# ME 201 Thermodynamics

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1 ME 0 Thermodynamics Second Law Practice Problems. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F The maximum work a substance can do is given by its availablity. We will assume that we have a closed system so that ψ = u - u o - T o (s - s o ) We take the dead state to be at STP or 5 C and 00 kpa or 76.4 F and 4.7 psia. Then using the appropriate table we have ψ air = (537) ln = Btu / lb m and = (537) ψ steam = Btu / lb So the steam can do more work m ( ). A heat pump provides 30,000 Btu/hr to maintain a dwelling at 68 F on a day when the outside temperature is 35 F. The power input to the pump is hp. If electricity costs 8 cents per kilowatt-hour, compare the actual operating cost per day with the minimum theoretical operating cost per day.

2 We sketch our device interactions Dwelling High Temperature Heat Reservoir at T H Q H Heat Pump W net Q L Outside Low Temperature Heat Reservoir at T L The cost is given by Cost = (0.08)W net For the actual cost we have (Cost) act = (0.08)()( kw / hp)(4 hr / day) = \$.43 To calculate the minimum cost we will allow the heat pump to operate as a Carnot cycle, so that COP Carnot = = = 6 T 495 L TH 58 Then the minimum possible power input is ( & Q& W ) = H = 30,000 net = 875 Btu / hr min COPCarnot 6 = kw

3 and the minimum cost is (Cost) = (0.08)(0.5495)(4 hr / day) min = \$ A cylinder/piston system contains water at 00 kpa, 00 C with a volume of 0 liters. The piston is moved slowly, compressing the water to a pressure of 800 kpa. The process is polytropic with a polytropic exponent of. Assuming that the room temperature is 0 C, show that this process does not violate the second law. To determine if this violate the nd law we will want to calculate the entropy change of the universe and compare it to zero. We have ( S ) = m(s - s ) + - Q sys universe Tsurr We now work this as a first law problem Working Fluid: Water(compressible) System: Closed System Process: Polytropic with n=.0 State State T = 00 C T = 4.7 C P = 00 kpa P = 800 kpa u = kj/kg u = 655.5kJ/kg V = 0.00 m 3 V = m 3 v =.0803 m 3 /kg v = m 3 /kg s = kj/(kg K) s = 6.88 kj/(kg K) phase: sup.vap. phase: sup.vap. italicized values from tables, bold values are calculated Initial State: Fixed Final State: Unknown W sh = 0 Q =???? W bnd =???? We begin by calculating the mass m = V = 0.00 = kg v

4 To fix the final state we use the polytropic relationship P V V = n / n (00)(0.00) / = P = m The specific volume at state is then v = V m = = kg / m which gives us superheated vapor. The boundary work can be shown to be W = P V ln V = (00)(0.00)ln bnd V 0.00 = kj We use the first law to determine the heat transfer Q sys = m(u - u ) + W = (0.085)( ) + (-5.55) Then = kj ( S ) = - Q sys surrounds T surr and for the system ( ) = - (-5.53) 93 = kj / K S = m(s - s ) = (0.085)( ) system = kj / K So that ( S ) = = kj / K universe Since this is greater than zero the second law is not violated. 4. When a man returns to his well-sealed house on a summer day, he finds that the house is at 3 C. He turns on the air conditioner which cools the entire house to 0 C in 5 minutes. If the COP of the heat pump system is.5, determine the power drawn by the heat pump. Assume the entire mass within the house is equivalent to 800 kg of air. 4

5 We begin by sketching our device interactions House High Temperature Heat Reservoir at T H Q H Heat Pump W net Q L AC System Low Temperature Heat Reservoir at T L By definition we have COP = Q & H W& net So if the required heat transfer can be determined the power can be determined. From a first law analysis on the house, we can write &Q = m u - u = (800) t (5)(60) = kw and Q & = - Q & H = 7.65 kw Then the power required is & W & = Q H COP = 7.65 net = 3.06 kw.5 5

6 5. An innovative way of power generation involves the utilization of geothermal energy, the energy of hot water that exists naturally underground (hot springs), as the heat source. If a supply of hot water at 40 C is discovered at a location where the environmental temperature is 0 C, determine the maximum thermal efficiency a geothermal plant built at that location can have. If the power output of the plant is to be 5 MW, what is the minimum mass flow rate of hot water needed? We begin by sketching our device interactions Geothermal Source High Temperature Heat Reservoir at T H Q H Heat Engine W net Q L Environment Low Temperature Heat Reservoir at T L The maximum thermal efficiency will occur when the heat engine operates as a Carnot cycle, ηth = η L Carnot = - T (0 + 73) = - = 0.9 T H ( ) The minimum mass flow rate of hot water corresponds to the maximum thermal efficiency or (& W& Q ) = net = 5000 H = 7,08 kw min η 0. 9 Carnot 6

7 Performing a first law analysis on the hot water stream we have Q = m& ( hout - h in) For the minimum flow rate we will assume that the hot water is cooled down to the environment temperature, then Q -7,08 m & = = cp( Tout Tin ) ( )( 0 40) = 34. kg / s 6. Air enters an adiabatic non-ideal nozzle at 9 m/s, 300 K, and 0 kpa and exits at 00 m/s and 00 kpa. Determine the irreversibility and the reversible work on a per mass basis. We first solve this as a first law problem Working Fluid: Air(ideal gas) System: Control Volume System Process: Nozzle State State T = 300K T = 95 C P = 0 kpa P = 00 kpa h = kj/kg h = kj/kg φ =.7003 kj/(kg K) φ =.6855 kj/(kg K) r r v = 9 m/s v = 00 m/s italicized values from tables, bold values are calculated Initial State: Fixed Final State:??? W sh = 0 Q = 0 We use the first law to fix the final state h + v r = h + v r Then solving for h h = h + v r v r = (9) ( 00) -3 (0 ) = kj / kg which allows us to determine the temperature and φ. Then the reversible work is 7

8 w = h - h - T - - R ln P rev HR φ φ P = (98) )ln 0 ( 00 = 5.7 kj / kg Since the actual work is zero the irreversibility is i = w rev = 5.7 kj / kg 7. Determine if a tray of ice cubes could remain frozen when placed in a food freezer having a COP of 9, operating in a room where the temperature is 3 C. We begin by sketching our device interactions Surroundings High Temperature Heat Reservoir at T H Q H Refrigerator W net Q L Freezer Compartment Low Temperature Heat Reservoir at T L 8

9 Assuming that the refrigerator operates on the Carnot cycle, we have COP = COP Carnot = TH TL Solving for T L T T = H = 305 L = 74.5 K + + COP 9 and since this is greater than 0 C the ice cubes will not remain frozen. 8. Air is compressed in a closed system from a state where the pressure is 00 kpa and the temperature is 7 C to a final state at 500 kpa and 77 C. Can this process occur adiabatically? If yes, determine the work per mass. If no, determine the direction of the heat transfer. To determine if the process can occur, we must calculate ( S ) = m(s - s ) + - Q sys universe Tsurr and compare it to zero. Since the process is adiabatic ( s ) = s - s = - - R ln P universe φ φ P Going to the air tables we find ( s ) = s - s = )ln 500 universe ( 00 = kj / kg Since this is less than zero, the process cannot be adiabatic. To make s universe greater than zero will require Q sys to be negative, so that the direction of heat transfer is out of the system. 9. The pressure of water is increased by the use of a pump from 4 to 40 psia. A rise in the water temperature from 60 F to 60. F is observed. Determine the irreversibility, the second law efficiency, and the isentropic efficiency of the pump. 9

10 We first solve this as a first law problem Working Fluid: Water (incompressible) System: Control Volume System Process: Pump State State s (ideal) State a (actual T = 60 F T s = T a = 60. F P = 4 psia P = 40 psia P = 40 psia bold values are calculated Initial State: Fixed Final State: fixed W sh =???? Q = 0 To calculate the irreversibility, we use i = T (s - s ) - q HR = (537)c ln T p,avg T - 0 = (537)(.004)ln = Btu / lb m To determine the second law efficiency we need both the actual work and the ideal work. Starting with the actual work we have w act = h - h + q = c p,avg (T - T ) + v avg (P - P ) - 0 = (.004)(60-60.) + ( )(4-40) / ( psia ft / Btu) = Btu / lb m The reversible work is given by w rev = i + w act = (0.068) + (-0.774) = Btu / lbm which allows us to determine the second law efficiency as η rev II = w = ( ) = 0.55 w act (-0.774) 3 0

11 To determine the isentropic efficiency, we must first calculate the ideal work. Recognizing that in a isentropic process, the water will not change temperature, we can write w = v (P - P ) ideal avg 3 = ( )(4-40) / ( psia ft / Btu) = Btu / lb m Then our isentropic efficiency is η ideal s = w = (-0.077) w act (-0.774) = Carbon dioxide undergoes an isothermal reversible process from 50 kpa and 300 C to 500 kpa. Determine the heat transfer per mass by using the first law and evaluating the boundary work from Pdv. Compare this to the heat transfer per mass calculated from the entropy change and the second law. We first solve this as a first law problem Working Fluid: CO (ideal gas) System: Closed System Process: Isothermal, Reversible State State T = 300C = 573K T = T = 573K P = 50 kpa P = 500 kpa u = kj/kg u = kj/kg φ = kj/(kg K) φ = kj/(kg K) v = m 3 /kg v = 0.65 m 3 /kg italicized values are from ideal gas relationships Initial State: Fixed Final State: fixed W sh = 0 Q =??? W bnd =????

12 We first go to the CO tables and get our properties. Our boundary work for an ideal gas undergoing an isothermal process is w = RT ln v bnd v = (0.889)(573)ln = Btu / lb m Using the first law our heat transfer is q = u - u + w bnd = ( ) = Btu / lb m From the second law we have q = T (s - s ) = T - - R ln P φ φ P = (573) )ln 500 ( 50 = Btu / lb m So the two calculations for heat transfer agree.

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