Question

In this exercise we will utilize the tables for superheated water vapor, (D-3). All graphs should be generated using plotting software, e.g. Excel, MatLab, etc. (If you use NIST, do NOT utilize the built-in plotting software to generate your graphs.) Plot the specific internal energy and the specific enthalpy Vs T(K) at 1 bar. (Use a reasonable number of available temperature values but include the maximum and minimum values.) The lower limit of the ordinate should be 2500kJ/kg and the lower limit of the abscissa should be 300K. Observe the nature of the relationships and based on these calculate the ratio of specific heats, gamma. Plot the compressibility factor as a function of temperature for 1 bar and 120 bar and comment on the significance of intermolecular forces. The compressibility factor of a gas is 0.94152 when the temperature, pressure and specific volume are -50 degree F, 30atm and 0.018317m^3/kg, respectively. Which gas is it?

Answer 1

Kindly upload different questions separately.(For compressibility factor) Thank you!

MATLAB CODE:

% Superheated water vapor graph at pressure of 1 Bar

% Temperature in Celcius
T_c=[50,100,150,200,250,300,400,500,600,700,800,900,1000,1100,1200,1300];

% Temperature in Kelvin
T=T_c+273;

% Specific internal energy in KJ/Kg
u=[2505.6,2506.2,2582.9,2658.2,2733.9,2810.7,2968.3,3132.2,3302.8,3480.4,3665,3856.7,4055,4259.8,4470.7,4687.2];

% Specific enthalpy in KJ/Kg
h=[2675,2675.8,2776.6,2875.5,2974.5,3074.5,3278.6,3488.7,3705.6,3929.4,4160.2,4398,4642.6,4893.6,5150.6,5413.3];

A=plot(T,u);
xlabel('Temperature (K)')
ylabel('KJ/Kg')
legend(A,'Specific internal energy')
xlim([300 1600])
ylim([2500 5500])
hold on

B=plot(T,h);
legend(B,'Specific enthalpy')
hold off

% Ratio of specific heats, gamma
r=mean(h./u)