#$&* course PHY 202 009. `query 9*********************************************
.............................................
Given Solution: The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it. Your Self-Critique: Duh, I am ashamed. I should have remembered that I needed to convert to Kelvins. I remember that from Thermal Applications.
.............................................
Given Solution: The energy to do the work and the energy transferred to the environment must be put into the engine. So this engine requires an input of 10 kJ + 8.5 kJ = 18.5 kJ of heat. It does 10 kJ of work, so its efficiency, which is the ratio of work done to energy input, is e = (work done) / (energy input) = 10 kJ / 18.5 kJ = .53, or 53%, roughly. Your Self-Critique: Your Self-Critique Rating: Question: Openstax: (a) What is the work output of a cyclical heat engine having a 22.0% efficiency and 6.00×10^9 J of heat transfer into the engine? (b) How much heat transfer occurs to the environment? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The work done is 22% of the energy input, or `dW = .22 * 6 * 10^9 J = 1.3 * 10^9 J. The rest of the energy input goes to the environment. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: Openstax: Assume that the turbines at a coal-powered power plant were upgraded, resulting in an improvement in efficiency of 3.32%. Assume that prior to the upgrade the power station had an efficiency of 36% and that the heat transfer into the engine in one day is still the same at 2.50×10^14 J . (a) How much more electrical energy is produced due to the upgrade? (b) How much less heat transfer occurs to the environment due to the upgrade? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: e_improvement = 3.32% e_prior = 36% e_total = 36% + 3.32% = 39.3% .36 * 2.5x10^14J = 9x10^13 (prior to turbine upgrade) .393 * 2.5x10^14J = 9.82x10^13 (since turbine upgrade) a) Electrical energy produced since turbine upgrade 9.82x10^13J - 9x10^13J = 8.2x10^12 J b) the percent difference of electrical energy is 8.2%. Therefore, it would be assumed that 8.2% less heat is transferred to the environment. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Previously the energy produced was .36 * 2.5 * 10^14 J = 9 * 10^13 J. After the improvement the efficiency is about 39.3% so the energy produces is .393 * 2.5 * 10^14 J = 9.8 * 10^13 J, roughly. The difference is about 8 * 10^12 J. This heat now goes into the electrical energy produced by the plant, and the heat transferred to the environment decreases by the same amount. Your Self-Critique: Your Self-Critique Rating: Question: Openstax: A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is 550ºC . What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is 20ºC .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Efficiency = 38%, T_h =550 C = 823 K, T_l = 20 C = 293 K Efficiency_carnot (max) = (823 - 293) / 823 = 64.4% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The maximum efficiency of an engine running between 550 C and 20 C is e_max = (T_h - T_c) / (T_h) = (823 K - 293 K) / (823 K) = 63%, roughly. Operating at 38%, the station is at 38% / (63%) = 60%, roughly, of the maximum possible efficiency. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: eff_c = (T_h - T_l) / T_h => T_h * eff_c = T_h - T_l => T_l = T_h(1-eff_c) T_h = 350 C = 623 K, T_l = 623(1 - 0.28) = 449K T_h = T_l / (1 - eff_c) = 449K / (1 - 0.35) = 689K = 416 C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Carnot efficiency is eff = (Th - Tc) / Th. Solving this for Tc we multiply both sides by Th to get eff * Th = Th - Tc so that Tc = Th - eff * Th = Th ( 1 - eff). We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature) If Th = 550 C = 823 K and efficiency is 30% then we have Tc =823 K * ( 1 - .28) = 592 K. Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th: Tc we multiply both sides by Th to get eff * Th = Th - Tc so that eff * Th - Th = -Tc and Tc = Th - eff * Th or Tc = Th ( 1 - eff) and Th = Tc / (1 - eff). If Tc = 576 K and eff = .35 we get Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx. This is (912 - 273) C = 639 C. ** Your Self-Critique: Your Self-Critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!