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Topic: Physics Problem please?**Question:**
A power plant has a power output of 1055 MW and operates with an efficiency of 33.0 percent. Excess energy is carried away as heat from the plant to a nearby river that has a flow rate of 1.1 x 10^6 kg/s. How much energy is transferred as heat to the river each second?

July 20, 2019 / By Colette

Total Work output of power plant: W_dot_net = 1055 MW If that's the total output, use efficiency definition to find heat input: eta = what we want/what we pay for eta = W_dot_net/Q_dot_in Solve for Q_dot_in: Q_dot_in = W_dot_net/eta By conservation of energy: Q_dot_in = W_dot_net + Q_dot_out Solve for Q_dot_out: Q_dot_out = Q_dot_in - W_dot_net Q_dot_out = W_dot_net*(1/eta - 1) Plug in data: W_dot_net :=1055 MW; eta:=0.33; Result for Heat Rejection: Q_dot_out = 2142 Megawatts Which means 2142 MegaJoules of heat are rejected to the river each second.

👍 272 | 👎 2

Did you like the answer? Find the velocity of the diver as the diver enters the water, then find the diver's momentum mv. Then use conservation of momentum: Force x time = change in momentum, and solve for the force. 2 a d = v² 2 ( 9.8 m/s² ) ( 6.1 m ) = v² v = 10.9 m/s so mv = 785 kg m/s = F t Solving for F yields 577 N.

use impulse-momentum theorem momentum = mass*velocity impulse = force * time momentum = impulse d = 0.5at^2 6.1 = 0.5 (9.8)t^2 t= 1.115749954 sec velocity = 1.115749954 * 9.8 = 10.93434955 momentum = 72 * 10.93434955 = 787.2731673 use impulse - momentum relationship 787.2731673 = force * 1.36 force = 578.8773289 so it takes about 579 Newtons of force to stop the diver hope this helps

Output Power = 1055e6 watt Input Power = 1055e6/0.33 = 3197e6 watt Power lost in river as heat = (3197 - 1055)e6 = 2142e6 watt = 2142e6 J/s Amount of water flowing = 1.1e6 kg/s Amount of heat removed by river per sec = 2142/1.1 = 1947 J/kg ANS Comment: add'l info: as the specific heat of water is: 4190 J/kg-C° The river water temperature would increase by: 1947/4190 = 0.5°C

👍 120 | 👎 -5

2142.66 MJ, or 511705.7 kcal. 1% of total power input is 31.979797........... MW which rounds up to 31.98. 67 times this is 2142.66 MW.

👍 119 | 👎 -12

Try this link, I think that it may help you... http://hypertextbook.com/facts/JianHuang.shtml

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