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ME 200 – Thermodynamics I – Spring 2020

Homework 17: System Integration

Given:

150.0 kW industrial waste water power generation with valve, flash chamber (5 bar) and turbine with the following known states.

State P (bar) T (C) x
1 15 180
2 5
3 5 0
4 5 1
5 0.1 0.95
In [2]:
# Given Inputs
P_1 = 15*100            # Pressure at state 1 [kPa]
P_2 = 5*100             # Pressure at state 2 [kPa]
P_3 = 5*100             # Pressure at state 3 [kPa]
P_4 = 5*100             # Pressure at state 4 [kPa]
P_5 = 0.1*100           # Pressure at state 5 [kPa]

T_1 = 180               # Temperature at state 1 [°C]

x_3 = 0                 # Quality at state 3 [-]
x_4 = 1                 # Quality at state 2 [-]
x_5 = 0.5               # Quality at state 2 [-]

W_turb = 150.0          # Power output of the turbine [kW] 

Find:

The mass flow rate at state 1 and sketch the processes on a T-v diagram

System Sketch:

Three different open systems are employed for the analysis: throttling valve, flash chamber, and turbine.

Assumptions:

1) steady flow, state state (SSSF), 2) negligible changes in kinetic energy (ke) and potential energy (pe) between the inlet and outlet flow streams for each control volume, 3) adiabatic devices (no heat transfer)

Basic Equations:

dEdt=˙Q˙W+in˙min(h+ke+pe)inout˙mout(h+ke+pe)outdmdt=˙min˙mouthχ=hf+χ(hghf)

Solution:

For a system involving multiple components, it is helpful to evaluate the states entering and leaving each compoment.

State 1:

The saturation pressure table for water is first used to find the corresponding saturation temperature at 15 bar: T_1=T_{sat,15bar} = 198.29 °C , which is higher than 180 °C. Therefore, the water is in a compressed liquid (CL) state. Since CL data cannot be found directly in the CL properties table at this pressure, we use saturated liquid properties from the Sat Water-Temp Table to estimate the specific enthalpy as h_{P,T} = h_{f,T} + v_{f,T} (P-P_{sat@T}).

In [3]:
h_f_180 = 763.05                             # Enthalpy for saturated liquid at 180 °C [kJ/kg]
v_f_180 = 0.0011274                          # Specific volume for saturated liquid at 180 °C [kJ/kg]
v_1 = v_f_180                                # Estimation of specific volume at state 1 [m^3/kg]
P_sat_180 = 1002.8                           # Saturation pressure at 180 °C [kPa]
h_1 = h_f_180 + (P_1 - P_sat_180)*v_f_180    # Estimation for enthalpy at state 1 [kJ/kg]

print('h_1 = ',round(h_1,1),'kJ/kg')
h_1 =  763.6 kJ/kg

State 2:

Valve Analysis:

\dfrac{dE}{dt} = \dot Q- \dot W + \dot m_{1} (h+ke+pe)_{1}- \dot m_{2}(h+ke+pe)_{2}

Applying a SSSF energy balance to the throttling valve with no work, no heat transfer, and no changes in ke and pe leads to h_2 = h_1.

With a saturated liquid entering the valve, the exit state (2) that enters the flash chamber is at a lower pressure and in the two-phase region (SLVM). The temperature can be found as the saturation temperature at 500 kPa: T_2 = 151.83 °C. Then, the quality can be determined from the known enthalpy and saturation properties as:

\chi_2= \dfrac{h_2 - h_{f,2}}{h_{g,2} - h_{f,2}}
In [4]:
T_2 = 151.83                          # Temperature at state 2 [°C ]
h_2 = h_1                             # Enthalpy at state 2 [kJ/kg]
h_f_2 = 640.09                        # Enthalpy for saturated liquid at 500 kPa from Table [kJ/kg]
h_g_2 = 2748.1                        # Enthalpy for saturated vapor at 500 kPa from Table[kJ/kg] 
x_2 = (h_2 - h_f_2)/(h_g_2 - h_f_2)   # Quality at state 2 [-]

print('x_2 = ',round(x_2,3))
x_2 =  0.059

States 3 and 4:

Phase separation of water happens in the flash chamber which is a very common device for lots of thermal systems. State 3 is saturated liquid and state 4 is saturated vapor. The temperature is the same as state 2 since they are all at the same pressure. The enthalpy of each exit state is the corresponding saturated liquid and vapor enthalpy value.

In [5]:
T_3 = T_2                                   # Temperature at state 3 [°C ]
T_4 = T_2                                   # Temperature at state 4 [°C ]
P_3 = P_2                                   # Pressure at state 3 [kPa]
P_4 = P_2                                   # Pressure at state 4 [kPa]
h_3 = h_f_2                                 # Enthalpy of saturated liquid at 500 kPa [kJ/kg]
h_4 = h_g_2                                 # Enthalpy of saturated vapor at 500 kPa [kJ/kg] 

print('h_3 = ',round(h_3,3), 'kJ/kg')
print('h_4 = ',round(h_4,3), 'kJ/kg')
print('T_3 = T_4 =',round(T_3), '°C')
print('P_3 = P_4 =',round(P_4), 'kPa')
h_3 =  640.09 kJ/kg
h_4 =  2748.1 kJ/kg
T_3 = T_4 = 152 °C
P_3 = P_4 = 500 kPa

State 5:

Pressure and quality are given at state 5. These are two independent properties of the water with quality of 0.95. It is a SLVM and all the properties can be determined using the Sat Water-Pressure Table in SLVM_Tables.

h_5 = h_{f,5} +\chi_5(h_{g,5} - h_{f,5})
In [6]:
T_5 = 45.81                             # Saturation temperature at state 5 (10 kPa) [°C]
P_5 = 0.1*100                           # Pressure at state 5 [kPa]
h_f_5 = 191.81                          # Enthalpy of saturated liquid at 10 kPa [kJ/kg]
h_g_5 = 2583.9                          # Enthalpy of saturated vapor at 10 kPa [kJ/kg]
x_5 = 0.95                              # Mixture quality at state 5 [-]
h_5 = h_f_5 + x_5*(h_g_5 - h_f_5)       # Mixture enthalpy at state 5 [kJ/kg] 

print('h_5 = ',round(h_5,3), 'kJ/kg')
h_5 =  2464.296 kJ/kg

Flash Tank Analysis:

This flash tank is an open system with a single inlet at state 2 and two outlets at states 3 and 4. For this case, the steady flow mass conservation equation: \sum \dot m_{in} = \sum \dot m_{out} , reduces to \dot m_{2} = \dot m_{3} + \dot m_{4}

Steady flow mass conservation for the valve reduces to: \dot m_{2} = \dot m_{1}

A general energy balance for the flash tank:

\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}

reduces to the following for the stated assumptions:

0 = 0 - 0 + \dot m_{2} h_2 - \dot m_{3} h_3 - \dot m_{4}h_4

Substituting \dot m_{3} = \dot m_{2} - \dot m_{4} gives:

\dot m_{2} h_2 = (\dot m_{2} - \dot m_{4}) h_3 + \dot m_{4}h_4

which can be rewritten as:

\dot m_{1} = \dot m_{2} = \dot m_4 \dfrac{h_4-h_3}{h_2-h_3}

In order to find the mass flow rate, we need to look at the turbine to determine \dot m_4.

Turbine Analysis:

For the turbine, there is only one inlet and one outlet and the energy balance reduces to:

0 = 0 - \dot W_{turb} + \dot m_{4} h_{4}- \dot m_{5} h_{5}

where \dot m_{5} = \dot m_{4}

The mass flow rate of the turbine can then be calculated as:

\dot m_{4} = \dfrac{\dot W_{turb}}{h_4-h_5}

Since all the enthalpy values are known and the turbine power output is given, we can determine the required mass flow rate as follows.

In [7]:
m_dot_4 = W_turb/(h_4-h_5)                  # mass flow rate through the turbine [kg/s]
m_dot_2 = m_dot_4*((h_4-h_3)/(h_2-h_3))     # mass flow rate through the valve [kg/s]
m_dot_1 = m_dot_2                           # mass flow rate of state point 1 [kg/s]

print('m_dot_4 = ',round(m_dot_4,2), 'kg/s')
print('m_dot_1 = ',round(m_dot_1,2), 'kg/s')
m_dot_4 =  0.53 kg/s
m_dot_1 =  9.02 kg/s

Table:

State P (bar) T (^{\circ}C) x
1 15 180 -
2 5 151.83 0.059
3 5 151.83 0
4 5 151.83 1
5 0.1 45.81 0.95

T-v diagram:

In [ ]: