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First Law of Thermodynamics for a Control Mass

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Prequisite Knowledge
Quality
First Law of Thermodynamics for a Change in State of a Control Mass

Problem
Five kilograms of water at 300 kPa with a quality of 20% has its temperature raised 45°C in a isobaric process. What are the heat transfer and work in the process?

Solution
Water is the control mass.

What is the energy equation for this problem?

Consider the first law of thermodynamics which takes into account internal energy, potential energy, and kinetic energy. There is no potential or kinetic energy, just internal energy. So the first law becomes:

m(u₂ − u₁) = ₁Q₂ − ₁W₂

The pressure is constant for the process. The work can be calculated from the following equation:

1W2 =

∫

PdV = mP(ν2 − ν1)

To find ν₁, use the saturated steam tables for water using the pressure value of 300 kPa. In the tables, find the corresponding values for ν_f and ν_g.

Then, ν_f = 0.001073 m³/kg and ν_g = 0.6058 m³/kg

ν_fg = ν_g − ν_f, so:

ν_fg = 0.6058 − 0.001073 = 0.6047 m³/kg

Then, using the quality:

ν₁ = 0.001073 + 0.2(.6047) = 0.1220 m³/kg

Similarly, using the same tables, for internal energy,

u_f = 561.15 kJ/kg and u_g = 2543.6 kJ/kg

u_fg = 2543.6 − 561.15 = 1982.5 kJ/kg

u₁ = 561.15 + 0.2(1982.5) = 957.65 kJ/kg

The corresponding temperature T₁ at 300 kPa is 133.55°C (from the same tables).

Then T₂ = 133.55 + 45 = 178.55°C

The water is a superheated vapor at state 2. Therefore, use the superheated tables to find ν₂ and u₂ using the pressure 300 kPa and temperature 178.55°C.

Because the temperature falls between 150°C and 200°C, you have to interpolate for the values. Then:

ν2 = 0.6339 +   178.55 − 150 200 − 150 (0.7163 − 0.6339) = 0.6810 m3/kg similarly: u2 = 2570.8 +   178.55 − 150 200 − 150 (2650.7 − 2570.8) = 2616.43 kJ/kg The work can then be calculated: 1W2 = 5(300)(0.6810 − 0.1220) = 838.5 kJ Then using the energy equation: 1Q2 = 5(2616.43 − 957.65) + 838.5 = 7455.4 kJ