Hydroelectric Power Generation from a Dam In a hydroelectric power plant, 100 m³/s of water flows from an elevation of 120 m to a turbine, where electric power is generated (Fig. 5–59). The total irreversible head loss in the piping system from point 1 to point 2 (excluding the turbine unit) is

Question

Hydroelectric Power Generation from a Dam

In a hydroelectric power plant, 100 m³/s of water flows from an elevation of 120 m to a turbine, where electric power is generated (Fig. 5–59). The total irreversible head loss in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbine– generator is 80 percent, estimate the electric power output.

Accepted Answer

The available head, flow rate, head loss, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Water levels at the reservoir and the discharge site remain constant.
Properties We take the density of water to be 1000 kg/m³.
Analysis The mass flow rate of water through the turbine is

[latex]\dot{m} = \rho \dot{V} = (1000 kg/m^3)(100 m^3/s) = 10^5 kg/s[/latex]

We take point 2 as the reference level, and thus [latex]z_2 = 0[/latex]. Also, both points 1 and 2 are open to the atmosphere ([latex]P_1 = P_2 = P_{atm}[/latex]) and the flow velocities are negligible at both points ([latex]V_1 = V_2 = 0[/latex]). Then the energy equation for steady, incompressible flow reduces to

[latex] \cancel{\frac{P_1}{\rho g} } +\alpha _1 \cancel{\frac{V^2_1}{2g} } + z_1 + \overset{0}{\cancel{h_{pump, u}}} = \cancel{\frac{P_2}{\rho g} } + \alpha _2 \cancel{\frac{V^2_2}{2g} } + \overset{0}{\cancel{z_2}} + h_{turbine, e} + h_L[/latex]

or

[latex]h_{turbine, e} = z_1 − h_L[/latex]

Substituting, the extracted turbine head and the corresponding turbine power are

[latex]h_{turbine, e} = z_1 - h_L = 120 - 35 = 85 m[/latex]

[latex] \dot{W}_{turbine, e} = \dot{m}gh_{turbine, e} = (10^5 kg/s)(9.81 m/s^2)(85 m)\left(\frac{1 kJ/kg}{1000 m^2/s^2} \right)= 83,400 kW[/latex]

Therefore, a perfect turbine–generator would generate 83,400 kW of electricity from this resource. The electric power generated by the actual unit is

[latex] \dot{W}_{electric} = \eta _{turbine–gen} \dot{W}_{turbine, e} = (0.80)(83.4 MW) = 66.7 MW[/latex]

Discussion Note that the power generation would increase by almost 1 MW for each percentage point improvement in the efficiency of the turbine– generator unit. You will learn how to determine [latex]h_L[/latex] in Chap. 8.

Question 5.13: Hydroelectric Power Generation from a Dam In a hydroelectric...

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