1. Unit 61: Engineering Thermodynamics
Lesson 13: General Formulation for Control Volumes

2. Objective
The purpose of this lesson is to examine a number of Heat Engine Cycles.

3. Control Volume
In the applications of the various laws of thermodynamics we restrict our consideration to systems where there is no mass crossing between boundaries.
for a more complete analysis we must relate Win, Qin, Wout and Qout to the pressure and temperature changes for a pump, boiler, turbine and condenser.

4. Schematic of a Power Plant
Boiler
Steam high energy)
Qin
Turbine
Wout
Steam
Low-energy
water
System Boundary
water
Condenser
Win
Pump
Qout

5. Control Volume
We need to consider each device in the power plant in turn as a control volume into which and from which a fluid flows.
Water for instance flows into the pump at low pressure and leaves the pump at a high pressure
The work input into the pump is obviously related to the pressure rise

6. Control Volume
Consider a steady flow (i.e. the flow variables, change in velocity, pressure, etc do not change with time) and a uniform flow (velocity, pressure, density are constant over the cross sectional area.

7. Control Volume
Control surface A1 V1
Control Volume A2 V2

8. Control Volume
Consider a general control volume with an area A1 where fluid enters and an area A2 where the fluid leaves.
It could have any shape and any number of entering and existing areas.
Conservation of mass requires that…
mass entering mass leaving change in mass
control volume control volume within control volume _ =
m m = Δ mc.v.

9. Control Volume
The mass that crosses an area A over a time increment Δt can be expressed as ρAVΔt, where VΔt is the distance the mass particles travel and AVΔt is the volume swept out by the mass particles.
Thus…
ρ1A1V1Δt – ρ2A2V2Δt = Δmc.v.
Where the velocities V1 and V2 are perpendicular to A1 and A2 respectively

10. Control Volume . Dividing by Δt and let Δt  0, then…
ρ1A1V1 = ρ2A1V2 = Δmc.v/dt
Thus for steady flow…
ρ1A1V1 = ρ2A1V2
The quantity of mass crossing an area each second is termed the mass flux (kg/s)
m = ρAV
AV is termed the flow rate m3/s .

11. Control Volume https://www.youtube.com/watch?v=VA03j6t5F-8
Water is flowing in a pipe that changes diameter from 20 to 40 mm. If the water in the 20 mm section has a velocity of 40 m/s, determine the velocity in the 40 mm section. Also calculate the mass flux.

12. Control Volume
A1V1 =A2V2 [π(0.02)2/4] x 40 = [π(0.04)2/4] x V V2 = 10 m/s Mass flux m = ρA1V1 = 1000 x [π(0.02)2/4] x 40 = kg/s .

13. Energy Equation
The 1st law of thermodynamics for control volume can be stated as…
The work is composed of two parts: the work due to the pressure needed to move the fluid sometime called work flow and the work that results from a rotating shaft called shaft work Ws
Net energy energy energy change of mass
transferred to entering leaving energy in
the c.v. the c.v. the c.v the c.v. _ + =
Q - W E E = Δ Ec.v.

14. Energy Equation

15. Energy Equation
W = P2A2V2Δt – P1A1V1Δt + Ws
Where PA is the pressure force and VΔt is the distance it moves during the time Δt. The negative sign results when work done on the system is negative when moving the fluid into the control volume.
The energy is composed of kinetic, potential and internal energy

16. Energy Equation . . . . Thus energy… E = (1/2)mV2 + mgz + mu
Thus the 1st law can be written as…
Q – Ws - P2A2V2Δt + P1A1V1Δt + ρ1A1V1[V12 + gz1 + u1]Δt –
ρ2A2V2[V22 + gz2 + u2]Δt = ΔEc.v.
Dividing by Δt gives…
Q – Ws = m2[(V22/2) + gz2 + u2 + (P2/ρ2)] – m1[(V12/2) + gz1 + u1 + (P1/ρ1)] + dEc.v/dt . . . .

17. Energy Equation . . . . . . Note Q = Q/Δt; Ws = W/Δt; m = ρAV
For a steady flow, a very common situation, the energy equation becomes…
Q – Ws = m[h2 – h1 + g(z2 – z1) + (V22 – V12)/2]
This is the flow most often used when a gas or a vapour is flowing. . . .

18. Energy Equation
Quite often the kinetic and potential energy changes are negligible. The first law then takes the simplified form of…
Q – Ws = m[h2 – h1] or q – ws = h2 – h1
Where q = Q/m and ws = Ws/m . . . . . . .

19. Energy Equation
For a control volume through which a liquid flows, it is most convenient to use…
– Ws = m[(P2 – P1)/ρ + (V22 – V12)/2 + g(z2 – z1)]
This is the form to use for a pump or hydroturbine.
If Q and Δu are not zero then simply add them to the equation . .

20. Applications of the Energy Equation
Its important to chose a control volume such that the flow variables are uniform or are known over the areas where the fluid enters and exits the control volume

21. Applications of the Energy Equation
Its also necessary to specify the process by which the flow variables change: is it compressible? Isothermal? Constant pressure? Adiabatic?
If the working substance behaves like an ideal gas, then appropriate equations may be used; if not tabulated values must be used such as those for steam.

22. Applications of the Energy Equation
Often heat transfer from a device or an internal energy change across a device such as flow through a pump is not desired.
For such situations the heat transfer and internal energy change may be lumped together as losses.
In a pipeline losses occur because of friction
In a centrifugal pump, losses occur because of poor fluid motion around the blades

23. Applications of the Energy Equation
For many devices losses are included as an efficiency of the device
Kinetic and potential energy changes can often be neglected in comparison with other items in the energy equation.
Potential energy changes are usually included where liquid is involved and where the inlet and exit areas are separated by a large vertical distance.

24. Applications of the Energy Equation – throttling devices
A throttling device involves a steady flow adiabatic process that provides a sudden pressure drop with no significant potential or kinetic energy changes
The process occur relatively rapidly with the result that negligible heat transfer occurs.

25. Applications of the Energy Equation – throttling devices
Orifice Plate
Globe Valve

26. Applications of the Energy Equation – throttling devices
If the energy equation is applied to such a device, with no work done and neglecting kinetic and potential energy change we have for this adiabatic non-quasi-equilibrium process….
h1 = h2
Most valves are throttling devices for which this holds true, including refrigeration units where the pressure drop causes a change of phase of the working substance – throttling is analogous to sudden expansion

27. Compressors, Pumps and Turbines
A pump is a device which transfers energy to a liquid flowing through the pump with the result that the pressure is increased.
Compressors and blowers also fall into this category but have the primary purpose of increasing the pressure in a gas.
A turbine on the other hand is a device in which work is done by the fluid on a set of rotating blades – as a result there is a pressure drop from the inlet to the outlet of the turbine

28. Compressors, Pumps and Turbines
In some circumstances there may be heat transferred from the device to the surroundings, but often the het transfer can be assumed to be negligible.
Kinetic and potential energy changes are usually neglected, thus…
- Ws = m(h2 – h1) or - ws = h2 – h1 . .

29. Compressors, Pumps and Turbines.
Ws is negative for a compressor and positive for a gas or steam turbine
In the event that heat transfer does occur from perhaps a high temperature fluid, it must be included in the equation
For liquids such as water, the energy equation neglecting kinetic and potential energy changes become…
- ws = (P2 – P1)/ρ .

30. Compressors, Pumps and Turbines
Steam enters a turbine at 4000kPa and 500oC and leaves at 80kPa. Given that the inlet velocity is 200m/s calculate the turbine output power.

31. Solution wT System Boundary v1 = 200m/s h1 = 3445.2kJ/kg P1 = 4000kPa
T1 = 500oc
Diameter = 50mm
h2 = kJ/kg
P2 = 80kPa
T2 = 338K
Diameter = 250mm wT

32. Compressors, Pumps and Turbines
The energy equation is…
-WT = m(h2 – h1)
Given that the specific volume is… m3
m = ρ1A1V1 = (1/v1)A1V1 = π(0.025)2(200) = kg/s
Thus max output power is…
WT = - ( – )(4.544) = 3542kJ/s = 3.54MW . . .

33. Compressors, Pumps and Turbines
Handout pg 81 Thermodynamics for Engineers Nozzles and Diffusers & Heat Exchangers