ME 200 – Thermodynamics I – Spring 2020

Homework 38: Otto Cycles

Part (a): Net work

Part (b): Thermal efficiency

Part (c): T-s diagram

Given:

An air-standard Otto cycle has a compression ratio of 8:1. At the beginning of the compression process, the pressure is 105 kPa, the temperature is 310 K (State 1). After the constant volume heat addition process, the air temperature is 1620 K (State 3).

In [27]:
#Given Inputs:
P_1 = 105          # pressure at the begining of the compression process (state 1) [kPa]
T_1 = 310          # temperature at the begining of the compression process (state 1) [K]
T_3 = 1620         # temperature after the constant volume heat addition process (state 3) [K]
r = 8              # compression ratio of 8:1 [-]
R = 0.287          # air gas constant [KJ/kgK]

Find:

Part (a): Net work (kJ/kg)

Part (b): Thermal efficiency

Part (c): T-s diagram

System Diagram:

The system diagram shows all the processes for the Otto cycle. We select the air in the piston-cylinder assembly as the closed system.

Assumptions:

1) closed system,

2) all the processes are internally reversible,

3) adiabatic compression and expansion processes,

4) negligible changes in kinetic and potential energy,

5) ideal gas for air

Basic Equations:

$$\dfrac{dE}{dt} = \dot Q- \dot W + \displaystyle\sum_{in} \dot m_{in} (h+ke+pe)_{in}-\displaystyle\sum_{out} \dot m_{out}(h+ke+pe)_{out}$$$$Pv = RT$$$$\bigg (\dfrac{v_2}{v_1}\bigg )_s=\bigg (\dfrac{v_{r2}}{v_{r1}} \bigg)_s$$$$\eta _{th} = \dfrac{W_{net,out}}{Q_{in}}$$

Solution:

Part (a): Net work

States 1 and 2:

The analysis begins by determining the properties at each state point of the cycle. The pressure and temperature at state 1 are known and the ideal gas law can be used to determine the specific volume. The ideal gas table is used to determine the internal energy and relative volume ($v_r$) at state 1. From state 1 to 2, the process is isentropic (adibatic + internaly reversible). Therefore, the relative volume at state 2 can be calculated from isentropic relations for an ideal gas along with the specified compression ratio and used to find the temperature and internal energy at state 2. The pressure at state 2 can then be found using the ideal gas law.

$$v_1 = \dfrac{RT_1}{P_1}$$$$\dfrac{v_1}{v_2}=r=\dfrac{v_{r1}}{v_{r2}}$$$$v_{r2}=\dfrac{v_{r1}}{r}$$
In [28]:
# State 1:
v_1 = R*T_1/P_1                             # specific volume at state 1 [m^3/kg]
v_r1 = 572.3                                # relative volume at state 1 from the ideal gas table
u_1 = 221.2                                 # internal energy of air at state 1 from the ideal gas table [kJ/kg]

print('v_1 = ',round(v_1,4),'m^3\kg')

# State 2:
v_2 = v_1/r                                 # specific volume at state 2 [m^3/kg]
v_r2 = v_r1/r                               # relative volume at state 2
T_2 = 693.6                                 # temperature at state 2 from the ideal gas table, interpolation used [K]
u_2 = 507.3                                 # internal energy of air at state 2 from the ideal gas table, interpolation used [kJ/kg]
P_2 = P_1*T_2*v_1/T_1/v_2                   # pressure at state 2 [kPa]

print('v_2 = ',round(v_2,4),'m^3\kg')
print('v_r2 = ',round(v_r2,2))
print('P_2 = ',round(P_2,1),'kPa')

v_1 =  0.8473 m^3\kg
v_2 =  0.1059 m^3\kg
v_r2 =  71.54
P_2 =  1879.4 kPa

State 3:

For the Otto cycle, the heat addition process from 2 to 3 occurs at constant volume. This is meant to represent a "fast" combustion process. Since the mass and volume are constant, then the specific volume is constant and the pressure at state 3 can be determined using the ideal gas law.

$$v_2 = v_3$$$$\dfrac{RT_2}{P_2} = \dfrac{RT_3}{P_3}$$$$P_3 = \dfrac{P_2T_3}{T_2}$$

The internal energy and relative volume for state 3 can also be determined using the ideal tables for the specified temperature.

In [29]:
# State 3:
P_3 = P_2*T_3/T_2                           # pressure at state 3 [kPa]
v_3 = v_2                                   # specific volume at state 3 [m^3/kg]
v_r3 = 5.574                                # relative volume at state 3 from the ideal gas table
u_3 = 1317                                  # internal energy of air at state 3 find in the ideal gas table [kJ/kg]
print('v_3 = ',round(v_2,4),'m^3\kg')
print('P_3 = ',round(P_3,1),'kPa')

v_3 =  0.1059 m^3\kg
P_3 =  4389.7 kPa

State 4:

The process 3-4 is an isentropic expansion process. Thus, similar to the process from 1 to 2, the relative volume at state 4 can be determined using the isentropic relation for an ideal gas along with the compression ratio.

$$\dfrac{v_4}{v_3}=r=\dfrac{v_{r4}}{v_{r3}}$$$$v_{r4}=v_{r3}r$$

This can then be used to determine the temperature at state 4 using the ideal gas tables. Also, since $v_4 = v_1$, then the pressure at state 4 can be calculated using the ideal gas law.

In [30]:
# State 4:
v_4 = v_1                                   # specific volume at state 4 [m^3/kg]
v_r4 = v_r3*r                               # relative volume at state 4
T_4 = 821.6                                 # temperature at state 4 from the ideal gas table, interpolation used [K]
u_4 = 609.9                                 # internal energy of air at state 4 from the ideal gas table, interpolation used [kJ/kg]
P_4 = P_3*T_4*v_3/T_3/v_4                   # pressure at state 4 [kPa] 

print('v_4 = ',round(v_4,4),'m^3/kg')
print('v_r4 = ',round(v_r4,3))
print('P_4 = ',round(P_4,1),'kPa')

v_4 =  0.8473 m^3/kg
v_r4 =  44.592
P_4 =  278.3 kPa

With all the states known, the following state table can be used as a summary for the cycle:

State T (K) P (kPa) u (kJ/kg) v (m^3/kg) v_r
1 310 105 221.2 0.8473 572.3
2 693.7 1879.7 507.3 0.1059 71.54
3 1620 4389.7 1317 0.1059 5.574
4 821.6 278.3 609.9 0.8473 44.592

Processes 1-2 and 3-4 have work transfer but no heat transfer. So, the first law for a closed system can be used to determine the work per unit mass under the assumptions of negligible changes in kinetic and potential energy.

$$ w_{12} = u_1 - u_2$$$$w_{34} = u_3 - u_4$$

The net work per unit mass is then

$$w_{net} = w_{12} + w_{34} = (u_1 - u_2) + (u_3 - u_4) $$
In [31]:
w_12 = u_1 - u_2                             # specific work for the process 1-2 [kJ/kg]
w_34 = u_3 - u_4                             # specific work for the process 3-4 [kJ/kg]
w_net = w_12 + w_34                          # net work output of the cycle per unit mass [kJ/kg]
print('w_12 = ',round(w_12,3),'kJ/kg')
print('w_34 = ',round(w_34,3),'kJ/kg')
print('w_net = ',round(w_net,3),'kJ/kg')

w_12 =  -286.1 kJ/kg
w_34 =  707.1 kJ/kg
w_net =  421.0 kJ/kg

Part(b): Thermal Efficiency

Using the nomenclature of this problem, the thermal efficiency of the cycle is

$$\eta _{th} = \dfrac{w_{net}}{q_{in}}$$

where $w_{net}$ and $q_{in}$ are per unit mass of air within the system. The heat addition process occurs from 2 to 3, such that

$$q_{in} = u_3 - u_2$$

This is where combustion would occur in a real engine.

In [32]:
q_in = u_3 - u_2
eta_th = w_net/q_in
print('q_in = ',round(q_in,3),'kJ/kg')
print('eta_th = ',round(eta_th,3))

q_in =  809.7 kJ/kg
eta_th =  0.52

Part(c): T-s diagram