MAE 11. Homework 8: Solutions 11/30/2018

MAE 11 Homework 8: Solutions 11/30/2018

MAE 11 Fall 2018 HW #8 Due: Friday, November 30 (beginning of class at 12:00p) Requirements:: Include T s diagram for all cycles. Also include p v diagrams for Ch 9 gas power cycles. Problems with asterisk *, do problem but do not submit (submit onlyh8.4andh8.6) H8.1* (Ch 8) Consider an ideal Rankine cycle. Water enters the turbineat10mpa,480 C(state1),andthe condenser pressure is 6 kpa. Determine: a. heat addition [kj/kg steam flowing] at steam generator (boiler) b. thermal efficiency c. heat rejection [kj/kg steam flowing] at condenser H8.2* (Ch 8) Now consider the effect of adding reheat to the ideal Rankine cycle in H8.1. The inlet to the turbine (state 1) remains at 10 MPa, 480 C, and the condenser pressure is still 6 kpa. Steam expands through the first-stage turbine to 0.7 MPa and is then reheated to 480 C. Determine: a. heat addition [kj/kg steam flowing] before first-stage turbine b. thermal efficiency c. heat rejection [kj/kg steam flowing] at condenser H8.3* (Ch 8) Now consider the effect of irreversibilities in the cycle of H8.2. Assume each stage of the turbine and the pump have an isentropic efficiency of 80%. Determine items a.-c. as in H8.2. H8.4 (Ch 9) A four cylinder, four-stroke spark ignition engine runs at 2800 rpm. The processes within each cylinder are modeled as an air-standard Otto cycle. At the beginning of the compression process(state 1),the air is at 14.7 psia, 80 F, and a volume of 0.035 ft 3. The compression ratio is 9 and the maximum pressure in the cycle is 890 psia. Determine: a. net work output per cylinder (per cycle) [Btu] b. thermal efficiency c. power output of engine [horsepower] H8.5* (Ch 9) Consider a cold air-standard Diesel cycle. The pressureand temperature at the beginning of compression are 95 kpa amd 300 K (state 1), respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2150 K. Specific heats are evaluated at 300 K, Determine: a. compression ratio b. cutoff ratio c. thermal efficiency d. mean effective pressure [kpa] H8.6 (Ch 10) Consider a vapor compression refrigeration cycle operating at steady-state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor (state 1) and saturated liquid exits the condenser. The temperature of the cooled region is T c =4 C, which is 10 Cgreaterthantheevaporatortemperature. The temperature of the environment is T h =40 C, which is 10 Clowerthanthetemperatureoftherefrigerant leaving the condenser. The compressor isentropic efficiency is 85%. The mass flow rate of refrigerant is 7.5 kg/min. Determine: a. evaporator and condenser pressures [bar] b. compressor power inout [kw] c. refrigeration capacity [tons] d. coefficient of performance

H8.4 (Ch 9) A four cylinder, four-stroke spark ignition engine runs at 2800 rpm. The processes within each cylinder are modeled as an air-standard Otto cycle. At the beginning of the compression process (state 1), the air is at 14.7 psia, 80, and a volume of 0.035 ft 3. The compression ratio is 9 and the maximum pressure in the cycle is 890 psia. Determine: a. net work output per cylinder (per cycle) [Btu] b. thermal efficiency c. power output of engine [horsepower] Given Model Diagram ω = 2800 rpm Closed system p 1 = 14.7 psia Air standard Otto Cycle T 1 = 80 Ideal gas 1 = 0.035 ft 3 ΔKE = ΔPE = 0 r = 9 p 3 = 890 psia Basic Equations ΔE = ΔU + ΔKE + ΔPE = Q in Q out + W in W out v 2 = v r2 (T 2 ) v 1 v r1 (T 1 ) r = max / min, η = W net,out /Q in Analysis Part a: Determine the net work output from a single cycle [Btu]. To apply conservation of energy needed to evaluate W net properties at each state need to be found. out State 1: Fixed by p 1 and T 1. From A22E: u 1 (T 1 ) = 92.04 Btu and v r1(t 1 ) = 144.32 State 2: Fixed by v 2 (found from r) and s 2 = s 1. Using the relative volume, v r2 (T 2 ) = v 2 = 1 v r1 (T 1 ) v 1 r Solving for v r2 v r2 (T 2 ) = v r1 (T 1 ) = 144.32 = 16.035 r 9 From Table A22E: T 2 = 1265 R and u 2 (T 2 ) = 221.3 Btu State 3: Fixed by v 3 = v 2 and p 3 = 890 psia Apply the ideal gas law twice - at state 3 and at state 1. p 3 v 3 Note v 3 = v 2 and v 2 /v 1 = 1/r p 1 v 1 = RT 3 RT 1 T 3 = p 3v 3 p 1 v 1 T 1 T 3 = p 3 1 T 890 psia p 1 r 1 = ( ) 14.7 psia (1 ) (540 R) = 3632.7 R 9 From A22E: u 3 (T 3 ) = 728.95 Btu and v r3(t 3 ) = 0.6258 State 4: Fixed by v 4 = v 1 and s 4 = s 3 v r4 (T 4 ) v r3 (T 3 ) = v 4 = r v 3 v r4 (T 4 ) = rv r3 (T 3 ) = 9(0.6258) = 5.632 From Table A22E at v r4 (T 4 ) = 5.632: T 4 = 1823 R and u 4 (T 4 ) = 331.1 Btu

Work is done on the system during the compression stroke process 1-2 ΔE = ΔU + ΔKE + ΔPE = Q in Q out + W in W out m = W in = m(u 2 u 1 ) p 1 1 = mrt 1 m = p 1 1 RT 1 lbf (14.7 in 2)(0.035 ft3 ) (53.353 ft lbf R )(540 R) [12 in 1 ft ]2 = 0.002572 m = 0.002572 W in = (0.002572 ) (221.3 Btu Work is done by the system during the power stroke (process 3-4) Btu 92.04 ) = 0.3324 Btu ΔE = ΔU + ΔKE + ΔPE = Q in Q out + W in W out m(u 4 u 3 ) = W out W out = m(u 3 u 4 ) W out = (0.002572 ) (728.95 Btu Part b: Determine the thermal efficiency, η. Btu 331.1 ) = 1.026 Btu Wnet = W out W in = 1.026 Btu 0.3324 Btu out Wnet = 0.6933 Btu out η = W net,out Q in Q in = m(u 3 u 2 ) = (0.002572 ) (728.95 Btu η = W net,out Q in = η = 53.1 % 6.926 Btu 13.09 Btu Btu 221.3 ) = 1.305 Btu Part c: Determine the power produced by the engine in [hp]. The power produced by the engine is the work per cycle multiplied by the rate at which the engine executes a cycle. Note the crank mechanism (see diagram) must undergo 2 revolutions to complete 1 thermodynamic cycle. There are also 4 cylinders at work. Btu rev cyc min hp W eng = (0.6933 ) (4 cyl) (2800 ) [1 ] [1 ] [1.415 cyl cyc min W eng = 91.56 hp 2 rev 60 s btu/s ] Discussion Using the relative volume ratio aids quick calculations. Although convenient it is important to understand the physics being employed. v r is derived from the Tds equation for an ideal gas undergoing an isentropic process. If these conditions are not met, or if v r data is unavailable, one can always revert back to the general Tds property relations

H8.6 (Ch 10) Consider a vapor compression refrigeration cycle operating at steady-state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor (state 1) and saturated liquid exits the condenser. The temperature of the cooled region is T c = 4, which is 10 greater than the evaporator temperature. The temperature of the environment is T h = 40, which is 10 lower than the temperature of the refrigerant leaving the condenser. The compressor isentropic efficiency is 85%. The mass flow rate of refrigerant is 7.5 kg/min. Determine: a. evaporator and condenser pressures [bar] b. compressor power input [kw] c. refrigeration capacity [tons] d. coefficient of performance Given Data Model Diagram State 1: Saturated vapor (x 1 = 1) Refrigerant 134a State 3: Saturated liquid (x 3 = 0) Vapor compression refrigeration cycle T c = 4 = 277.15 K Steady state T h = 40 = 313.15 K Uniform 1D fluid properties T 4 = T 1 = 6 = 267.15 K Nelect KE and PE effects T 3 = 50 = 323.15 K η c = 0.85 T m = 7.5 kg/min = 0.125 kg/s T h Q out W in Governing Equations dm cv dt = i m i e m e η c = (W /m ) s (W /m ) = h 2s h 1 h 2 h 1 β = Q in W in de cv dt T h T c = Q in Q out + W in W out + i m i(h i + 1 V 2 i 2 + gz i ) i m e(h e + 1 V 2 e 2 + gz e ) s T c Q in Analysis a. Determine the evaporator and condenser pressures [bar] Evaporators and condensers are modeled as constant pressure heat exchangers. States 1 and 3 are on the saturation curve where pressure and temperature are dependent properties. From Table A10 p evap = p 4 = p 1 = p sat (T 1 ) and p cond = p 2 = p 3 = p sat (T 3 ) p evap = p sat ( 6 ) = 2.3489 bar and p cond = p sat (50 ) = 13.1885 bar b. Determine the compressor power input [kw] State 1: Fixed by saturated vapor at T 1 = 267.15 K From Table A10: s 1 = s g1 = 0.9226 kj/kg K and h 1 = h g1 = 243.72 kj/kg

State 2s: Fixed by the isentropic process (s 2s = s 1 ) to p 2 From Table A12 (2D interpolation in p and s) h 2s 283 kj/kg State 3: Fixed by saturated liquid at T 3 = 50 = 323.15 K From Table A10 h 3 = 121.465 kj/kg The refrigerant undergoes an expansion process as it flows through the valve. Consider conservation of energy across the expansion valve. de cv dt = Q in Q out + W in W out + i m i(h i + 1 V 2 i 2 + gz i ) i m e(h e + 1 V 2 e 2 + gz e ) 0 = m (h 3 h 4 ) h 4 = h 3 State 4: Fixed by h 4 = h 3 and p 4 = p 1 The isentropic efficiency is a measure of the irreversibilities present in the compressor and can be used to relate the actual state 2 to the ideal isentropic state 2s. η c = (W in/m ) s (W in/m ) = h 2s h 1 h 2 h 1 W in/m = (W in/m ) s η c W in/m = h 2s h 1 η c W in = m (h 2s h 1 ) η c = (0.125 W in = 5.776 kw kg kj kj )(283 243.72 s kg kg ) 0.85 c. Determine the refrigeration capacity, Q in, in [tons]. Apply conservation of energy to the evaporator, de cv dt = Q in Q out + W in W out + i m i(h i + 1 V 2 i 2 + gz i ) i m e(h e + 1 V 2 e 2 + gz e ) Q 0 = Q in + m 4h 4 m 1h 1 in = m (h 1 h 4 ) = (0.125 kg kj kj ) (243.72 121.465 ) [ s kg kg Q in = 4.34 tons 1 ton ] [ 60 s ] 211 kj/min 1 min

d. Coefficient of performance β = Q in 15.28 kw = W in 5.776 kw = 2.65 β = 2.65 Discussion Irreversibilities are present in the compressor. These irreversibilties are accounted for by the isentropic efficiency and reduce the overall performance β.