h = (p2-p1)/pg + 7.611 - 3.665

h = ( 148 × 10³ )/(10³×9.81) + 7.611 - 3.665

h = 15.0866 + 3.946

h = 19.0326 ≈ 19.033 m

therefore the h_turbine is 19.033 m

b)

now to find the work done in the turbine W_turbine

W_turbine = y × mgh

we know that mass flow rate m = pQ

so

W_turbine = y × pQgh

we substitute

W_turbine 0.83 × 10³ × 0.6 × 9.81 × 19.0326

W_turbine = 92981.48 J

answer:

an example of solubility is salt or sugar dissolving in water.

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