The First Law of Thermodynamics Closed System

So far… We’ve considered various forms of energy Q (heat) W (work) E (Energy) We haven’t tried to relate them during a process They are related through the FIRST LOW OF THERMODYNAMICS (CONSEVATION OF ENERGY PRINCIPLE)

Recall… Energy can be neither created nor destroyed; it can only change forms ( a rock falling off the cliff) Consider a system undergoing a series of adiabatic processes from a specified state 1 to another specified state 2. Obviously the processes can not involve any heat transfer but they may involve several kinds of work interactions

Recall… Experimentally observed: all adiabatic processes between two specified states of a closed system, the net work done is the same regardless of the nature of the closed system and the details of the process. This principle is called the first law A major consequences of the first law is the existence and the definition of the property total energy E

Recall… The net work is the same for all adiabatic processes of a closed system between two specified states, The net work must depend on the end states of the system only. Therefore it must correspond to a change in property of the system (i.e. the total energy) Simply the change in the total energy during an adiabatic process must be equal to the net work done.

First low of thermodynamics for Closed Systems Reminder of a Closed System. Closed system = Control mass It is defined as a quantity of matter chosen for study. No mass can cross its boundary but energy can.

Experience 1: W= 0 If we transfer 5 kJ of heat to a potato, its total energy will increase by 5 kJ.

Experience 2: W=0 Heat a water in a pan by transferring 15 KJ from the range There is 3 KJ losses This means E2-E1= 12 KJ

Experience 3: Q=0 Heat an insulated room with electric heater., W= 5KJ (electric) means E2-E1= 5KJ. Replace the electric heater with a paddle wheel.

Experience 4: Q=0 If we do a 10 kJ of boundary work on a system, the system’s internal energy will increase by 10 kJ. This is because (in the absence of any heat transfer (Q = 0), the entire boundary work will be stored in the air as part of its total energy.

Experience 5: Qin, Qout, Win If we do 6 kJ of shaft work on system and Transfer 15 kJ of heat in. Loose 3 kJ of heat out. Doing 4 kJ of Work out. Then the system internal energy will increase by 14 kJ.

Before proceeding further, let us fix the direction of Energy Transfer Heat transferred in to the system is positive Qin is +ve Heat transferred out of the system is negative Qout is -ve Work done on the system increases energy of the system. Win is +ve Work done by the system decreases energy of the system. Wout is -ve

Based on the previous experimental observations, the conservation of energy principle may be expressed as: On a rate basis Rate of net energy transfer by heat, work and mass Rate of change in total energy of the system

Usually the KE and PE are small Let us discuss the right hand side of the first low equation, i. e.  E Usually the KE and PE are small

Let us turn to the Left hand side of the 1st low equation Emass=0 for closed system If we rearrange, we get

The First low of thermodynamics Assuming Qin>Qout and Wout> Win usually we drop the subscripts, hence General Form (KJ) per unit time (or on a Rate basis) KJ/s= Watt per unit mass basis (KJ/kg) differential form

KEY CONCEPT

Energy Change for a system undergoing a cycle Q-W =U But U = U2-U1=0 (initial state and final states are the same) Hence : Q=W

Example (4-1): Cooling of a Hot fluid in a Tank A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800kj. During the cooling process, the fluid loses 500kj of heat, and the paddle wheel does 100 kj of work on the fluid. Determine the final internal energy of the fluid. Assumptions: Tank is stationary and thus KE=PE=0. Analysis: Q-W=U2-U1

Example (4-2): Electric Heating of a Gas at Constant Pressure A piston-cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. Show that for a closed system the boundary work Wb and the change in internal energy U in the first-law relation can be combined into one term, H, for a constant pressure process. Determine the final temperature of the steam.

Unrestrained Expansion of water into an Evacuated Tank Example (4-3): Unrestrained Expansion of water into an Evacuated Tank A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25oC, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25oC. Determine a) the volume of the tank, b) the final pressure, and c) the heat transfer for this process. Expansion a against a vacuum involves no work and thus no energy transfer

Example (4-6): Heating of a Gas at Constant Pressure A piston-cylinder device initially contains air at 150 kPa and 27oC. At this state, the piston is resting on a pair of stops, as shown in the figure above, and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa pressure is required to move it. The air now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and c) the total heat transferred to the air. <Answers: a) 1400 K, b) 140 kJ, c) 766.9 kJ> Sol: