ME 200 – Thermodynamics I – Spring 2020

Homework 8: Properties Tables for Real Fluids

Part (i): Short answer questions

Part (ii): Property evaluations for R134a

Part (iii): Property evaluations for H2O

Part (i): Short answer questions

a) Can quality be expressed as the ratio of the volume occupied by the vapor phase to the total volume?

Answer:

No. Quality is defined as the ratio of mass of vapor to total mass of mixture.

$$\chi=\dfrac{m_{vapor}}{m_{total}}$$

Mass (m) and volume (V) can be related with:

$$m=\rho V = \dfrac{V}{v}$$

Substituting this into the definition of quality for a two-phase mixture:

$$\chi=\dfrac{m_{vapor}}{m_{total}}=\dfrac{\rho_{vapor} V_{vapor}}{\rho_{total} V_{total}} = \dfrac{v_{total} V_{vapor}}{v_{vapor} V_{total}}$$

Since the density of the vapor phase is much smaller than the liquid/total density, the ratio of the volume occupied by the vapor phase to the total volume would be much larger than the quality.

b) An ME 200 student is asked to determine the specific volume and internal energy of a compressed liquid at a given P and T. However, no compressed liquid tables exist for the substance. As a result, the student looks up the values for saturated liquid specific volume and internal energy at the given P using the saturation pressure table. Is this a good approximation in general? If not, then what would be a better approach?

Answer:

In general,this is not a good approximation. Specific volume and internal energy depend strongly on temperature for a compressed liquid, but very weakly on pressure. Using the saturated liquid properties at the given pressure (from the pressure table) could result in very significant errors if the pressure is different than the saturation pressure because the saturation temperature could be very different than the actual temperature. The appropriate approximation would be to use the saturation temperature table at the given T to find the saturated liquid specific volume and internal energy.

c) Water is contained within a tank at a temperature of 160°C and volume of 2 $m^3$. If there are equal volumes of liquid and vapor, then what is the quality?

Given:

#Given Inputs
V_tank = 2                 # volume of the tank [m^3]
V_liquid = 1               # volume of the liquid is half the tank volume [m^3]
V_vapor  = 1               # volume of the vapor is the other half of the tank volume [m^3]
T_tank = 160               # temperature of the liquid and vapor mixture [°C]

Find:

Quality ($\chi$)

System

Water: liquid and vapor mixture

Basic Equations:

$$v=V/m$$ $$\chi=\dfrac{m_{vapor}}{m_{total}}$$

Solution:

As stated, the mixture of the liquid and vapor exists in eqilibrium, the liquid phase is a saturated liquid and the vapor phase is a saturated vapor. With the given temperature, the specific volumes of the saturated liquid water and vapor can be found.

The saturated temperature table for water (Sat Water-Temp Table) in the file SLVM_Tables is used to look up the specific volume of a saturated liquid and vapor at a temperature of $160°C$ resulting in the following values.

v_f = 0.001102          # specific voloume of saturated liquid at 160 °C for water  [m^3/kg]
v_g = 0.30678           # specific voloume of saturated vapor at 160 °C for water  [m^3/kg]

The masses of the liquid and vapor in the tank are calculated with $m=V/v$. These masses are then used to determine the mixture quality from the definition: $\chi=m_{vapor}/m_{total}$.

The following code cell determines and prints the quality.

m_liquid = V_liquid/v_f       # mass of liquid in the tank [kg]
m_vapor = V_vapor/v_g         # mass of vapor in tank [kg]
x = (m_vapor) / (m_vapor + m_liquid) 

print('The quality of the water is',round(x,5))

The quality of the water is 0.00358

Part(ii): Property evaluations for R134a

Given:

Numerical tables of values for temperature (T), pressure (P), and internal energy (u) for four different states.

State	T ($^{\circ}$C)	P (kPa)	u (kJ/kg)	Phase Description
1	20		95
2	-12			Saturated Liquid
3		400	300
4	8	800

#Given Inputs
T_1 = 20         # [°C]
u_1 = 95         # [kJ/kg]
T_2 = -12        # [°C] 
x_2 = 0          # saturated liquid indicates the quality is 0
P_3 = 400        # [kPa]
u_3 = 300        # [kJ/kg]
T_4 = 8          # [°C] 
P_4 = 800        # [kPa]

Find:

Values of unknown temperature, pressure, internal energy and phase for each state

System

R134

Solution:

State 1

The internal energy is given for a specific temperature. First, check the phase of R134a by comparing the given valve of internal energy with values associated with a saturated vapor and liquid.

Using the R-134a-Temp Table within the SLVM_Table file at $20°C$:

u_g_1 = 241.02              # internal energy of saturated vapor at 20°C for R134a  [kJ/kg]
u_f_1 = 78.857              # internal energy of saturated liquid at 20°C for R134a  [kJ/kg]

The given valve of $u_1 = 95 kJ/kg$ is between the internal energy for saturated vapor and saturated liquid at that temperature, which indicates that it is a saturated liquid-vapor mixture (SLVM). The pressure is the correponding saturated pressure at 20°C, which can also be found in the R-134a-Temp Table (file SLVM_Table) at $20°C$:

$$P_{1} = 5.7171 bar = 572 kPa$$

The quality of the state 1 could then be calculated as (optional):

$$\chi_1= \dfrac{u_{1} - u_{g,1}}{u_{g,1} - u_{f,1}}$$

x_1 = (u_1 - u_f_1)/(u_g_1 - u_f_1)
print('x_1 =',round(x_1,5))

x_1 = 0.09955

State 2

This state is a saturated liquid at a temperature of -12 °C. Using the R-134a-Temp Table from the SLVM_Table file at -12 °C:

$$P_{2} = 1.8524 bar = 185 kPa$$ $$u_{2} = 35.783 kJ/kg$$

State 3

Similiar to State 1, the internal energy is specified but with pressure this time. The phase of R134a is determined by comparing this specified internal energy with the values for saturated vapor and liquid determined using the R-134a-Pressure Table from SLVM_Table at 400 kPa (4 bar):

u_g_3 = 235.09            # internal energy of saturated vapor at 400 kPa for R134a  [kJ/kg]
u_f_3 = 63.651            # internal energy of saturated liquid at 400 kPa for R134a  [kJ/kg]

The specified value of 300 kJ/kg is larger than the internal energy at the saturated vapor state, which indicates it is a superheated vapor. In order to find the temperature, we need to use the Superheated R-134a Vapor Table in the SHV_Tables file with P = 4 bar. Since there is no exact value of internal energy associated with 300 kJ/kg for this pressure, linear interpolation is used to estimate the temperature. The first step is to find the internal energy values that are closest to 300 kJ/kg along with the corresponding temperatures:

u_3_L = 294.53            # internal energy of R134a at 400 kPa that is just below u_3 [kJ/kg]
T_3_L = 80                # temperature associated with u_3_L
u_3_H = 303.33            # internal energy of R134a at 400 kPa that is just above u_3 [kJ/kg]
T_3_H = 90                # temperature associated with u_3_H

The temperature at state 3 is then calculated from linear interpolation:

$$T_3 = T_{3,L} + (T_{3,H} - T_{3,L})\dfrac{u_3 - u_{3,L}}{u_{3,H} - u_{3,L}}$$

This is applied in the following computational block to find $T_3$ with the previous data.

#Linear interpolation computation block

T_3 = T_3_L + (T_3_H - T_3_L)*(u_3 - u_3_L)/(u_3_H - u_3_L)

print('T_3 = ',round(T_3,1),'°C')

T_3 =  86.2 °C

State 4

Both temperature and pressure are given. In this case, we can check the phase by comparing the temperature with the saturation temperature at the given pressure using the R-134a-Pressure Table within SLVM_Table.

$$T_{sat}\Bigr\rvert_{P_4 = 800kPa} = 31.327°C$$

Since the given temperature for state 4 is lower than the saturation temperature, then the R134a is a compressed liquid.

This could alternatively be assessed by comparing the pressure with the saturation pressure at the given temperature using the R-134a-Temp Table within SLVM_Table.

$$P_{sat}\Bigr\rvert_{T_4 = 8°C} = 3.8761 bar$$

Since the saturation pressure is lower than the given pressure at state 4, then R134a is a compressed liquid.

Since we don't have compressed liquid data for R134a at this relatively low pressure, the saturated liquid internal energy at the given temperature represents an excellent approximation. Using the R-134a-Temp Table from the SLVM_Table with T = 8 °C:

$$u_{4} = 62.388 kJ/kg$$

Finally, the complete table is given as follows:

State	T ($^{\circ}$C)	P (kPa)	u (kJ/kg)	Phase Description
1	20	572	95	SLVM
2	-12	185	36	SL
3	86.2	400	300	SHV
4	8	800	62.4	CL

Part (iii): Property evaluations for H2O

Given:

Numerical table of values for temperature (T), pressure (P) and internal energy (u) for four different states are given below.

State	T ($^{\circ}$C)	P (kPa)	h (kJ/kg)	x	Phase Description
1		200		0.7
2	140		1800
3	80	500
4		1000	3179.4

#Given Inputs
P_1 = 200        #[kPa]
x_1 = 0.7
T_2 = 140        #[°C]
h_2 = 1800       #[kJ/kg]
T_3 = 80         #[°C] 
P_3 = 500        #[kPa]
P_4 = 1000       #[kPa]
h_4 = 3179.4     #[kJ/kg]

Find:

Values of unknown temperature, pressure, internal energy, quality (if applicable), and phase condition for each state

System:

H20

Solution:

State 1

The quality is given as 0.7, which indicates it is a liquid-vapor mixture. The pressure can be found directly using the saturation pressure table and the enthalpy can be calculated based on the quality.

$$h = h_f + \chi(h_g - h_f)$$

Using the Sat Water-Pressure Table from SLVM_Table at 200 kPa:

T_1 = 120.21                        # Saturation temperature of water at 200 kPa [°C]
h_g_1 = 2706.2                      # Enthalpy of saturated vapor at 200 kPa for water [kJ/kg]
h_f_1 = 504.70                      # Enthalpy of saturated vapor at 200 kPa for water [kJ/kg]
h_1 = h_f_1 + x_1*(h_g_1 - h_f_1)   # Enthalpy of state 1 for water [kJ/kg]

print('T_1 = ',round(T_1,1),'°C')
print('h_1 = ',round(h_1,1),'kJ/kg')

T_1 =  120.2 °C
h_1 =  2045.8 kJ/kg

State 2

The enthalpy is specified along with temperature. First, check the phase of the water by comparing the enthalpy to the saturated vapor and liquid values at the given temperature

Using the Sat Water-Temp Table within SLVM_Table at 140°C:

h_g_2 = 2733.4           # enthalpy of saturated vapor at 140 °C for water [kJ/kg]
h_f_2 = 589.16           # enthalpy of saturated liquid at 140 °C for water [kJ/kg]

The given value of 1800 kJ/kg is between the saturated vapor and liquid enthalpies which indicates that it is saturated liquid-vapor mixture (SLVM). The pressure is the corresponding saturated pressure at 140 °C, which can also be found in SLVM_Table for water:

$$P_{2} = 3.6154 bar = 361.5 kPa$$

.

The vapor quality at state 2 is then calculated as:

$$\chi_{2}= \dfrac{h_{2} - h_{f,2}}{h_{g,2} - h_{f,2}}$$

x_2 = (h_2 - h_f_2)/(h_g_2 - h_f_2)
print('x_2 = ',round(x_2,3))

x_2 =  0.565

State 3

In this case, temperature and pressure are given. First, we check the phase by comparing the temperature with the saturated temperature at the given pressure (use pressure table) or the pressure with the saturation pressure at the given temperature (use temperature table).

$$T_{sat}\Bigr\rvert_{P_3 = 500kPa} = 151.83°C$$ $$P_{sat}\Bigr\rvert_{T_3 = 80°C} = 0.47414 bar$$

Since the temperature is lower than the saturation temperature at state 3 (and pressure is greater than the saturation pressure at the given temperature), then the water is a compressed liquid. In this case, quality is not defined.

Since compressed liquid data doesn't exist in the tables at the given pressure, then the enthalpy can be approximated as

$$h_3 = u_{f,3} + v_{f,3}P_3 = h_{f,3} + v_{f,3} (P_3 - P_{sat,3})$$

Using the Sat Water-Temp Table from SLVM_Table with T = 80 °C:

P_sat_3 = 47.414                       # saturation pressure at T_3 [kPa]                   
h_f_3 = 335.01                         # enthalpy of saturated liquid at T_3 for water [kJ/kg]
v_f_3 = 0.0010291                      # enthalpy of saturated liquid at T_3 for water [m^3/kg]
h_3 = h_f_3 + v_f_3*(P_3-P_sat_3)      # enthalpy at state 3 [kj/kg]
print('h_3 = ',round(h_3,1),'kJ/kg')

h_3 =  335.5 kJ/kg

State 4

Enthalpy and pressure are specified, so we check the phase by comparing enthalpy with the saturated vapor and liquid enthalpies at the given pressure.

h_g_4 = 2777.1           # enthalpy of saturated vapor at 1000 kPa for water  [kJ/kg]
h_f_4 = 762.52           # enthalpy of saturated liquid at 1000 kPa for water  [kJ/kg]

Since the given enthalpy of 3179.4 kJ/kg is larger than the saturated vapor enthalpy, the water is a superheated vapor where quality is not defined. Looking in the superheated vapor tables for water at P = 10 bar, the specific enthlapy occurs at a temperature $T_4$ of $360°C$.

Finally, the complete table is given as follows:

State	T ($^{\circ}$C)	P (kPa)	h (kJ/kg)	x	Phase Description
1	120.2	200	2045.8	0.7	SLVM
2	140	361.5	1800	0.565	SLVM
3	80	500	335.5	-	CL
4	360	1000	3179.4	-	SHV