#$&* course Phy 232 If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
.............................................
Given Solution: The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it. Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Using the same Carnot efficiency equation, we can rearrange it to solve for Tcold. This is by by taking efficiency = 1 - (Tcold / Thot) and turning it into Tcold = Thot*(1 - efficiency). Thot is equal to (550 C + 273) = 823 K, with an efficiency of 28 %. Using these numbers we can find Tcold. Tcold = 823 K*(1 - .28) = 592.6 K Now we want to solve this equation for a new Thot, with the same Tcold as we just got and a new efficiency of 35%. Given that, the formula, efficiency = 1 - (Tcold / Thot) turns into Thot = Tcold / (1 - efficiency). Plugging in the new numbers we get, Thot = 592.6 K / (1 - .35) = 911.6 K. Bringing this temperature back to Celsius gives us (911.6 K - 273) = 638.6 C. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: OK Your Self-Critique Rating: OK ********************************************* Question: univ phy problem 20.45 11th edition 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The efficiency of a system is the amount of work done divided by the total energy used. The first temperature would be 6 C (279 K) and the second temperature would be 27 C (300 K). The efficiency would then equal (300 K - 279 K) / 300 K = 7% efficiency. The extraction rate is also known as the power that was put into the system. And this can be calculated by (210 kW) / (.07) = 3,000 kW. The rate of absorption is known by the heat put into the system minus the work done in the system. (3,000 kW - 210 kW) = 2,790 kW. To determine the flow rate, we can use the expression Q = (m)*(c)*(deltaT). Dividing this equation by time on both sides gives us Q/t which can be used as power, and also m/t which gives us the rate we are looking for. Applying these changes and plugging in the values gives us the equation, (2,790,000 J / s) = (m / t)*(4,190 J*kg/K)*(4 K). Solving this for m/t we get 166.4 kg / s. To turn the rate from kg/s to kg/hr we multiply by 3,600 s because there are 3,600 in an hour. (166.4 J/s)*(3,600 s) = 6.0 * 10^5 kg/hr. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** work done / thermal energy required = .07 so thermal energy required = work done / .07. Translating directly to power, thermal energy must be extracted at rate 210 kW / .07 = 3,000 kW. The cold water absorbs what's left after the 210 kW go into work, or 2,790 kW. Each liter supplies 4186 J for every degree, or about 17 kJ for the 4 degree net temp change of the water entering and exiting the system. Needing 3,000 kJ/sec this requires about 180 liters / sec, or about 600 000 liters / hour (also expressible as about 600 cubic meters per hour). Comment from student: To be honest, I was surprised the efficiency was so low. Efficiency is low but the energy is cheap and environmental impact in the deep ocean can be negligible so the process can be economical. ** Your Self-Critique: OK Your Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!