IB: 2P4: Thermofluids Flashcards

Question

Whats the difference between a gas turbine and a jet engine?

Answer 1

It allows us to examine where the available power transferred in the engine as heat goes

Answer 2

* First law efficiency focuses on **total** energy input * Second law efficiency focuses on **available** energy input

Answer 3

A fluid whose function is to provide a medium for the transfer of heat and work

Answer 4

In the absence of external effects, the state of a pure substance is fixed by the values of 2 independent properties. This allows us to say p = f(T,v) etc

Answer 5

From experiments it is known that the temeprature and specific volume can be regarded as independent and pressure determined as a function of these two. p = p(T,v). The graph of such a function is a surface, the p-v-T surface. * The saturation state is a state at which phase change starts and ends * The vapour dome is the region composed of the two-phase liquid-vapour states. The saturated liquid and vapour lines are the lines bordering the vapour dome * The critical point is the point where the saturated liquid and vapour lines meet. The critical temperature ofa. pure substance is the maxiumum temperature at which liquid and vapour

Answer 6

It is a p-v-T surface projected onto the p-v plane:

Answer 7

A liquid at the saturation temperature and pressure, ready to vaporize and enter the 2 phase region. It is sat on the saturated liquid line.

Answer 8

A gas at the saturation temperature and pressure, ready to condense and enter the 2 phase region. It is sat on the saturated vapour line.

Answer 9

Vapour heated above the saturation temperature for a given pressure.

Answer 10

Liquid cooled below the saturation temperature for a given pressure.

Answer 11

The point with the maximum temperature at which liquid and vapour phases can exist. Above this point no distinction can be made between liquid and vapour and it is known as a supercritical fluid.

Answer 12

It is constant

Answer 13

It is a p-v-T surface projected onto the T-v plane:

Answer 14

It is a p-v-T surface projected onto the p-T plane, it is also known as a phase diagram as all 3 phases are separated by lines:

Answer 15

A substance in the two-phase liquid-vapour region consists of both liquid and vapour coexisting in equilibrium. x is known as the dryness fraction, where each 1kg of mixture contains xkg of vapour and (1-x)kg of liquid.

Answer 16

* Wet saturated: x = 0 * Dry saturated: x = 1

Answer 17

Lots, almost all information can be obtained from these: * Properties of saturated water & steam * Specific enthalpy of (not saturated) water & steam * Specific entropy of (not saturated) water & steam * Density of (not saturated) water & steam * Specific internal energy of (not saturated) water & steam * Transport properties of (not saturated) common gases * Properties of R-134a The key point is to use the tables and graphs in the data book!!!

Answer 18

**Interpolation** (required only once for saturated fluid, potentially twice if not saturated)

Answer 19

* Significant practical problems with developing pumps that operate in the two-phase region * Turbine blades operating in a two-phase region can be eroded by impingement of liquid droplets on the blade * The heat additions is limited to the critical point value, this limits the maximum temperature of the heat addition and thus the carnot efficiency

Answer 20

* 1→ 2: Pump * 2 → 3: Heat addition * 3 → 4: Turbine * 4 →1 : Condenser

Answer 21

Advantages: * Takes advantage of heat transfer at lower temperature * Pump in single phase region Problems: * Turbine in two phase region results in high blade erosion

Answer 22

* 1→ 2: Pump * 2 → 3: Heat addition * 3 → 4: Turbine * 4 →1 : Condenser

Answer 23

Boiler heats into the superheated region: * Turbine operates in less of the two phase region * Average temperature of heat input rises and so efficiency rises

Answer 24

* Raise the average temperature of heat addition (T higher for Qin) * Lower the average temperature of heat rejection (T lower for Qout)

Answer 25

Key role of reheat: * Allows the boiler pressure to be far higher, increasing the temperature of Qin, and reheating vastley reduces the amount that the turbine has to operate in the 2 phase region

Answer 26

* The average T at which Qin occurs is raised by multiple reheats However: * The benefit of the 2nd reheat is around half of the first reheat and so on... * Therefore, the use of more than two reheats is not practical

Answer 27

**The heat rejection (Qout) from a gas turbine is used to heat (Qin) a rankine cycle.** Gas turbines have a high average T of Qin but problem is the exhaust temperature is very high, by using this rejected heat as Qin for the rankine cycle it can greatly increase the efficiency.

Answer 28

**HRSG (Heat Recovery Steam Generator)** An HRSG is a heat-exchanger which exchanges heat from the exhaust gas of the gas turbine to the boiler of the Rankine cycle. The HRSG is a counter flow heat-exchanger. This means that the gas (hot side), supplied by the gas turbine exhaust, enters at one end and the water (cold side), in a steam boiler, enter at the opposite end. The aim of the heat-exchanger designer is usually to minimise the temperature difference between the hot and cold flow at each point through the device. This lowers the "finite temperature difference" over which the heat transfer occurs and thus lower the irreversible entropy generation.

Answer 29

A T-x diagram is a diagram used to analyse heat-exchangers. It can almost be through of as a temperature distance graph through the HRSG (although strictly speaking rather than "distance" it is the fraction of heat transferred). The pinch point is the closest temperature between the two streams and it is the point at which water becomes saturated.

Answer 30

You can equate the heat flows between the two streams of the heat exchanger. ## Footnote NOTE: Final equation should be "... cp Ln (T9 / T8) ....", not s9/s8

Answer 31

* For a refrigerator you are concerned with Qcold * For a heat pump you are concerned with Qhot

Answer 32

Worst case scenario: * COPʀ = 0 * COPʜᴘ = 1

Answer 33

The pressure of a component of a gas mixture if it was to singly occupy the container at a volume equal to that of the mixture and a temperature equal to that of the mixture. The sum of all the partial pressures is the total pressure of the mixture.

Answer 34

* Molar Fraction: p, v * Mass Fraction: u, h, s, cv, cp, R

Answer 35

**Evaporation of liquid water** * If there’s a liquid water layer in the container, water will evaporate into the air, raising the partial pressure of the water vapour * Evaporation continues until the partial pressure of water vapour equals the saturation pressure at the given temperature T. * At this point, equilibrium is reached, and the air is saturated. **Increasing the total pressure** * If you compress the mixture (increase total pressure) without changing temperature, the partial pressure of water vapour rises. * When the water vapour partial pressure reaches the saturation pressure at that temperature, the mixture becomes saturated. **Lowering the temperature** * The saturation pressure of water vapour decreases with decreasing temperature. * If you cool the mixture at constant total pressure, eventually the partial pressure of water vapour equals the (lower) saturation pressure, and condensation begins. * The air becomes saturated as it cools. ## Footnote This is conceptually identical to just the saturation of steam! However, instead of considering the pressure we must look at the partial pressure as it is a mixture of gases.

Answer 36

The temperature at which air becomes saturated with water vapour if it is cooled at constant pressure.

Answer 37

**ω = mᵥ / mₐ** Where: mᵥ = mass of water vapour mₐ = mass of air

Answer 38

**φ = nᵥ / nᵥ,ₛₐₜ** The ratio of the moles of vapour to the moles of vapour necessary to saturate

Answer 39

The balanced chemical equation where a fuel reacts with exactly the right amount of oxygen required for complete combustion, there is no excess oxygen and no unburned fuel.

Answer 40

Air is assumed to be 21% oxygen and 79% Nitrogen. You must use this expression in the equation, simply using a multiple outside the brackets to establish how many moles are needed.

Answer 41

## Footnote Defined as ratio of masses (not moles)

Answer 42

You simply have unburnt oxygen in the products

Answer 43

* λ > 1: Combustion is lean * λ = 1: Combustion is stoichiometric * λ < 1: Combustion is rich

Answer 44

Excess oxygen

Answer 45

Excess fuel

Answer 46

Carbon Monoxide (CO) and Carbon Dioxide (CO₂) are formed. The amount of moles can be calculated by simultaneous equations formed from equating the moles of carbon and oxygen on each side of the equation

Answer 47

* Wet basis: Water vapour is formed, therefore you **do** include H₂O in any gas calculations * Dry basis: Liquid water is formed, therefore you **do NOT** include H₂O in any gas calculations

Answer 48

Split the combustion into three processes: 1, Heat removed so reactants brought from initial state (R1) to 25°C (R0). * This must account for the heat removed to change the temperature (ṁ x cp x ΔT) * and the latent heat if there is a change of state (ṁ x latent heat). 2, Heat removed so that combustion occurs at constant 25°C (R0 to P0). * If this is a stoichiometric reaction you can determine the change in enthalpy from the "combustion" section of the data book * if it is not stoichiometric, but is oxygen rich, the enthalpy change would be the same as that of the stoichiometric reaction * If it is not stoichiometric, but is fuel rich, this is more difficult!! Using the combustion section in the databook, you must construct the equation by adding or subtracting the equations of formation of carbon monoxide and carbon dioxide to the stoichiometric equation to form the real reaction equation. The corresponding enthalpy change will then be the sum (or difference) of the enthalpy values. 3, Heat added so combustion products raised to final T (P0 to P2). * This can be calculated from the "molar enthalpies at low pressures" table in the databook as it contains all the products of combustion. * Alternatively, you could the same method as part 1 and account for the heat removed to change the temperature (ṁ x cp x ΔT) , and also the latent heat if there is a change of state (ṁ x latent heat).

Answer 49

1kmol of fuel

Answer 50

The average distance a molecule moves before a collision with another molecule. This is far larger than the molecule diameter.

Answer 51

Its molecular nature, you can derive equations such as ideal gas equation of state or the specficic heat capacity by considering the gas at its molecular level.

Answer 52

The ratio of the mean free path (λ) to the size of the region we are considering (L) Kn = λ / L

Answer 53

The continuum model of a gas is the assumption that a gas can be treated as a continuous medium, rather than as discrete molecules. This is done by averaging out all the molecular velocities around a point in space and saying the fluid has a certain velocity at this point in space. It has a certain velocity field

Answer 54

**Kn << 1** When we average over very many molecules and very many collisions. This requires the Knudsen number to be much less than 1. For example, the continuum model would break down for gases at extrememly low pressures, or for flows in extremely small domains.

Answer 55

Each layer of the brick experiences the same shear stress and hence can support shear in static equilibrium. This is because its molecules are held togther by rigid bonds.

Answer 56

The fluid would flow and form a new shape. In this new shape, at static equilibrium, there is no shear stress. This is because the molecules in the fluid do not have defined positions, when one layer is displaced with respect to another, the two flow over each other to accomodate the displacement. During motion, there is a shear stress until statical equilibrium is reached. **A viscous fluid cannot support shear in static equilibrium as its molecules are not held by rigid bonds.**

Answer 57

τ = shear stress μ = dynamic viscosity dvₓ/dy = velocity gradient normal to the flow direction (flow in x direction, gradient in y direction)

Answer 58

Temperature μ = f(T)

Answer 59

* In gases, viscocity increases with temperature as the average molecular speed increases and the momentum transfer per unit time therefore increases. * In liquids, it decreases with increasing temperature, as molecules in a liquid do not simply bounce off each other but form temporary bonds with one another which enhance the transfer of momentum.

Answer 60

A fluid where the rate of momentum transfer is proportional to the velocity gradient. i.e. it obeys the law "τ = μ dvₓ/dy"

Answer 61

A fluid where the rate of momentum transfer is **not** proportional to the velocity gradient. i.e. it does not obeys the law "τ = μ dvₓ/dy". This often occurs if the fluid has molecules that are long chains or the fluid contains small suspended solids, these can align or distort with the flow direction so the viscocity depends on the velocity gradient.

Answer 62

A non-newtonian fluid where the viscocity increases with the velocity gradient.

Answer 63

A non-newtonian fluid where the viscocity decreases with the velocity gradient.

Answer 64

Orthogonal

Answer 65

Flow where the fluid is of constant density

Answer 66

Stagnation points are created from the combination of the sink and sources, and then a constant velocity stream to the right. The streamlines joining these two stagnation points behave like a solid obstacle against the flow.

Answer 67

The curl of the velocity field. Its magnitude is equal to twice the angular velocity of the fluid and it is orientated along the local axis of rotation of the flow

Answer 68

When the vorticity is zero

Answer 69

A scalar operator that gives the variation of a scalar field (such as height) following **v**

Answer 70

A "blob" of fluid small enough that it can be considered as a point moving through the fluid on a macroscopic level, but large enough that it contains a large number of molecules on the microscopic scale.

Answer 71

* This measures the change at a fixed point in space. * It is the Eulerian viewpoint. * For steady flow, this would be zero. * It does not represent the acceleration of a fluid particle by itself.

Answer 72

* This measures the change following a specific fluid particle * It is the Lagrangian viewpoint * It gives the actual acceleration of a particle

Answer 73

* This measures the change following a particle, but computed using field data * It is the Eulerian form of the Lagrangian derivative * It is numerically equal to dv/dt for particle motion

Answer 74

The navier stokes equations describe the motion of viscous fluids by applying Newton’s second law to a fluid element. For an incompressible fluid they simplify to the continuity equation and the momentum equation: ## Footnote They are essentialld F = ma for a viscous fluid

Answer 75

The navier stokes equations but ignoring viscous effects

Answer 76

The Euler equation is the same thing but in vector notation and differential form

Answer 77

For steady, incompressible, inviscid flow, if the Euler equation is applied along a streamline and integrated, it becomes Bernoullis equation.

Answer 78

It has a viscocity of zero, μ = 0, and the flow is free to slip. It cannot support any shear stress at all, and the tangential velocity of the fluid in contact with the surface is completely independent from the velocity of the surface itself.

Answer 79

When a molecules collides with a surface, it sticks to the surface long enough to reach thermal equilibrium before returning to the gas. Consequently, when they leave the surface, they have, on average, the same x-velocity and same temperature as the surface.

Answer 80

Once a molecule leaves the surface, it will collide with neighbouring molecules. After several collisions, the extra x-momentum of the molecules coming from the top surface has been diffused into adjacent layers of fluid. These in turn jostle with the molecules adjacent to them, transferring x-momentum deeper into the fluid. Eventually x-momentum diffuses right down to the bottom plate and, averaging over all the molecules' velocities, you obtain a linear velocity profile.

Answer 81

* When one plate moves parallel to the other with velocity V, the flow is known as "Couette flow" * In Couette flow, the pressure gradient is zero so the balance of forces reduces to dτ/dy = μ d²vₓ/dy² = 0 * This has a solution of the form vₓ = By + C, where B and C are constants determined from the boundary conditions

Answer 82

* When flow between stationary pates is driven by a consant pressure gradient, the flow is known as "Poiseuille" flow * In Poiseuille flow, there is a constant pressure gradient and so the balance of forces becomes dτ/dy = dp/dx, and so μ d²vₓ/dy² = dp/dx (as dτ/dy = μ d²vₓ/dy²) * This has a solution of the form: vₓ = (1/2μ dp/dx) y² + By + C, where B and C are constants determined from boundary conditions

Answer 83

1. From continuity (∇⋅v=0), the velocity is uniform along the flow direction (vₓ = independent of x, vᵧ = 0) 2. Hence the flow acceleration is then zero everywhere 3. Therefore balancing forces on a small fluid element must equal zero 4. The force balance along x gives us the result: **dτ/dy = dp/dx** 5. This can become: dτ/dy = d/dy (μ dvₓ/dy) = μ d²vₓ/dy², Hence **dτ/dy = dp/dx = μ d²vₓ/dy²** 6. **dp/dx = μ d²vₓ/dy²** can then be solved to find an expression for vₓ (Note: For pure Couette flow dp/dx = 0, and for pure Poiseuille flow dp/dx = constant) Note: The force balance along y gives us the result: ∂p/∂y = ∂τ/∂x. Since τ is a function of the gradient of velocity, it is uniform along x (as the velocity is uniform along x), hence: ∂p/∂y = ∂τ/∂x = 0 → p = p(x)

Answer 84

If a given physical phenomenon can be represented by a mathematical equation that involves a certain number N of physical variables, and the number of fundamental dimentsions involved is K, the original equation can be rewritten merely in terms of a set of P = N - K dimensionless variables

Answer 85

1. Identify your variables (Separate by independent and dependent) 2. Identify the number of fundamental dimensions (usually 3: M, L, T) 3. Use the buckingham pi theorem to identify the number of dimensionless variables 4. Create your dimensionless variables (dependent and independent). Always start by using standard non-dimensional groups from the databook! The dependent variable must only occur in the dependent dimensionless group, and each independent variable must appear in at leasy one independent dimensionless group 5. Dependent dimensionless group = fn(independent dimensionless groups) 6. Solve question as required ## Footnote Note: There would be multiple dependent dimensionless variables if there are multiple dependent variables, they would both be functions of all the independent variables and so 2 equations must be constructed

Answer 86

For 2 **geometrically similar** scenarios, but of different sizes, they would be dynamically similar as long as: * They have the same dimensionless boundary and initial conditions * They have the same Reynolds number This scenario is only try for incompressible flow, if it was compressible they must also have the same Mach number

Answer 87

The relative importance of viscous effects compared to advective effects ((u⋅∇)u this term in the navier stokes equation)

Answer 88

The volumetric flowrate divided by the cross sectional area. It is the average velocity in the pipe

Answer 89

The flow experiences zero mean acceleration, and the pressure loss can be directly related to the friction force done by the walls

Answer 90

The flow takes some length from the entrance to fully develop, but from there on it becomes independent of x and adopts the familiar Poiseuille velocity profile

Answer 91

If the flow were to remain laminar, the friction coefficient would always decrease with increasing Reynolds number. However, as Re increases (in a given pipe and for a given fluid, by increasing V), the flow is not viscous enough to dissipate all the energy put into it merely by viscous friction, and breaks down into smaller and smaller eddies, down to sizes where viscosity can eventually act. The flow is then highly fluctuating and chaotic: it has become turbulent.

Answer 92

By using the time-averaged quantities as these has steady values

Answer 93

* In laminar flow, molecular diffusion is the only transport process between layers of fluid, this makes them very hard to mix, they must be folded rather than stirred. * In turbulent flow, packets of fluid move between layers of fluid in turbulent eddies, this occurs on a much larger scale than molecular diffusion so the mixing rate is greatly increased. There is a higher rate of momentum transfer from the fluid to the pipe walls, hence a higher shear stress and a greater pressure drop.

Answer 94

In laminar flow the transport of momentum is due to molecular mixing only. However, in turbulent flow whole eddies move between layers, greatly increasing the rate of transport of momentum. Therefore we can model the effect of turbulent flow by increasing the viscosity by the "eddy viscosity", μᴛ. There is no universal model for μᴛ, the eddy size and intensity vary throughout the flow.

Answer 95

The ratio of the average bump size, k, to the diameter of the pipe, D. k/D

Answer 96

## Footnote k/D << 1

Answer 97

Hydraulically smooth flow is turbulent flow in which the wall roughness has no effect on the flow resistance. The flow breaks down into large eddies that are too large to notice the details of the surface roughness.

Answer 98

Flow when the friction becomes independent of the viscosity, and thus the reynolds number, and becomes entirely dependent on the pressure drag from roughness bumps. This is caused when the smallest eddies (for which viscosity is important) are so small that the viscosity becomes irrelevant at the length scale of the bumps, k.

IB: 2P4: Thermofluids Flashcards

(148 cards)