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+clear ;
+clc;
+// Example 22.5
+printf('Example 22.5\n\n');
+//page no. 670
+// Solution 
+
+// Take pipe between initial and final level of water
+// Given
+h_in = -20 ;// Depth of water below ground-[ft]
+h_out = 5 ;// Height of water level above ground-[ft]
+h = h_out - h_in ;// Total height to which water is pumped-[ft]
+V = 0.50 ;// Volume flow rate of water - [cubic feet/s]
+ef = 100; // Efficiency of pump - [%] 
+g = 32.2; // Acceleration due to gravity -[ft/square second] 
+gc = 32.2 ;//[(ft*lbm)/(second square*lbf)]
+
+M = V * 62.4 ;// mass flow rate - [lbm/s]
+PE_in = 0 ;// Treating initial water level to be reference level
+PE_out = (M*g*h*1.055)/(gc*778.2) ;// PE of discharged water -[lbm*(square feet/square second)]
+
+// Assumptions to simplify Equation 22.8 are:
+//1. Change in KE = 0.
+//2. Q = 0 -By given assumption
+//3. Let us assume that temperature of water is same in well and when it is discharged, therefore del_H = 0
+// Reduced equation is W = del_PE, using this:
+W = PE_out - PE_in ;//Work done on system = power delivered by pump, (since we are using mass flow rate and pump efficiency is 100 % , so W = Power) -[kW]
+
+printf('The electric power required by the pump is %.2f kW. \n', W);
+\ No newline at end of file