MCE 317 Lecture 5 - Engineering Material PDF

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THE STEADY FLOW ENERGY EQUATION Introduction A very large class of devices of interest to engineers such as turbines, compressors, nozzles, boilers, condensers etc. operate under long-term steady-state condition. After initial startup period, they operate in a way that there is no variation of properties with time. A system in a steady state has numerous properties (velocity, pressure, cross section) which may differ from point to point but do not change in/with time. This means that for those or any of the properties p of the system, the partial derivative with respect to time is zero:

If a system is in steady state, then the recently observed behavior of the system will continue into the future. In many systems, steady state is not achieved until sometime after the system is started or initiated. This initial situation is known as a transient state, start-up or warm-up period. When a periodic force is applied to a mechanical system, it will typically reach steady state after going through some transient behavior. The length of the transient state will depend on the initial conditions of the system. Given certain initial conditions a system may be in steady state from the beginning. Steady Flow Energy Equation The Steady Flow Energy Equation (SFEE) is used for open systems (a material system in which mass or energy can be lost to or gained from the environment) to determine the total energy flows. Assumptions  It is assumed that the mass flow (flow of fluids in e.g. a tube per unit of time) through the system is constant.  It is also assumed that the total energy input to the system is equal to the total energy output.  The streams of material crossing the control surface must not change their state or flow rate with time.  Each point within the control volume must not change its state with time or only cyclic state variation occurs.  The heat and work transfer rates must not change with time or the mean rates in the case of cyclic behaviour must not change.

The energies that are included are; (a) internal, (b) flow, (c) kinetic, (d) potential, (e) heat and work. The equation is shown below where suffix 1 is the entrance and suffix 2 the exit from the system u1 + p1V1 + v12/2 + gZ1 + Q = u2 + p2V2 + v22/2 + gZ2 + W where: u = p = V = v = g = Z = Q = W =

internal energy (J) pressure (N/m2) volume (m3) velocity (m/s) acceleration due to gravity (m/s2) height above a datum (m) heat flow (J) work (J)

The term pV represents flow energy. The term v2/2 represents the kinetic energy. The term gZ represents the potential energy.

In thermodynamics the changes in potential energy are usually small, thus omitting the potential energy term simplifies the equation to; u1 + p1V1 + V12/2 + Q = u2 + p2V2 + V22/2 + W Also the term (u + pV) is also known as specific enthalpy (h), so the equation is now written; h1 + V12/2 + Q = h2 + V22/2 + W

The SFEE is generally written as;

Where gZ tends to 0 Power from the system (kW) = Work output (kJ/kg) ∗ mass flow rate of fluid (kg/s)

Applications of the SFEE (i) Heat Exchangers Consider an evaporator of a refrigeration plant. Liquid Freon enters a coil in contact with the air in the refrigerator cabinet and leaves as vapour.

The velocities are small from SFEE

Thus by finding the change in enthalpy, we find the heat transfer.

(ii) Adiabatic Nozzles

A nozzle is used to produce a jet of high velocity by allowing a gas at high pressure and temperature to expand examples in rockets and jet-engines. For small velocity changes the nozzle is convergent:

While for high velocity its convergent divergent. Writing SFEE:

I.e. the increase in kinetic energy is equal to the decrease in enthalpy.

(iii) Adiabatic Throttling When a fluid flows through a restriction, such as a half-closed valve in a pipeline, the pressure downstream is always appreciably lower than upstream. This is called

throttling, in which one reduces pressure without increasing KE e.g.:

SFEE can be written between (2) and (1) if we choose (2) at a position where outgoing stream is fairly uniform:

(iv) Turbine

SFEE

The quantity:

is known as the Stagnation enthalpy

Worked Examples Example 1 In a steady flow open system a fluid flows at a rate of 4 kg/s. It enters the system at a pressure of 6 bar, a velocity of 220 m/s, internal energy is 2200 kJ/kg and specific volume 0.42 m3/kg. It leaves the system at a pressure of 1.5 bar, a velocity of 145 m/s, internal energy is 1650 kJ/kg and specific volume 1.5 m 3/kg. During its passage through the system, the fluid has a loss by heat transfer of 40kJ/kg to the surroundings. Determine the power of the system, stating whether it is from or to the system. Note: Neglect any change in potential energy. Solution Work output (kJ/kg) can be found by rearranging the SFEE. We can work in kilojoules (kJ). This means that the kinetic energy section of the SFEE will be divided by 103 to put it in kJ. W = (u1 - u2) + (P1v1 - P2v2) + (C12 – C22) + Q 2 W = (2200 - 1650) + (600 x 0.42 – 150 x 1.5) + (2202-1452) – 40 2 x 10 3 W = 550 + 27 + 13.69 - 40 W = 550.69 kJ/kg This is positive so it is energy output from the system.

Power from the system (kW) = W x m = 550.69 x 4 kg/s = 2202.76 kW Example 2 Air enters a gas turbine system with a velocity of 105 m/s and has a specific volume of 0.8 m3/kg. The inlet area of the gas turbine is 0.05m2. At exit the air has a velocity of 135 m/s and has a specific volume of 1.5 m3/kg. In its passage through the turbine system, the specific enthalpy of the air is reduced by 145 kJ/kg and the air also has a heat transfer loss of 27 kJ/kg. Determine: (a) The mass flow rate of the air through the turbine. (b) The exit area of the turbine (c) The power developed by the turbine system in kW. (a) The mass flow rate at inlet equals the mass flow rate at the outlet. Mass flow rate (m) = volume flow rate (V) x density (r) Also, Mass flow rate (m) = volume flow rate (V) / specific volume Volume flow rate (V) = velocity (v) x CSA (cross sectional area) So, Mass flow rate (m) = velocity (v) x CSA / specific volume Or m = v x CSA / specific volume At inlet m1 = 105 x 0.05 / 0.8 m1 = 6.5625 kg/s (b) Mass flow rate (m) = m2 = 6.5625 CSA2 = CSA2 =

velocity (v) x CSA / specific volume v2 x CSA2 / specific volume = 135 x CSA2 / 1.5 6.5625 x 1.5 / 135 0.0729 m2.

(c) First find the work output. u1 + P1v1 + C12/2 + Q = u2 + P2v2 + C22/2 + W since the term specific enthalpy is used we can adopt the following equation; h1 + C12/2 + Q = h2 + C22/2 + W (h1 - h2) = 145 kJ/kg or 145000 (J/kg), therefore; 145000 + (1052 / 2) + - 27000 = (1352 / 2) + W 145000 + 5512.5 - 27000 = 9112.5 + W W = 114400 J/kg W = 114.4 kJ/kg Power = W∗m P = 114.4 ∗ 6.5625 P = 750.75 kW...