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// Theory and Problems of Thermodynamics
// Chapter 4
// Energy Analysis of Process
// Example 2
clear ;clc;
//Given data
V = 0.3 // Volume of container in m^3
P1 = 0.2 // Initial Pressure in MPa
power = 200 // Power of electric motor in watts
t = 15 // electric motor run time in min
// at 0.2 MPa
h_1 = 2706.7 // Specific volume of vapor in kJ/kg
v_1 = 0.8857 // Specific volume of vapor in m^3/kg
// Units conversion
t = t*60 // From min to seconds
P1 = P1*1e3 // From MPa to kPa
// Calculations
W = power*t // work done in joules
W = W*1e-3 // units conversion J to kJ
// at 0.2 MPa
u_1 = h_1 - P1*v_1 // internal energy of stream in kJ/kg
m = V/v_1 // Mass of stream in the tank in kg
del_U = W // Amount of heat energy transferred
u_2 = del_U/m + u_1
v_2 = v_1
// Final state of stream
P = 0.4 // Assumed temperature for trail an error method
// read value of Temperature by interpolation of P assumed
T = 496.92 // From super heated steam tables at P = 0.4MPa, v = 0.8857m^3/kg
h = 3478.39 // at P = 0.4MPa and T = 496.92 C
u = h - P*T // calculated internal energy from assumed value
// Output Results
mprintf('The final state of steam is: Superheated steam')
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