Hydropower tutorial(1)
- 1. Hydropower: Tutorial questions and solutions. 1. Estimate the volume flow rate and mass flow rate of water flowing in a river whose channel is roughly rectangular with a depth of 0.6 m and a width of 4 m, and which has a flow speed of 0.85 m/s. A weir on the river provides a working head of 1.4 m for a turbine/generator. If one third of the river flow could be diverted through the turbine which has an efficiency of 75%, estimate the electrical power that could be generated. 2. The Hoover Dam lies across the Colorado river and holds back a huge reservoir of water producing a head of water for power production of 180 m. All of the river flow passes through the turbines. In a typical year 4.2 TWh (4.2 x 1012 Wh) of electricity is produced. (a) Assuming an overall efficiency of 85% for the turbine/generator, calculate the corresponding average mass flow rate of water. (b) When the river is running high a total of 1333 tonnes/second of water flows through the turbines. Calculate the power generated in these conditions.
- 2. 1. Volume flow 𝑉̇ = 𝐴𝑈 where A is the flow area and U the flow speed. A = depth * width = 0.6 * 4 = 2.4 m2 so 𝑉̇ = 𝐴𝑈 = 2.4 ∗ 0.85 = 2.04 𝑚3 /𝑠. Mass flow rate 𝑚̇ = density * Volume flow rate = 𝜌𝑉̇ = 1000 ∗ 2.04 = 2040 𝑘𝑔 𝑠 . One third of this can be diverted into the turbine: 𝑚̇ = 2040 3 = 680 𝑘𝑔 𝑠 . The working head is 1.4 m. 𝑃 = 𝜂𝑚̇ 𝑔ℎ = 0.75 ∗ 680 ∗ 9.81 ∗ 1.4 = 7004 𝑊; 𝑎𝑏𝑜𝑢𝑡 7𝑘𝑊. 2. (a) Average power = energy / time. 4.2 TWh = 4.2x1012 W* 3600s = 1.512x1016 J. Time = 365*24*3600 s (in a year) = 31.54x106 s. Power = 1.512x1016 / 31.54x106 = 479.5x106 W (480 MW). 𝑃 = 𝜂𝑚̇ 𝑔ℎ 𝑠𝑜 𝑚̇ = 𝑃 𝜂𝑔ℎ = 479.5𝑥106 0.85∗9.81∗180 = 319.5𝑥103 𝑘𝑔/𝑠. (b) 𝑃 = 𝜂𝑚̇ 𝑔ℎ = 0.85 ∗ 1333𝑥103 ∗ 9.81 ∗ 180 = 2.001𝑥109 𝑊; 𝑎𝑏𝑜𝑢𝑡 2000 𝑀𝑊.