Part (i): Property evaluations for R134a¶
Part (ii): Property evaluations for H2O¶
#Given Inputs
T_1 = 20 # [°C]
s_1 = 0.60005 # [kJ/kg-K]
T_2 = -12 # [°C]
x_2 = 0 # saturated liquid indicates the quality is 0
P_3 = 400 # [kPa]
s_3 = 1.0701 # [kJ/kg-K]
T_4 = 8 # [°C]
P_4 = 800 # [kPa]
Values of unknown temperature, pressure, entropy and phase for each state
R134
s_g_1 = 0.92243 # entropy of saturated vapor at 20°C for R134a [kJ/kg-K]
s_f_1 = 0.30063 # entropy of saturated liquid at 20°C for R134a [kJ/kg-K]
The given valve of $s_1 = 0.60005 kJ/kg-K$ is between the entropy 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):
$$\chi_1= \dfrac{s_{1} - s{g,1}}{s_{g,1} - s_{f,1}}$$x_1 = (s_1 - s_f_1)/(s_g_1 - s_f_1)
print('x_1 =',round(x_1,4))
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:
$$P_{2} = 1.8524 bar = 185 kPa$$$$s_{2} = 0.14505 kJ/kg-K$$Similiar to State 1, the entropy is specified but with pressure this time. The phase of R134a is determined by comparing this specified entropy with the values for saturated vapor and liquid determined using the R-134a-Pressure Table from SLVM_Table at 400 kPa (4 bar):
s_g_3 = 0.92698 # entropy of saturated vapor at 400 kPa for R134a [kJ/kg-K]
s_f_3 = 0.24771 # entropy of saturated liquid at 400 kPa for R134a [kJ/kg-K]
The specified value of 1.0701 kJ/kg-K is larger than the entropy 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 specific entropy associated with 1.0701 kJ/kg-K for this pressure, linear interpolation is used to estimate the temperature. The first step is to find the specific entropy values that are closest to 1.0701 kJ/kg-K along with the corresponding temperatures:
s_3_L = 1.0528 # entropy of R134a at 400 kPa that is just below s_3 [kJ/kg-K]
T_3_L = 50 # temperature associated with u_3_L
s_3_H = 1.0814 # entropy of R134a at 400 kPa that is just above s_3 [kJ/kg-K]
T_3_H = 60 # temperature associated with u_3_H
The temperature at state 3 is then calculated from linear interpolation:
$$ T_3 = T_{3,L} + (T_{3,H} - T_{3,L})\dfrac{s_3 - s_{3,L}}{s_{3,H} - s_{3,L}} $$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)*(s_3 - s_3_L)/(s_3_H - s_3_L)
print('T_3 = ',round(T_3,2),'°C')
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 entropy at the given temperature represents an excellent approximation. Using the R-134a-Temp Table from the SLVM_Table with T = 8 °C:
$$s_{4} = 0.24323 kJ/kg-K$$Finally, the complete table is given as follows:
| State | T ($^{\circ}$C) | P (kPa) | s (kJ/kg-K) | Phase Description |
|---|---|---|---|---|
| 1 | 20 | 572 | 0.60005 | SLVM |
| 2 | -12 | 185 | 0.14504 | SL |
| 3 | 56.05 | 400 | 1.0701 | SHV |
| 4 | 8 | 800 | 0.24323 | CL |
#Given Inputs
P_1 = 200 #[kPa]
x_1 = 0.7
T_2 = 140 #[°C]
S_2 = 4.9887 #[kJ/kg-K]
T_3 = 80 #[°C]
P_3 = 500 #[kPa]
P_4 = 1000 #[kPa]
T_4 = 360 #[°C]
Values of unknown temperature, pressure, entropy, quality (if applicable), and phase condition for each state
H20
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 entropy can be calculated based on the quality.
$$ s = s_f + \chi(s_g - s_f) $$Using the Sat Water-Pressure Table from SLVM_Table at 200 kPa:
T_1 = 120.21 # Saturation temperature of water at 200 kPa [°C]
s_g_1 = 7.1269 # Entropy of saturated vapor at 200 kPa for water [kJ/kg-K]
s_f_1 = 1.5302 # Entropy of saturated vapor at 200 kPa for water [kJ/kg-K]
s_1 = s_f_1 + x_1*(s_g_1 - s_f_1) # Entropy of state 1 for water [kJ/kg-K]
print('T_1 = ',round(T_1,1),'°C')
print('s_1 = ',round(s_1,4),'kJ/kg-K')
The entropy is specified along with temperature. First, check the phase of the water by comparing the entropy to the saturated vapor and liquid values at the given temperature
Using the Sat Water-Temp Table within SLVM_Table at 140°C:
s_g_2 = 6.9293 # entropy of saturated vapor at 140 °C for water [kJ/kg-K]
s_f_2 = 1.7392 # entropy of saturated liquid at 140 °C for water [kJ/kg-K]
The given value of 4.9887 kJ/kg-K is between the saturated vapor and liquid entropies 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:
$$\chi_{2}= \dfrac{s_{2} - s_{f,2}}{s_{g,2} - s_{f,2}}$$x_2 = (S_2 - s_f_2)/(s_g_2 - s_f_2)
print('x_2 = ',round(x_2,3))
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 entopy can be approximated as
$$ s_3 = s_f(T) $$Using the Sat Water-Temp Table from SLVM_Table with T = 80 °C:
P_sat_3 = 47.414 # saturation pressure at T_3 [kPa]
s_f_3 = 1.0756 # entropy of saturated liquid at T_3 for water [kJ/kg-K]
print('s_3 = ',round(s_f_3,4),'kJ/kg-K')
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).
$$T_{sat}\Bigr\rvert_{P_4 = 1000kPa} = 179.88°C$$$$P_{sat}\Bigr\rvert_{T_4 = 360°C} = 186.66 bar$$Since the temperature is higher than the saturation temperature at state 4 (and pressure is lower than the saturation pressure at the given temperature), then the water is a super heated vapor. In this case, quality is not defined.
The entropy at state 4 is then taken from the super heated vapor table at T=360 °C and P=10 bar.
s_4 = 7.3367 # entropy of saturated vapor at 1000 kPa and for 360 °C water [kJ/kg-K]
Finally, the complete table is given as follows:
| State | T ($^{\circ}$C) | P (kPa) | s (kJ/kg-K) | x | Phase Description |
|---|---|---|---|---|---|
| 1 | 120.2 | 200 | 5.4479 | 0.7 | SLVM |
| 2 | 140 | 361.5 | 4.9887 | 0.6261 | SLVM |
| 3 | 80 | 500 | 1.0756 | - | CL |
| 4 | 360 | 1000 | 7.3367 | - | SHV |