Part (i): One-dimensional linear interpolation between rows in a table¶
Part (ii): One-dimensional linear interpolation between columns in a table¶
Part (iii): Two-dimensional interpolation using intermediate results from Part (ii)¶
Part (iv): Integration of equations used in evaluating boundary work¶
Part (v): Simple application of the Ideal Gas Model¶
Numerical tables of values for temperature T, pressure P and specific volume v
| P = 1.00 MPa | P = 1.50 MPa | ||
|---|---|---|---|
| T ($^{\circ}$C) | v (m^3/kg) | T ($^{\circ}$C) | v (m^3/kg) |
| 200 | 0.20602 | 200 | 0.13245 |
| 240 | 0.22756 | 240 | 0.14831 |
The specific volume $v$ for $P=1.00$ MPa and $T=213$ $^{\circ}$C.
Consider linear interpolation between two adjacent points $(x_L,y_L)$ and $(x_H,y_H)$ in a table.
| x | y |
|---|---|
| $x_L$ | $y_L$ |
| $x_H$ | $y_H$ |
Given a value of $x_M$ such that $x_L < x_M < x_H$, then linear interpolation implies that
$$ \dfrac{y_M - y_L}{x_M - x_L} = \dfrac{y_H - y_L}{x_H - x_L}\\ $$Then
$$ y_M = y_L + (x_M - x_L)*\dfrac{y_H - y_L}{x_H - x_L}\\ $$For this problem, (x,y) corresponds to (T,v). Then, for $P=1.00$ MPa
$$ v_{213} = v_{200} + (T_{213} - T_{200})*\dfrac{v_{240} - v_{200}}{T_{240} - T_{200}}\\ $$The following computation block implements this result for linear interpolation to determine $v$ for $P=1.00$ MPa and $T=213$$^{\circ}$C using the given data table.
#Linear interpolation computation block for T = 213 C and P = 1 Mpa
v_200C_1000kPa = 0.20602 #value of specific volume at 200 C and 1 MPa, m^3/kg
v_240C_1000kPa = 0.22756 #value of specific volume at 240 C and 1 MPa, m^3/kg
T = 213 #C
v_213C_1000kPa = v_200C_1000kPa + (T - 200)*(v_240C_1000kPa - v_200C_1000kPa)/(240 - 200)
print('v @(T=213 C, P=1 MPa) = ',round(v_213C_1000kPa,5),'m^3/kg')
Table from Part (i)
Values for $v$ at P = 1.2 MPa, T =200 $^{\circ}$C and P = 1.2 MPa, T =240$^{\circ}$C.
For this problem,(x,y) in the linear interpolation relation above corresponds to (P,v) for each value of T such that
$$ v_M = P_L + (P_M - P_L)*\dfrac{v_H - v_L}{P_H - P_L}\\ $$This is applied in the following computational block to find $v$ at P = 1.2 MPa for both T =200$^{\circ}$C and T =240 $^{\circ}$C.
P = 1.2 #MPa
# Interpolate specific volume at 1.2 MPa and 200 C:
v_200C_1500kPa = 0.13245 #value of specific volume at 200 C and 1.5 MPa, m^3/kg
v_200C_1200kPa = v_200C_1000kPa + (P - 1)*(v_200C_1500kPa - v_200C_1000kPa)/(1.5 - 1)
# Interpolate specific volume at 1.2 MPa and 240 C
v_240C_1500kPa = 0.14831 #value of specific volume at 240 C and 1.5 MPa, m^3/kg
v_240C_1200kPa = v_240C_1000kPa + (P - 1)*(v_240C_1500kPa - v_240C_1000kPa)/(1.5 - 1)
print('v @(T=200 C, P=1.2 MPa) = ',round(v_200C_1200kPa,5),'m^3/kg')
print('v @(T=240 C, P=1.2 MPa) = ',round(v_240C_1200kPa,5),'m^3/kg')
Results from Part (ii)
Value for v at P = 1.20 MPa, T = 213$^{\circ}$C
Using the new table entries at 1.2 MPa as determined in (ii):
# Interpolate specific volume at 1.2 MPa and 213 C:
v_213C_1200kPa = v_200C_1200kPa + (T - 200)*(v_240C_1200kPa - v_200C_1200kPa)/(240 - 200)
print('v @(T=213 C, P=1.2MPa) = ',round(v_213C_1200kPa,5),'m^3/kg')
An alternative two-dimensional interpolation approach would be to first use the original data and linear interpolation to determine values for v at (1 MPa, 213 $^{\circ}$C) and (1.5 MPa, 213 $^{\circ}$C) and then use those values to determine an interpolated value for v at (1.2 MPa, 213 $^{\circ}$C). This would give the nearly same result.
A process involving compression of a gas in a closed system where the volume (V) is decreasing and the pressure (P) is increasing. The work input to the gas from an initial state 1 to a final state 2 is determined by evaluating the following integral:
$$ W_{\rm 12} = \int \dfrac{C}{V^n} dV $$where $ C = P_1 V_1^n = P_2 V_2^n = PV^n $
and n is a constant that depends on the process. In addition, the following data are provided and assigned to variables for calculations.
V_1 = 0.005 #m^3
P_1 = 100 #kPa
P_2 = 800 #kPa
a) For the case of $n > 1$, then
$$ W_{\rm 12} = P_1 V_1^n \int\limits_{V_1}^{V_2} \dfrac{dV}{V^n} = P_1 V_1^n \dfrac{{V_2}^{1-n}-{V_1}^{1-n}}{1-n} = \dfrac{P_1 V_1^n {V_2}^{1-n}-{P_1 V_1}}{1-n} $$However, we also have $ P_1 V_1^n = P_2 V_2^n $. So,
$$ W_{\rm 12} = \dfrac{P_2 V_2-P_1 V_1}{1-n} $$where $V_2$ is determined as $ V_2 = V_1 {\left( \dfrac{P_1}{P_2} \right)}^{1/n} $
The following code evaluates the work for $n = 1.4$.
# Determine V_2 for part a with n = 1.4:
n_a = 1.4 #-
V_2_a = V_1*(P_1/P_2)**(1/n_a)
# For n > 1, the integral for work is:"
W_12_a = -(P_2*V_2_a - P_1*V_1)/(n_a - 1) #units are kN-m which is kJ
print('Work, W_12_a = ',round(W_12_a,3),'(kJ)')
b) For the case of $n = 1$, then
$$ W_{\rm 12} = P_1 V_1^n \int\limits_{V_1}^{V_2} \dfrac{dV}{V} = P_1 V_1 ln\left( \dfrac{V_2}{V_1} \right) $$where $V_2$ is still determined as $ V_2 = V_1 {\left( \dfrac{P_1}{P_2} \right) }^{1/n} $
The following code evaluates the work for $n = 1$.
from math import log
# Determine V_2 for part b with n = 1:
n_b = 1 #-
V_2_b = V_1*(P_1/P_2)**(1/n_b)
# For n = 1, the integral for work is"
W_12_b = P_1*V_1*log(V_2_b/V_1)
print('Work, W_12_b = ',round(W_12_b,3),'(kJ)')
A system that contains a gas whose properties can be represented using the ideal gas model according to
$$ P V = n \bar{R} = m R T $$with $R = \dfrac{\bar{R}}{M}$
Also, the following specific data is provided and assigned to variables for calculations. Note that $T$ and $P$ must be absolute temperatures and pressures when using the ideal gas equation.
R_bar = 8.314 #kJ/(kmol K)
V_gas = 0.1 #m^3
m_gas = 7 #kg
P_gas = 2000 #kPa
T_gas = 80 + 273.15 #K
Molecular weight of the gas
Solving for the specific gas constant using the ideal gas model results in
$$ R = \dfrac{P V}{m T} $$Then $ M = \dfrac{\bar{R}}{R} $
R_gas = P_gas*V_gas/(m_gas*T_gas)
MW_gas = R_bar/R_gas
print('Molecular Weight, MW = ',round(MW_gas,2),'(kg/kmol)')