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// Theory and Problems of Thermodynamics
// Chapter 4
// Energy Analysis of Process
// Example 1
clear ;clc;
//Given data
V = 2 // Volume of container in m^3
P1 = 0.2 // Initial Pressure in MPa
P2 = 0.1 // Final Pressure in MPa
// at 0.2 MPa
h_1 = 2706.7 // Specific enthalpy of vapor in kJ/kg
v_g1 = 0.8857 // Specific volume of vapor in m^3/kg
// at 0.1 MPa
v_f2 = 0.00104 // Specific volume of liquid in m^3/kg
v_g2 = 1.694 // Specific volume of liquid in m^3/kg
h_f2 = 417.46 // Specific enthalpy of vapor in kJ/kg
h_fg2 = 2258.0 // Specific enthalpy of vapor in kJ/kg
v_1 = 0.8857 // Specific volume of vapor in m^3/kg
// UNits conversion
P1 = 0.2*1e3 // From MPa to kPa
P2 = 0.1*1e3 // From MPa to kPa
// Calculations
// at 0.2 MPa
m = V/v_g1 // Mass of stream to be filled in kg
// at 0.1 MPa
X = (v_1 - v_f2)/(v_g2 - v_f2) // Fraction of steam
// The final state of stream: wet steam quality Fraction of Steam at 0.1 MPa
h_2 = h_f2 + X*h_fg2
// Amount of heat energy transferred as heat to surroundings
del_U = m* ((h_2-h_1)-v_1*(P2-P1))
// Output Results
mprintf('The final steam quality at 0.1 MPa = %4.4f', X)
mprintf('\n Energy transferred as heat to the surroundings = %6.1f kJ' ,-del_U)
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