ASEN 3113 Thermodynamics And Heat Transfer

ASEN 3113 Thermodynamics and Heat Transfer Homework #4 Assigned: September 28, 2006 Due: October 10, 2006 (class time) (Total points noted in each section; must clearly show equations with values and units, drawings, assumptions, etc.)

1. (20 points) A cold air standard Otto cycle has a compression ratio of 8. The cylinder state at the end of the expansion process is T = 500 K and P = 620 kPa. The heat rejection per unit mass Q 41  163kJ / kg Determine the following: m a) (10 points) Net work per unit mass of air. b) (5 points) Thermal efficiency. c) (5 points) Mean effective pressure.

Solution:

Q41 Using the given heat rejection,  u  u . Thus, with u  c T and cv = 0.721 kJ/kg K: m 4 1 v

Q41 163kJ / kg T1  T4   500K   273.93K m  cv 0.721kJ / kgK k 1 For the isentropic compression T2 /T1  (V1 /V2 ) . Therefore

1.41 T2  (8) (273.93K)  629.35K

Also for the isentropic expansion

k 1 k1 1.41 T3  (V4 /V3 ) T4  (V1 /V2 ) T4  (8) (500K)  1148.7K a) The net work per unit mass of air is

W Q Q Q cycle  23  41  c (T  T )  41  0.721kJ / kgK(1148.7K  629.35K) 163kJ / kg  211.45kJ / kg m m m v 3 2 m b) The thermal efficiency is

W / m 211.45kJ / kg   cycle   0.565 Q23 / m 374.45kJ / kg

c) The mean effective pressure is given by W W mep  cycle  cycle V1 V2 V1 (11/ r)

Evaluating the specific volume V1=V4

RT4 (0.2870kJ / kgK)(500K) 3 Where V4    0.2315m / kg P4 620kPa

211.45kJ / kg Thus mep   1043.86kPa (0.2315m3 / kg)(11/ 8)

2. (20 points) An air standard gas turbine engine has a compressor pressure ratio of 12. It operates between 290 K and 1400 K. The turbine and compressor each have isentropic efficiencies of 90%.Determine the following: a) (10 points) Net work per unit mass of air flow. b) (5 points) Heat rejected per unit mass flow of air. c) (5 points) Thermal efficiency.

Solution: Determine the unknown temperatures:

(k1) / k  P   2  (1.41) /1.4 T2  T1    290K12  589.84K  P1 

(k 1) / k  P   4  (1.41) /1.4 T4  T3    1400K1/12  688.32K  P3  Determine the work input to the compressor and the work output of the turbine:

Win  c p (T2  T1 )  (1.005kJ / kgK)(589.84K  290K)  301.34kJ / kg

Wout  c p (T3  T4 )  (1.005kJ / kgK)(1400K  688.32K)  715.24kJ / kg a) Net work:

Wnet  Wout Win  715.24kJ / kg  301.34kJ / kg  413.9kJ / kg b) Heat rejected:

qout  c p (T4  T1 )  1.005kJ / kg(688.32K  290K)  400.31kJ / kg c) Thermal efficiency:

qin  c p (T3  T2 )  1.005kJ / kg(1400K  589.84K)  814.21kJ / kg

Wnet 413.9kJ / kg th    0.508 qin 814.21kJ / kg

3. (30 points) A four-cylinder, four-stroke internal combustion engine has a bore of 95.25 mm. and a stroke of 87.63 mm. The clearance volume is 17% of the cylinder volume at bottom dead center and the crankshaft rotates at 2600 RPM. The total volume of the cylinder is the volume of bore and stroke volume plus the clearance volume. The processes within each cylinder are modeled as an Otto cycle with a pressure of 1 atm and a temperature of 288 K at the beginning of compression. The maximum temperature in the cycle is 2888 K. Use the following constants in your analysis: k = 1.4 for air/fuel mixture (compression) and k = 1.285 for combustion products (expansion) Cv = 0.926 kJ/kg*K between steps 2 and 3, Cv = 0.825 kJ/kg*K between steps 4 and 1

(a) (4 points) Draw the P-v diagram; label Pressures, Temperatures, Qin, and Qout (b) (2 points) Calculate the mass of air at the beginning of the cycle (c) (12 points) Calculate the Pressure in kPa and Temperature in K at each step in the cycle (d) (6 points) Calculate the net Work per cycle in Joules (e) (3 points) Calculate the power developed by the engine in kW (f) (3 points) Calculate the thermal efficiency of this engine

Solution (a) See diagrams in Cengal pg 292-293.

 d 2    0.09525m2  V  V V    h  0.17V    0.08763m  0.17V Total bore&stroke clearance    total    total  4   4  3 3 VTotal  0.000624m  0.17VTotal  VTotal  0.000752m

N 3 101,000 2 0.000752m  P1V1 m m    0.92g (b solution) RT   1 0.2870 J 288K   g * K  1-2 Isentropic compression: k 1.4  v   0.000752m3  P  P  1  101kPa   P 1205kPa 2 1    3  2  v2   0.000128m  k 1 1.41  P  k 1205kPa  1.4  2  T2  T1    288K   T2  585K  P1   101kPa 

2-3 Heat addition: T  2888K  Q  mCvT  T   0.00092kg0.926 kJ 2888K  585K  3 in 3 2  kg * K 

Qin 1.96kJ P V P V V  T  2888K 2 2 3 3  2  3      P3  P2     1205kPa1   P3  5950kPa RT2 RT3 V3  T2   585K 

3-4 Isentropic Expansion: k 1.285  v   0.000128m3  P  P  3   5950kPa   P  611kPa 4 3    3  4  v4   0.000752m  k 1 1.2851  P  k 611kPa 1.285  4    T4  T3    2888K   T4 1743K  P3   5950kPa 

4-1 Heat rejection Q  mCvT  T   0.00092kg0.825kJ 288K 1743K  out 1 4  kg * K 

 Qout  1.10kJ

Net work per cycle:

Wnet  Qin  Qout 1.96kJ 1.10kJ Wnet  856J

Power developed by the engine:  2600 rev   RPM   1min   min   1min  Power  # cylinders Wnet    4cylinders 856J    Power  74.2kW  2   60sec   2   60sec    Thermal efficiency:

Wnet 856J  th   th  43.6% Qin 1960J 4. (30 points) The conditions at the beginning of compression in an air-standard Diesel cycle are fixed by p1= 200 kPa, T1 = 380 K. The compression ratio is 20 and the heat addition per unit mass is 900 kJ/kg. Use k = 1.4 for air/fuel mixture (compression) and k = 1.34 for combustion products (expansion).

Hint: To determine T3, use the averaged temperature value between T3 and T2 to determine the temperature corresponding to a value for average Cp and iterate until you calculate values for average Cp and T3 that work in the equations. A first guess for the value of T3 will be between 1600 and 2000 K. Use three decimal places only for average Cp in kJ/kg*K and also round temperature values to the nearest value in four significant digits for this problem (i.e. No decimal places for temperatures). Table A-19 has values for air and the units for Cp should be J/kg *K.

(Total points noted in each section; must clearly show equations with values and units, drawings, assumptions, etc.) (a) (4 points) Draw the P-v diagram and label Pressures, Temperatures, Qin, and Qout (b) (14 points) Calculate the Pressure in kPa and Temperature in K at each step in the cycle (c) (4 points) Calculate the net Work per cycle in kJ/kg. (d) (4 points) Calculate the cutoff ratio (e) (4 points) Calculate the thermal efficiency of the cycle

Solution: (a) See pg 298 in Cengal for diagrams.

1-2 Isentropic compression k  v   1  0.4 T2  T1    38020  T2 1259K  v2  k  v   2  1.4 P2  P1    200kPa20  P2 13300kPa  v1 

2-3 Heat addition q  C T  T  900 kJ  C T 1259K in p  3 2  kg p  3 

Choose T3 = 1983K, Calculate a Cp q  C T  T  900 kJ  C 1983 1259K  C 1.243 in p  3 2  kg p   p

Check average temperature of T3 and T2 in Cengal Table A-19

T  T  3 2 1621K  TableA 19, Air  C  1.243  Temp _ Guess _ OK 2 p Calculate cutoff ratio P V P V V   T  1983K 2 2 3 3  3   3    Const _ P        rc  1.575 RT2 RT3 V2   T2  1259K

3-4 Isentropic expansion k 1 0.34  v    1   T  T  3 r  1983K 1.575  T  836K 4 3   c     4  v4    20   k 1.34  v    1   P  P  3 r  13300kPa 1.575  P  442kPa 4 3   c     4  v4    20   q  CvT  T   0.764 kJ 380K  836K  out 1 4  kg * K   q  348kJ out kg

Net work per cycle: wnet  qin  qout  900kJ  348kJ  wnet  552kJ

Thermal efficiency:

wnet 552kJ th   th  61.3% qin 900kJ