Determining the Electrical Energy of a Small Hydroelectric and the Water Velocity The reservoir height of a small hydroelectric is 5 m. The water passes through the penstock at the rate of 100 kg/s. The efficiency of the penstock is ηp = 95%, the power coefficient is Cp = 47%, the turbine

Question

Determining the Electrical Energy of a Small Hydroelectric and the Water Velocity
The reservoir height of a small hydroelectric is 5 m. The water passes through the penstock at the rate of [latex]100 kg/s[/latex]. The efficiency of the penstock is [latex]η_p = 95\%[/latex], the power coefficient is [latex]C_p = 47\%[/latex], the turbine efficiency is [latex]η_t = 85\%[/latex], and the efficiency of the generator is [latex]η_g = 90\%[/latex]. Calculate (a) the generated energy in one month and the income assuming an energy cost of [latex]$0.15/kW h[/latex], and (b) the speed of the water exiting the penstock.

Accepted Answer

[latex]h = 5 m, f = 100 kg/s, g = 9.807 m/s^2, t = 30 days, c = 0.15/kW-h, C_p = 0.47, \eta _p = 0.95,\eta _t = 0.85, n_g = 0.90[/latex].

a. Substituting the values in Eq. (16.69)

[latex]E_g = fgHt(C_p\eta _p\eta _t\eta_g)[/latex] (16.69)

gives the generated energy in 30 days

[latex]E_g = 100 imes 9.807 imes 5 imes 30 imes 24 imes 0.47 imes 0.95 imes 0.85 imes 0.90 = 1.206 imes 10^6 J[/latex]

The amount of energy cost or saving is

[latex]Income(Saving) = E_g c = 1.206 imes 10^6 imes 0.15 imes 10^{-3} = \$ 180.883[/latex]

b. For a flow of 1 s, the mass of water [latex]m = f imes 1 = 100 kg[/latex]
The input potential energy to the penstock is

[latex]PE_{p-in} = mgh = 100 imes 9.807 imes 5 = 4.903 kJ [/latex]

Equation (16.65)

[latex]\eta _p=\frac{KE_{p-out}}{PE_{p-in}}[/latex] (16.65)

gives the kinetic energy of the penstock

[latex]KE_{p-out} = \eta _pPE_{p-in} = 0.95 imes 4.903 imes 10^3 = 4.658 kJ[/latex]

Substituting the values in Eq. (16.64)

[latex]KE_{p-out}=\frac{1}{2} A_pvt ho v^2[/latex] (16.64)

gives the speed of the water

[latex]\upsilon =\sqrt{\frac{2KE_{p-out}}{m} }=\sqrt{\frac{2 imes 4.658 imes 10^3}{1000} }=9.652 m/s[/latex]

Question 16.8: Determining the Electrical Energy of a Small Hydroelectric a...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions