import math # Variables m = 1. #[lbm] Mass of the steam T_1 = 300. #[F] Initial temperature P_1 = 14.7 #[psia] Initial pressure P_sorronding = 14.7 #[psia] Q = 50. #[Btu] Amount of the energy added to the system as heat # This is a closed system and we can apply the following equations # delta_U_system = sum(dQ_in_minus_out) + sum(dW_in_minus_out) (A) # dS_system = (m*ds)_system = sum((dQ)/T)_in_minus_out + dS_reversible (B) # From the steam tables, we look up the properties of steam at temperature 300F and pressure 14.7 psia and find u_initial = 1109.6 #[Btu/lbm] Internal energy of the steam h_initial = 1192.6 #[Btu/lbm] Enthalpy of the steam s_initial = 1.8157 #[Btu/(lbm*R)] Entropy of the steam # The work here is done by the system, equal to # -delta_w = P*A_piston*delta_x = P*m*delta_v # Calculations # Substituting this in the equation (A) and rearranging, we have # m*delta_(u + P*v) = m*delta_h = delta_Q # From which we can solve for the final specific enthalpy h_final = h_initial + Q #[Btu/lbm] # Now, by the linear interpolation we find that at h = 1242.6 Btu/lbm and P = 1 atm, temperature of the steam is given T_2 = 405.7 #[F] Final temperature # At this final temperature and pressure we have the steam properties u_final = 1147.7 #[Btu/lbm] s_final = 1.8772 #[Btu/(lbm*R)] # Thus, increase in the internal energy, enthalpy and entropy are delta_u = u_final - u_initial #[Btu/lbm] delta_s = s_final - s_initial #[Btu/(lbm*R)] delta_h = Q #[Btu/lbm] # The work done on the atmosphere is given by w = delta_h - delta_u #[Btulbm] # Results print "The increase in internal energy of the steam by adding the heat is %0.2f Btu/lbm"%(delta_u) print "The increase in enthalpy of the steam by adding the heat is %0.2f Btu/lbm"%(delta_h) print "The increase in entropy of the steam by adding the heat is %0.4f Btu/lbm"%(delta_s) print "Work done by the piston expanding against the atmosphere is %0.2f Btu/lbm"%(w)
The increase in internal energy of the steam by adding the heat is 38.10 Btu/lbm The increase in enthalpy of the steam by adding the heat is 50.00 Btu/lbm The increase in entropy of the steam by adding the heat is 0.0615 Btu/lbm Work done by the piston expanding against the atmosphere is 11.90 Btu/lbm
import math # Variables T_in = 600. #[F] Input steam temperature P_in = 200. #[psia] Input steam pressure P_exit = 50. #[psia] # Because this is a steady-state, steady-flow process, we use # (work per pound) = W/m = -( h_in - h_out ) # From the steam table we can read the the inlet enthalpy and entropy as h_in = 1322.1 #[Btu/lbm] s_in = 1.6767 #[Btu/(lb*R)] # Now, we need the value of h_out # For a reversible adiabatic steady-state, steady-flow process, we have # sum(s*m_in_minus_out) = ( s_in - s_out ) = 0 # Which indicates that inlet and outlet entropies are same # We can find the outlet temperature by finding the value of the temperature in the steam table # For which the inlet entropy at 50 psia is the same as the inlet entropy, 1.6767 Btu/(lb*R). # By linear interpolation in the table we find T_in = 307.1 #[R] # and by the linear interpolation in the same table we find that h_out = 1188.1 #[Btu/lb] # Calculations # Thus, we find W_per_pound = (h_in - h_out) #[Btu/lb] # Results print " The work output of the turbine of steam is %0.1f Btu/lb"%(-W_per_pound)
The work output of the turbine of steam is -134.0 Btu/lb
import math # Variables T = 500. #[F] P = 680. #[psi] # Calculations # It is reported in the book in the table A.1(page 417) that for water # We know that T_r = T/T_c and P_r = P/P_c, so T_c = 647.1*1.8 #[R] P_c = 220.55*14.51 #[psia] w = 0.345 T_r = (T+459.67)/T_c P_r = P/P_c z_0 = 1+P_r/T_r*(0.083-0.422/T_r**(1.6)) z_1 = P_r/T_r*(0.139-0.172/T_r**(4.2)) z = z_0+w*z_1 # Results print "The compressibility factor of steam at the given state is %0.3f"%(z) # Based on the steam table (which may be considered as reliable as the experimental data, the value of z is 0.804.
The compressibility factor of steam at the given state is 0.851