ME 200 – Thermodynamics I – Spring 2020

Homework 31: Isentropic Processes and Efficiencies

Part (i): Nozzle

Part (ii): Turbine

Part (iii): Compressor

Part (i): Nozzle

Given:

Air at 200 kPa and 950 K enters an adibatic nozzle and leaves at 80 kPa.

#Given Inputs: 
P_1 = 200                   # air pressure entering the nozzle [kPa]
T_1 = 950                   # air temperature entering the nozzle [K]
V_1 = 0.                    # neglect inlet velocity 
P_2 = 80                    # air pressure leaving the nozzle [kPa]
eta_N = 0.92                # nozzle efficiency 

Find:

a) The maximum possible exit velocity

b) The minimum possible exit temperature

c) The exit velocity and temperature for an isentropic efficiency of 92%, depict the real and ideal processes on a T-s diagram

System Diagram:

The system is the air inside the nozzle.

Assumptions:

1) open system, 2) steady state, steady flow (SSSF) 3) adiabatic (no heat transfer to the surroundings), 4) air is an ideal gas, 5) negelect changes in potential energy, 6) no work, 7) internally reversible process for parts a and b

Basic Equations:

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$ $$\dfrac{dS}{dt} = \displaystyle\sum_{j} \dfrac{\dot Q_j}{T_{j,boundary}}+ \displaystyle\sum_{in} \dot m_{in} s_{in}-\displaystyle\sum_{out} \dot m_{out}s_{out}+\dot \sigma _{generation}$$ $$\dfrac{dE}{dt} = \dot Q- \dot W + \displaystyle\sum_{in} \dot m_{in} (h+ke+pe)_{in}-\displaystyle\sum_{out} \dot m_{out}(h+ke+pe)_{out}$$ $$\eta _N=\dfrac{V_2^2}{V_{2s}^2}$$ $$\bigg (\dfrac{P_2}{P_1}\bigg )_s=\bigg (\dfrac{P_{r2}}{P_{r1}} \bigg)_s$$

Solution:

A mass balance is applied to the flow stream with the assumption of steady flow ($dm/dt=0$).

$$\dot m_{2} = \dot m_{1} = \dot m$$

Entropy and energy balances are applied to the entire system with the assumption of SSSF ($dS/dt=0=dE/dt$), no change in potential energy, no work, adiabatic, and internally reversible (parts a and b). The simplified energy and entropy balances become

$$0 = \dot m (h_1+\dfrac{V_1^2}{2})-\dot m(h_2+\dfrac{V_2^2}{2})$$ $$s_{2s}=s_1$$

Thus, the isentropic relations for ideal gases can be used to determine the minimum exit temperature

$$\bigg (\dfrac{P_2}{P_1}\bigg )_s=\bigg (\dfrac{P_{r2}}{P_{r1}} \bigg)_s$$ $$P_{r2}=P_{r1}\bigg (\dfrac{P_2}{P_1}\bigg )_s$$

Pr_1 = 93.08         # relative pressure for air at T1

Pr_2 = Pr_1*(P_2/P_1)

print('Pr_2 = ',round(Pr_2,5))

Pr_2 =  37.232

The temperature associated with $P_{r2}$ can be found in the ideal gas tables ($T_{2s}$) and represents the minimum possible temperature at the exit of the nozzle as this assumes there are not irreversibilities. From the tables,

$$T_{2s}=749.4K$$

Using the isentropic temperature, the isentropic enthalpy can also be found in the ideal gas tables ($h_{2s}$) and used to determine the isentropic velocity using the energy balance with negligible inlet velocity

$$\dfrac{V_{2s}^2}{2} =h_1-h_2$$

$V_{2s}$ represents the maximum possible exit velocity for this adiabatic nozzle.

h_1 = 989.2                     # specific enthalpy of air at state 1 [kJ/kg]

T_2s = 749.4                    # temperature of air at state 2s [K]
h_2s = 766.6                    # specific enthalpy of air at state 2s [kJ/kg]

V_2s = (2*(h_1-h_2s)*1000)**.5  # maximum possible exit velocity [m/s]

print('V_2s = ',round(V_2s,1),' m/s')

V_2s =  667.2  m/s

The actual velocity can be determined using an isentropic efficiency along with the insentropic velocity.

$$\eta _N=\dfrac{V_2^2}{V_{2s}^2}$$

so that

$$V_2=(\eta _N)^{0.5} V_{2s}$$

Using the actual velocity, the actual enthalpy value for state 2 can then be determined using the energy balance.

$$h_2 = h_1-\dfrac{V_2^2}{2}$$

Finally, the ideal gas tables and the actual enthalpy value can be used to determine the actual temperature of state 2.

V_2 = V_2s*(eta_N)**.5      # Actual exit velocity [m/s]
h_2 = h_1-(V_2)**2/2000     # Actual exit enthalpy [kJ/kg]
T_2 = 765.7                 # Actual temperature of air at state 2 [K]   

print('V_2 = ',round(V_2,1),' m/s')
print('h_2 = ',round(h_2,1),' kJ/kg')

V_2 =  640.0  m/s
h_2 =  784.4  kJ/kg

The exit air temperature assocated with this enthalpy is $T_{2}=765.7K$.

Part(ii): Turbine

Given:

Steam enters an adiabtic turbine at 8 MPa and 520°C with a mass flow rate of 3 kg/s and leaves at 30 kPa. The isentropic efficiency is 90%.

#Given Inputs:
P_1 = 8000               # turbine inlet pressure [kPa]
T_1 = 520+273.15         # turbine inlet temperature [K]
m_dot = 3                # turbine inlet mass flow rate [kg/s]
P_2 = 30                 # turbine outlet pressure [kPa]
eta_T = 0.9              # turbine isentropic efficiency

Find:

a) The temperature at the turbine exit

b) The power output of the turbine

Depict the process on a T-s digram.

System Diagram:

Assumptions:

1) open system, 2) steady state, steady flow (SSSF) 3) adiabatic (no heat transfer to the surroundings), 4) negelect changes in kinetic and potential energy

Basic Equations:

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$ $$\dfrac{dS}{dt} = \displaystyle\sum_{j} \dfrac{\dot Q_j}{T_{j,boundary}}+ \displaystyle\sum_{in} \dot m_{in} s_{in}-\displaystyle\sum_{out} \dot m_{out}s_{out}+\dot \sigma _{generation}$$ $$\dfrac{dE}{dt} = \dot Q- \dot W + \displaystyle\sum_{in} \dot m_{in} (h+ke+pe)_{in}-\displaystyle\sum_{out} \dot m_{out}(h+ke+pe)_{out}$$ $$\eta _T=\dfrac{w_{act}}{w_{s}}$$

Solution:

A mass balance is applied to the flow stream with the assumption of steady flow ($dm/dt=0$).

$$\dot m_{2} = \dot m_{1} = \dot m$$

An energy balance is applied to the entire system with the assumption of SSSF ($dE/dt=0$), no changes in potential andkinetic energy, and adiabatic. The simplified energy balance becomes

$$\dot W_{12}= \dot m (h_1-h_2)$$

The entropy balance for an internally reversible and adiabatic tubiine leads to an isentropic process such that

$$s_{2s}=s_1$$

The enthalpy for state 1 can be found in the SHV tables at the given conditions ($h_1=h(P_1,T_1)$). Using the entropy from state 1 and the pressure at states 2, the entahlpy for the isentropic state 2 can be found in the SLVM tables.

$$h_{2s} = h(P_2,s=s_1)$$

Using the value for the isentropic enthalpy at state 2, the isentropic work for the turbine can determined from the energy balance.

$$\dot W_{12s}= \dot m (h_1-h_{2s})$$

Finally, the turbine isentropic efficiency can be used to find the actual work and the energy balance can be applied to find the actual enthalpy at state 2.

$$\dot W_{12}=\eta _T \dot W_{12s}$$ $$h_2= h_1-\dfrac{\dot W_{12}}{\dot m}$$

Given the exit enthalpy and pressure, the temperature at state 2 can be determined. Since the enthalpy falls between the saturated liquid and saturated vapor values for the exit pressure, the temperature is the saturation temperature at $P_2$

import CoolProp.CoolProp as CP

h_1 = CP.PropsSI('H','T',T_1,'P',P_1*1000,'Water')/1000.         # enthalpy of steam enetering the turbine [kJ/kg]
s_1 = CP.PropsSI('S','T',T_1,'P',P_1*1000,'Water')/1000.         # entropy of steam entering the turbine [kJ/kgK]
h_2s = CP.PropsSI('H','S',s_1*1000,'P',P_2*1000,'Water')/1000.   # enthalpy of steam leaving an isentropic turbine [kJ/kg]

W_dot_12s = m_dot*(h_1-h_2s)                                     # isentropic power [kW]
W_dot_12 = W_dot_12s*eta_T                                       # actual power [kW]
h_2 = h_1-W_dot_12/m_dot                                         # enthalpy of steam leaving the turbine

T_2 = CP.PropsSI('T','H',h_2*1000,'P',P_2*1000,'Water')-273.15   # temperature leaving the turbine [C]

print('h_1 = ',round(h_1,1),'kJ\kg')
print('s_1 = ',round(s_1,5),'kJ\K')
print('h_2s = ',round(h_2s,1),'kJ\kg')
print('W_dot_12s = ',round(W_dot_12s,1),'kW')
print('W_dot_12 = ',round(W_dot_12,1),'kW')
print('h_2 = ',round(h_2,1),'kJ\kg')
print('T_2 = ',round(T_2,1),'C')

h_1 =  3448.7 kJ\kg
s_1 =  6.78948 kJ\K
h_2s =  2289.8 kJ\kg
W_dot_12s =  3476.7 kW
W_dot_12 =  3129.1 kW
h_2 =  2405.7 kJ\kg
T_2 =  69.1 C

Part(iii): Compressor

Given:

A steady state and adiabatic air compressor has inlet conditions of 27°C and 95 kPa and exit conditions of 277°C and 600 kPa.

#Given Inputs:
T_1 = 27+273     # compressor inlet temperature [K]
P_1 = 95         # compressor inlet pressure [kPa]
T_2 = 277+273    # compressor outlet temperature [K]
P_2 = 600        # compressor outlet pressure [kPa]

Find:

a) The isentropic efficiency of the compressor

b) The exit temperature of the air if the process is reversible

Depict the actual and reversible process on a T-s diagram.

System Diagram:

The system is the air inside the compressor.

Assumptions:

1) open system, 2) steady state, steady flow (SSSF) 3) adiabatic (no heat transfer to the surroundings), 4) air is an ideal gas, 5) neglect changes in kinetic and potnetial energy, 6) internally reversible process for part b

Basic Equation:

$$\dfrac{dm}{dt}=\Sigma \dot m_{in}-\Sigma \dot m_{out}$$ $$\dfrac{dS}{dt} = \displaystyle\sum_{j} \dfrac{\dot Q_j}{T_{j,boundary}}+ \displaystyle\sum_{in} \dot m_{in} s_{in}-\displaystyle\sum_{out} \dot m_{out}s_{out}+\dot \sigma _{generation}$$ $$\dfrac{dE}{dt} = \dot Q- \dot W + \displaystyle\sum_{in} \dot m_{in} (h+ke+pe)_{in}-\displaystyle\sum_{out} \dot m_{out}(h+ke+pe)_{out}$$ $$\eta _C=\dfrac{w_{s}}{w_{act}}$$ $$\bigg (\dfrac{P_2}{P_1}\bigg )_s=\bigg (\dfrac{P_{r2}}{P_{r1}} \bigg)_s$$

Solution:

A mass balance is applied to the flow stream with the assumption of steady flow ($dm/dt=0$).

$$\dot m_{2} = \dot m_{1} = \dot m$$

Entropy and energy balances are applied to the system with the assumption of SSSF ($dS/dt=0=dE/dt$), no changes in potential and kinetic energy, adiabatic, and internally reversible (part b). The simplified energy and entropy balances become

$$\dot W_{12}= \dot m (h_2-h_1)$$ $$s_{2s}=s_1$$

Thus, the isentropic relations for ideal cases can be used to determine the isentropic state.

$$\bigg (\dfrac{P_2}{P_1}\bigg )_s=\bigg (\dfrac{P_{r2}}{P_{r1}} \bigg)_s$$ $$P_{r2}=P_{r1}\bigg (\dfrac{P_2}{P_1}\bigg )_s$$

Pr_1 = 1.3860        # relative pressure of air at state at T_1

Pr_2 = Pr_1*(P_2/P_1)

print('Pr_2 = ',round(Pr_2,3))

Pr_2 =  8.754

For this value of $P_{r2}$, then the associated temperature for the isentropic process can be found in the ideal gas tables.

$$T_{2s}=505.5K$$

The isentropic efficiency can be determined from

$$\eta _C=\dfrac{w_{s}}{w_{act}}$$

which reduces to the following based on the simplified energy balances

$$\eta _C=\dfrac{h_{2s}-h_1}{h_{2}-h_1}$$

The inlet and exit enthalpies are determined from the ideal gas tables using the given temperatures and the isentropic enthalpy can be found using the calculated valve for $T_{2s}$.

h_1 = 300.1          # specific enthalpy of air at T_1 [kJ/kg]
h_2 = 555.0          # specific enthalpy of air at given value of T_2 [kJ/kg]
h_2s = 509.0         # specific enthalpy of air at state 2s [kJ/kg] 

eta_c = (h_2s-h_1)/(h_2-h_1)  # compressor isentropic efficiency

print('eta_c = ',round(eta_c,3))

eta_c =  0.82