Thermo Review 2024 - Otto Cycle

An ideal Otto cycle running with air rejects 327 kJ/kg of waste heat each cycle. What is the temperature of the gas before heat rejection if at the end of the heat rejection process the gas is 27C and 95 kPa.

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 482 C

• 27 C

• 352 C

• 1461 C

Answer explanation

q = Cv*ΔT allows you to calculate the change in temperature during constant volume heat rejection.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

An ideal Otto cycle running with air rejects 327 kJ/kg of waste heat each cycle. What is the pressure of the gas before heat rejection if at the end of the heat rejection process the gas is 27 C and 95 kPa.

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 239 kPa

• 755 kPa

• 534 kPa

• 4392 kPa

Answer explanation

Because this is a constant volume process,

Pv/T = R

allows you to relate both states (through R) where v cancels on both sides (constant volume process) creating P2/T2 = P1/T1. You have enough information to solve for P1 using this information if you also calculate T1.

Tags

NGSS.HS-PS3-4

An ideal Otto cycle running with air starts its cycle at 27 C and 95 kPa. After the compression stroke, the gas state is 416 C and 1745 kPa. What's the compression ratio (r) of this cycle?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 1.2

• 8

• 7

• 10

Answer explanation

You can arrange the ideal gas law as,

v = RT/P

giving you enough information to calculate the specific volume at both states. The compression ratio is then r =v1/v2

Tags

NGSS.HS-PS3-4

750 kJ/kg of heat (q_in) is added to an ideal Otto Cycle after the compression process to the compressed gas (at 689 K, 1745 kPa). What's the temperature of the gas after the heat addition?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 1435 K

• 1734 K

• 755 K

• 1745 K

Answer explanation

q = Cv ΔT allows you to calculate a temperature change during this constant volume heat addition process.

Tags

NGSS.HS-PS3-1

An ideal Otto Cycle with a compression ratio r=8 has 750 kJ/kg of heat addition. What's the net specific work done per cycle?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 423 kJ/kg

• 41 kJ/kg

• 709 kJ/kg

• 287 kJ/kg

Answer explanation

You can use the compression ratio to calculate the cycle efficiency. From there, q_in * eff_cycle gives you the net work done by the cycle.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

An ideal Otto Cycle with a compression ratio r=7 has 536 kJ/kg heat added per cycle. How much heat is rejected each cycle?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 246 kJ/kg

• 290 kJ/kg

• 501 kJ/kg

• 303 kJ/kg

Answer explanation

Use r to calculate the efficiency, and use the efficiency to calculate how much heat is rejected and/or added.

Tags

NGSS.HS-PS3-4

What is the specific volume of the compressed gas in an ideal Otto Cycle with a mean effective pressure (MEP) of 368 kPa, net work of 313 kJ/kg, and starting conditions of 27 C and 90 kPa?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 0.1063 m^3/kg

• 0.9567 m^3/kg

• 0.1367 m^3/kg

• 0.009567 m^3/kg

Answer explanation

Use the ideal gas law to calculate the initial specific volume and then use the MEP equation solved for v2.

Tags

NGSS.HS-PS3-4

How much mass is in an ideal Otto Cycle which starts at 27 C and 90 kPa, has a compression ratio of 7, a maximum cycle temperature of 1127 C, and 1.029 kJ of total heat rejection per cycle?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 0.004181 kg

• 0.04646 kg

• 0.00283 kg

• 0.002988 kg

Answer explanation

Use r to get efficiency, use efficiency to calculate Q_in based on Q_out, then use Q=m*Cv*ΔT to calculate mass based on the heat addition to the cycle.

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-4

What is the total compressed volume in an ideal Otto Cycle, which starts at 27 C and 90 kPa, has a compression ratio of 7, and 0.004181 kg of air?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 0.9567 m^3

• 0.1366 m^3

• 0.0005714 m^3

• 0.003999 m^3

Answer explanation

Use the ideal gas law to calculate the initial specific volume, the compression ratio to find the final specific volume, and then multiply the specific volume by mass to get total volume

Tags

NGSS.HS-PS3-4

How much work is done during the power stroke of an ideal Otto Cycle with a maximum temperature of 1127 C, compression ratio of 7, and 0.004181 kg of air?

Props: Cp=1.005 [kJ/kg K], Cv=0.718 [kJ/kg K], K=1.4, R= 0.287 [kJ/kg K]

• 1.831 kJ

• 2.273 kJ

• 2.241 kJ

• 1.029 kJ

Answer explanation

Use an isentropic process relationship to find the gas temperature at the end of the power stroke, then use the temperature difference, Cv, and the mass to calculate the work done.

Tags

NGSS.HS-PS3-1