Characterize the following algorithm in terms of Big-O notation. Also find the exact number of additions executed by the loop. (Assume that all variables are properly declared.) for (int i = 1; i <= n; i++) sum = sum + i * (i + 1);

The work for the process at constant pressure is 3.1 kj/kg and Heat transfer is 23.42 kj/kg.

Explanation:

We have a process with Refrigerant 22, which is a constant pressure P = 3.5 bar. We have to find the thermodynamic properties of Refrigerant 22 in Thermodynamic tables.  

According to the information presented in the problem, we can mention:  

1 - Refrigerant is a closed system.  

2- The process occurs at constant pressure

3- Work in this system will be a volume change mode.

4 - Variations for kinetic and potential energy are close to zero.  

To calculate the work in kJ/kg, we should study the constant pressure process. As we only will consider volume change mode work, it will be calculated as follows:

Values for V1 and V2 can be read from Thermodynamic tables for R-22.  

As the initial condition is P=3.5 bar saturated vapor. We should search the properties at P = 3.5 bar using saturated R-22 Table (See Table A-7 Attached).  

V1 is the specific volume for the initial condition, as it is satured vapor, V1 should be the specific volume for the gas phase in equilibrium.  

Final condition is the same pressure P=2.5 bar and T = 25°C. As Temperature is higher than Saturation Temperature at P = 2.5 bar. R-22 is superheated vapor. We can find the Volume (V2) in the table (See Table A-9 Attached).  

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After that, we can replace the equation for the work and convert to adequate units:  

The work for the process at constante pressure is 3.1 kj/kg.  

The second part of the problem is to evaluate the heat transfer. We can use an energy balance, where Ek is kinetic energy, Ep is gravitational potential energy, U is internal energy, W is the work previously calculated and Q is the heat transfer.  

ΔEk + ΔEp +ΔU =Q - W

Kinetic and potential variations are negligible. So: ΔEk = ΔEp = 0.  

Internal energies U2 and U1 are found in the table, as we did for the volume.  

U1 = Ug (P=3.bar) = 223.02 kj/kg.

U2 = U (P = 3.5 bar ; T = 25 C) = 243.34 kj/kg.

Replacing we can calculate the heat transfer Q.  

Heat transfer is 23.42 kj/kg.