SOLUTION: Consider the system to be the refrigerator (shown in the following schematic), which operates over a cycle in normal operation.

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Dayna Simmons
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1 Soln_21 An ordinary household refrigerator operating in steady state receives electrical work while discharging net energy by heat transfer to its surroundings (e.g., the kitchen). a. Is this a violation of the Kelvin-Planck statement of the Second Law of Thermodynamics? Explain your answer. b. Consider the same question, but now consider an electric motor operating in steady state. Consider the system to be the refrigerator (shown in the following schematic), which operates over a cycle in normal operation. 3 surrounding air in the kitchen refrigerator electrical power! There is one thermal reservoir: the surrounding air. The Kelvin-Planck statement of the Second Law states: It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving energy by heat transfer from a single thermal reservoir. or, mathematically,! "# %#%,'#'() 0 (single thermal reservoir). In the case of a refrigerator, although there is heat transfer with a single thermal reservoir, there is work being done on the system over the cycle, not by the system, i.e.,! "# %#%,'#'() 0 (zero only if the heat transfer is zero). Thus, there is no violation of the Kelvin-Planck statement of the 2 nd Law. Now consider the system to be an electric motor, which also operates cyclically in normal operation. surrounding air 3 electrical power motor! )()' shaft power! %,-./ Furthermore, from the 1 st Law,! %,-./! )()' (zero only if there s no heat transfer). As in the previous case, net work is done on the motor rather than by the motor, i.e.,! 0)/,"# %#%,'#'() 0. Thus, there is no violation of the Kelvin-Planck statement of the 2 nd Law. Page 1 of 1

2 Soln_22 A heat pump cycle is used to maintain the interior of a building at 21 C. At steady state, the heat pump receives energy by heat transfer from well water at 9 C and discharges energy by heat transfer to the building at a rate of 120,000 kj/h. Over a period of 14 days, an electric meter records that 1490 kwh of electricity is provided to the heat pump. Determine: a. the amount of energy that the heat pump receives over the 14-day period from the well water by heat transfer, in kj. b. the heat pump s coefficient of performance. c. the coefficient of performance of a reversible heat pump cycle operating between hot and cold reservoirs at 21 C and 9 C. A schematic of the problem is shown below with the system consisting of the heat pump, the well water serving as the cold reservoir, and the building interior serving as the hot reservoir. It is assumed that the heat pump is operating cyclically. hot reservoir, T H = 21 C = 294 K (building interior) Q H heat pump W on,cycle Q C cold reservoir, T C = 9 C = 282 K (well water) The energy received by the heat pump over the 14-day period from the well water may be found by applying the 1 st Law of Thermodynamics to the system. " #$#,&$&'( = 0 = +,-./ #$# + 2 /- #$#,&$&'(, (1) +,-./ #$# = , (2) ð 0 = /- #$#,&$&'(, (3) ð + 3 = /- #$#,&$&'(. (4) Using the given data, Q H = (120,000 kj/h)(24 h/day)(14 days) = 4.032*10 7 kj, (5) W on sys,cycle = 1490 kwh = (1490 kw)(3600 s/h) = 5.36*10 6 kj, (6) ð Q C = 3.50*10 7 kj. The coefficient of performance for a heat pump is defined as, < 678 9: =, (7) >?@ ABA,CBCDE Using the given data (Eqs. (5) and (6)), COP hp = If the heat pump operates reversibly between the hot and cold reservoirs, then the coefficient of performance may be found using the absolute temperature of the reservoirs, 678 9:,F(G = H = H = IH J, (8) Page 1 of 2

3 Soln_22 T H = 294 K and T C = 282 K ð COP hp,rev = Since the COP hp calculated using Eq. (7) is smaller than COP hp,rev (found using Eq. (8)), we conclude that the heat pump in the given problem is operating irreversibly. Page 2 of 2

4 Soln_23 A Carnot power cycle is executed on 1 kg of water. The cycle consists of isobaric expansion of saturated liquid at 160 C to a volume of 0.3 m followed by a reversible adiabatic expansion to 20 C and a quality of 76.0%. The water is then compressed isothermally to a quality of 19.7% and, finally, compressed reversibly and adiabatically back to the original state. a. Sketch the cycle on a p-v plot. b. Determine the heat added and net work, each in kj. c. Evaluate the thermal efficiency of this power cycle using the values found in part (b) and compare this value with the value based on the absolute temperatures of the thermal reservoirs. Let the system consist of the water. The cycle s path is sketched on the following p-v plot. p [bar (abs)] C 20 C water V [m 3 ] State 1 is in a saturated liquid phase at T 1 = 160 C => p 1 = p sat (T 1 ) = bar (abs), v 1 = m 3 /kg, and u 1 = kj/kg (values found from the SLVM Table for water). Note that since m = 1 kg, V 1 = mv 1 = m 3. At State 2, p 2 = p 1 = bar (abs) since the process from State 1 to State 2 is an isobaric expansion (and also isothermal). The specific volume at State 2 is! " = $ % &, (1) V 2 = 0.3 m 3 (given) and m = 1 kg (given) => v 2 = 0.3 m 3 /kg. Since v f (p 2 ) = m 3 /kg < v 2 < v g (p 2 ) = m 3 /kg, State 2 must be a saturated liquid-vapor mixture (SLVM) and, thus, T 2 = 160 C. The quality at State 2 is, ' " = ( %)( *% ( +% )( *%, (2) Using the previously found values, x 2 = The corresponding specific internal energy is,, " = 1 ' ", /" + ' ", 1", (3) where u f2 = kj/kg and u g2 = kj/kg => u 2 = 2530 kj/kg. Energy is added to the system via heat transfer only when going from State 1 to State 2. The added energy can be found by applying the 1 st Law to the system, = 6 789: 454 < =5 454 => 6 789: 454 = < =5 454, (4) 3 454,?" 454,?" + A3 454,?" + B3 454,?" 454,?" = C, 454,?" = C, ",?, (5) (no change in kinetic or potential energies) $ % < =5 454,?" = DEF = D $ G? F " F? (pressure is constant from 1 to 2). (6) Using the previously found values, ΔU sys,12 = 1850 kj, Page 1 of 2

5 Soln_23 W by sys,12 = 185 kj, ð Q into sys,12 = 2040 kj. To find the net work done by the system over the cycle, it s easiest to utilize the 1 st Law applied to the system over the cycle, 3 454,H5HIJ = 6 789: 454,H5HIJ < =5 454,H5HIJ => < =5 454,H5HIJ = 6 789: 454,H5HIJ, (4) since ΔE sys,cycle = 0. The energy added the system due to heat transfer is, 6 789: 454,H5HIJ = 6 789: 454,?" : 454,KL, since the processes from 2 to 3 and from 4 to 1 are adiabatic. We already found Q into sys,12 so now we must calculate Q into sys,34. Applying the 1 st Law from State 3 to State 4, = 6 789: 454 < =5 454 => 6 789: 454 = < =5 454, (4) 3 454,KL 454,KL + A3 454,KL + B3 454,KL 454,KL = C, 454,KL = C, L, K, (5) (no change in kinetic or potential energies) $ M < =5 454,KL = DEF = D $ N K F L F K = CD K! L! K = (pressure is constant from 3 to 4). (6) The properties of interest at States 3 and 4 are,, K = 1 ' K, /K + ' K, 1K, (7)! K = 1 ' K! /K + ' K! 1K, (8), L = 1 ' L, /L + ' L, 1L, (9)! L = 1 ' L! /L + ' L, 1L, (10) x 3 = (given), x 4 = (given), u f3 = u f4 = kj/kg (water saturation properties table at 20 C), u g3 = u g4 = kj/kg (water saturation properties table at 20 C), v f3 = v f4 = m 3 /kg (water saturation properties table at 20 C), v g3 = v g4 = m 3 /kg (water saturation properties table at 20 C), p 3 = p 4 = bar (abs) (water saturation properties table at 20 C), ð u 3 = 1850 kj/kg, v 3 = 43.9 m 3 /kg, u 4 = 541 kj/kg, v 4 = 11.4 m 3 /kg ð W by sys,34 = kj, ΔU sys,34 = kj, Q into sys,34 = kj. Thus, W by sys,cycle = 660 kj. The thermal efficiency of a power cycle is defined as, O Q RS TST,USUVW X Y, (1) Using the previously calculated data, η = (1) The thermal efficiency for a reversible power cycle may be found using, O ZJ( = [ Y)[ \ = 1 [ \, (1) [ Y [ Y Using the given temperatures, T H = 160 C = 433 K and T C = 20 C = 293 K, ð η rev = The efficiency calculated using Eq. (1) is equal to the efficiency calculated using Eq. (3) (to within rounding error) as expected since a Carnot cycle is reversible. Page 2 of 2

Spring_#7. Thermodynamics. Youngsuk Nam.

Spring_#7. Thermodynamics. Youngsuk Nam. Spring_#7 Thermodynamics Youngsuk Nam ysnam1@khu.ac.kr You can t connect the dots looking forward; you can only connect them looking backwards. So you have to trust that the dots will somehow connect in