General Physics I (aka PHYS 2013)

General Physics I (aka PHYS 2013) PROF. VANCHURIN (AKA VITALY) University of Minnesota, Duluth (aka UMD)

OUTLINE CHAPTER 12 CHAPTER 19 REVIEW

CHAPTER 12: FLUID MECHANICS Section 12.1: Density Section 12.2: Pressure in a Fluid Section 12.3: Buoyancy Section 12.4: Fluid Flow Section 12.5: Bernoulli s Equation Section 12.6: Viscosity and Turbulence

SECTION 12.1: DENSITY We have already seen that the local destiny of a material can be defined as ρ = dm dv. (1) When the object has uniform (i.e. position independent) density, then the local density is the same as average density defined as ρ = m V. (2) For the same substance this number does not change even if the mass and volume might be different. For example both a steel wrench and a steel nail have the same density which the density of steel.

SECTION 12.1: DENSITY In SI the units of density are given by kilogram per cubic meter 1 kg/m 3 = 1 kg 1 m 3 (3) but gram per unit centimeter are also widely used 1 g/cm 3 = 1000 kg/m 3. (4) Another useful (but dimensionless) measure of density is specific density also known as relative density. It is defined as a ratio of density of a given substance to density of water (at temperature 4.0 C), specific density = ρ substance ρ water. (5)

SECTION 12.1: DENSITY Densities of common substances:

SECTION 12.1: DENSITY Example 12.2. Find the mass and weight of the air (at 1 atm and 20 C) in a living room with 4.0m 5.0m floor and a ceiling 3.0m high, and the mass and weight of an equal volume of water.

SECTION 12.1: DENSITY Example. Rank the following objects in order from highest to lowest average density: (i) mass 4.00 kg, volume 1.60 10 3 m 3 ; (ii) mass 8.00 kg, volume 1.60 10 3 m 3 ; (iii) mass 8.00 kg, volume 3.20 10 3 m 3 ; (iv) mass 2560 kg, volume 0.640 m 3 ; (v) mass 2560 kg, volume 1.28 m 3.

SECTION 12.2:PRESSURE IN A FLUID Pressure. In fluids pressure might change from one place to another and thus it is convenient to define a local pressure as p = df da which reduces to average pressure introduced earlier p = F A for uniform (i.e. position independent) pressures. Units of pressure were already introduced in the previous chapter, (6) (7) 1 Pa 1 N/m 2 1 atm 1.013 10 5 Pa 1 psi 6900 Pa. (8)

SECTION 12.2:PRESSURE IN A FLUID Example 12.2. In the living room with 4.0m 5.0m floor what is the total downward force on the floor due to air pressure of 1.00 atm?

SECTION 12.2:PRESSURE IN A FLUID Pressure with depth. Consider an infinitesimal volume element of fluid dv = dxdydz, where y-axis points upwards. In the equilibrium all forces acting on the object must add up to zero and thus along y-axis we have (ρdv) g (p + dp) da + pda = 0 ρ (dxdydz) g (p + dp) (dxdz) + p (dxdz) = 0 ρgdy (p + dp) + p = 0 dp dy = ρg (9) and thus the pressure must change linearly with y. The above equation can be solved by direct integration, i.e. p(y2 ) p(y 1 ) dp = y2 y 1 ρgdy p(y 2 ) p(y 1 ) = ρg (y 1 y 2 ). (10)

SECTION 12.2:PRESSURE IN A FLUID In terms of depth and reference pressure pressure at arbitrary depth is given by d = y 2 y 1 (11) p 0 = p(y 2 ) (12) p = p 0 + ρgd. (13) Thus if p 0 (atmospheric pressure) and p(pressure at the bottom of liquid) is the same than d must be the same:

SECTION 12.2:PRESSURE IN A FLUID If we change pressure p at the surface of fluid, the pressure will change by the same amount everywhere in the fluid. Hydraulic lift uses the same principle to measure weights p = F 1 A 1 = F 2 A 2 F 2 = A 2 A 1 F 1. (14) Pascal s law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.

SECTION 12.2:PRESSURE IN A FLUID Gauge pressure. It often useful to measure relative pressure compared to atmospheric pressure, p atm = 1 atm 14.7 psi 1.01 10 5 Pa. (15) For example, if the absolute pressure of a car tire is then it is often said that the gauge pressure is p = 47 psi (16) p gauge = p p atm = 32 psi. (17)

SECTION 12.2:PRESSURE IN A FLUID Example 12.3. Water stands 12.0 m deep in a storage tank whose top is open to the atmosphere. What are the absolute and gauge pressure at the bottom of the tank?

SECTION 12.2:PRESSURE IN A FLUID To measure the gauge pressure directly, one can use an open-tube manometer, p gauge = p p atm = ρg(y 2 y 1 ) (18) To measure air-pressure one can use a barometer where the difference in heights tells you what the atmospheric is p atm = p 0 + ρg(y 2 y 1 ) = ρg(y 2 y 1 ). (19)

SECTION 12.2:PRESSURE IN A FLUID Example 12.4. A manometer tube is partially filled with water. Oil (which does not mix with water) is poured into the left arm of the tube until the oil-water interface is at the midpoint of the tube as shown. Both arms of the tube are open to the air. Find a relationship between the heights h oil and h water.

SECTION 12.3:BUOYANCY Any object placed in a fluid experiences a force (buoyant force) arising due to changes of the pressure inside fluid. To prove it consider an arbitrary element of fluid: If the fluid is in equilibrium then the sum of all forces must balance the force of gravity B = F gravity or B = Vρ fluid g (20)

SECTION 12.3:BUOYANCY Now if we fill the shape with some other material, then the equilibrium condition might not be satisfied, but the buoyant force due to water pressure would not change. B = Vρ fluid g (21) Archimedes s principle: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body.

SECTION 12.3:BUOYANCY Example 12.5. A 15.0 kg solid gold statue is raised from the sea bottom. What is the tension in the hosting cable (assumed massless) when the statues is a) at rest and completely underwater. b) at rest and completely out of water.

SECTION 12.3:BUOYANCY Example. You place a container of seawater on a scale and note reading on the scale. You now suspend the statue of Example 12.5 in the water. How does the scale reading change? (i) it increases by 7.84 N; (ii) it decreases by 7.84 N; (iii) it remains the same; (iv) none of these.

SECTION 12.4: FLUID FLOW Consider a simple model of fluid which is incompressible (density is constant) and inviscid (vanishing internal friction). Then one can follow trajectories (or flow lines) of small elements of water (or particles) which will flow along these trajectories. The flow is steady if the flow line do no change with time. This doesn t mean that velocities of particles do not change with time. When the flow becomes irregular the small scales and large scale interact with each other giving rise to turbulence.

SECTION 12.4: FLUID FLOW Consider a flow of fluid through a pipe If the fluid is incompressible (i.e. constant density), then ρa 1 ds 1 = ρa 2 ds 2 or A 1 v 1 dt = A 2 v 2 dt (22)

SECTION 12.4: FLUID FLOW Which give us the (1D) continuity equation for non-compressible fluid A 1 v 1 = A 2 v 2. (23) The continuity equation equates the volume flow rate across different cross-sectional areas dv dt = Av (24) and can be easily generalized to the case when densities do change ρ 1 A 1 v 1 = ρ 2 A 2 v 2. (25)

SECTION 12.4: FLUID FLOW Example 12.6. Incompressible oil of density 850 kg/m 3 is pumped through a cylindrical pipe at a rate of 9.5 liters per second. (a) The first section of the pipe has a diameter of 8.0 cm. What is the flow speed of the oil? What is the mass flow rate? (b) The second section of the pipe has a diameter of 4.0 cm. What is the flow speed and the mass flow rate in that section?

SECTION 12.4: FLUID FLOW Example. A maintenance crew is working on a section of a three-lane highway, leaving only one lane open to traffic. The result is much slower traffic flow (a traffic jam.) Do cars on a highway behave like: (i) the molecules of an incompressible fluid or (ii) the molecules of compressible fluid?

SECTION 12.5: BERNOULLI S EQUATION As fluid moves through pipe external forces such as gravitational force can do work on the fluid. Bernoulli s equation: p 1 + ρgy 1 + 1 2 ρv2 1 = p 2 + ρgy 2 + 1 2 ρv2 2 (26)

SECTION 12.5: BERNOULLI S EQUATION Example 12.7. Water enters a house through a pipe with an inside diameter of 2.0 cm at an absolute pressure of 4.0 10 5 Pa. A 1.0 cm diameter pipe leads to the second-floor bathroom 5.0 m above. When the flow speed at the inlet pipe is 1.5 m/s, find the flow speed, pressure and volume flow in the bathroom.

SECTION 12.5: BERNOULLI S EQUATION Example 12.8. A gasoline storage tank with cross-sectional area A 1, filled to a depth h. The space above the gasoline contains air at pressure p 0, and the gasoline flows out the bottom of the tank through a short pipe with cross-sectional area A 2. Derive expression for the flow speed in the pipe and the volume flow rate.

SECTION 12.5: BERNOULLI S EQUATION Example 12.9. Venturi meter is used to measure flow speed in a pipe. Derive an expression for the flow speed v 1 in terms of the cross-sectional areas A 1 and A 2 and the difference in height h of the liquid levels in the two vertical tubes.

SECTION 12.5: BERNOULLI S EQUATION Example. Lift of an airplane.

CHAPTER 19: THE FIRST LAW OF THERMODYNAMICS Section 19.0: Preliminaries Section 19.1: Thermodynamic systems Section 19.2: Work done by a system Section 19.3: Paths between states Section 19.4: Internal Energy Section 19.5: Kinds of processes Section 19.6: Internal Energy of ideal gas Section 19.7: Heat Capacities Section 19.8: Adiabatic Processes

SECTION 19.0: PRELIMINARIES Our main objective is to understand the behavior of systems with a very large number of degrees of freedom N 1 (e.g. N A = 6.02214 10 23 molecules in a box). For N 10 analytical methods may be useful; for N 10 10 computer may work; for N 1 googol = 10 100 statistical physics may be the only tool. There are two standard ways to study the large N limit: phenomenological (e.g. thermodynamics) and fundamental (e.g. statistical mechanics). We will only discuss a phenomenological approach. The main idea is that only a small number of measurable (or thermodynamics) parameters: volume V, pressure p, temperature T, etc. should be sufficient for describing so-called equilibrium states.

SECTION 19.0: PRELIMINARIES Thermodynamics parameters (i.e. p, V, T) are related to each other by the so-called equation of state, f (p, V, T) = 0. (27) The equation of state for the so-called ideal gas is given by pv = nrt (28) and of the van de Waals gas ) (p + an2 V 2 (V nb) = nrt (29) where n is the amount of substance described in moles, R = 8.31 J/mol K (30) and a and b constant which (roughly speaking) represent the attractive intermolecular force and the size of molecules respectively.

SECTION 19.1: THERMODYNAMICS STATES As thermodynamic systems evolves from one thermodynamic state to another the macroscopic parameters might only change in in such a way that the equation of state remains invariant. There are two different mechanisms that allow to change a thermodynamic state of the system from the outside: Heat: Q > 0 when the system is heated Q < 0 when the system is cooled (31) Work W > 0 when the system does work W < 0 when work is done on the system. (32) Note that the sign convention for work is different in mechanics.

SECTION 19.2: WORK DONE BY SYSTEM The work done by a system can be calculated by considering transfer of energy by gas molecules when the piston is moving By integrating both sides we obtain dw = Fdx = padx = pdv (33) W = V2 V 1 p(v) dv. (34) On pv diagram the work equals to area under p(v):

SECTION 19.2: WORK DONE BY SYSTEM When pressure does not change with volume an expression for work can be obtained by integrating Eq (34) W = p(v 2 V 1 ). (35) Expansion corresponds to positive dv and thus positive W Contraction corresponds to negative dv and thus negative W

SECTION 19.2: WORK DONE BY SYSTEM Example 19.1. As an ideal gas undergoes an isothermal (constant temperature) expansion at temperature T, its volume changes from V 1 to V 2. How much work does the gas do?

SECTION 19.3: PATHS BETWEEN STATES Work done by a thermodynamic system depends on the path it takes in (p, V, T) space. Speaking mathematically dw is not an exact differential and for this reason sometimes written as δw. For example, consider the following processes:

SECTION 19.3: PATHS BETWEEN STATES Similarly the heat transferred to a system Q depends on the path it takes and thus dq (or perhaps δq) is not an exact differential. Isothermal vs. free expansion:

SECTION 19.4: INTERNAL ENERGY The First Law of Thermodynamics There is, however, a thermodynamic quality whose change does not depend on the path it takes known as internal energy, U. In thermodynamics internal energy is defined not an absolute sense, but as a differential after integration du = dq dw (36) U = U 2 U 1 = Q W (37) Note that in mechanics potential energy corresponding to conservative forces is another physical quantity of the same type that we have already discussed in some details. Equation (37) is also known as the first law of thermodynamics.

SECTION 19.4: INTERNAL ENERGY Three Laws of Thermodynamics We have already seen that W = pdv and dq = SdT, where S is another thermodynamic quantity known as entropy which measures the amount of disorder in the system. Then the first law of thermodynamics can be written as du = SdT pdv (38) It is an important experimental fact that entropy almost always grows with time ds dt 0 (39) which is also known as the second law of thermodynamics. There is also a third law of thermodynamics stating that there is an absolute zero temperature T = 0 K at which entropy is zero S(0 K) = 0. (40)

SECTION 19.4: INTERNAL ENERGY There are two special kinds of the thermodynamic processes. The first one is a cyclic process U 2 = U 1 Q = W (41) The second one is for a system in isolation W = Q = 0 U 2 = U 1. (42) Cyclic process of human body:

SECTION 19.4: INTERNAL ENERGY Example 19.4. Consider the following pv- diagram. In process a b, 150 J of heat is added to the system; in process b d, 600 J of heat is added to the system. Find (a) the internal energy change in process a b (b) the internal energy change in process a b d (c) the total hear added in process a c d.

SECTION 19.5: KINDS OF PROCESSES Adiabatic: Q = 0 U = W Isochoric: dv = 0 dw = pdv = 0 U = Q Isobaric: dp = 0 dw = pdv U = Q p (V 2 V 1 ) Isothermal: dt = 0 du = 0 Q = W.

SECTION 19.6: HEAT CAPACITIES How much heat should be added to a system to change its temperature? This is described by dq = nc (43) dt where n is mass per unit mole of the material and C is a molar heat capacity (or simply heat capacity). The heat capacity depends on the material, but moreover it depends on the process by which the heat is added. If we consider isobaric and isochoric process, then the corresponding heat capacities are notated by C p and C V respectively.

SECTION 19.6: HEAT CAPACITIES OF AN IDEAL GAS These two constants can be related to each other by applying the first law to the following transformation Then 1. isochoric decrease in temperature and decrease in pressure 2. isobaric increase in volume and increase in temperature U = Q W 0 = nc V (T 1 T 2 ) + nc V (T 2 T 1 ) p 1 (V 2 V 1 ) (44) and for ideal gas 0 = nc V (T 1 T 2 ) + nc p (T 2 T 1 ) nr (T 2 T 1 ) (45) or C p = C V + R. (46)

SECTION 19.6: HEAT CAPACITIES OF AN IDEAL GAS Using the so-called kinetic theory of gases one can show that for the ideal monatomic gases and thus but from example for diatomic molecules C V = 3 2 R C p = 5 2 R (47) γ C p = 5 1.67 (48) C V 3 γ C p = 7 = 1.4. (49) C V 5

SECTION 19.6: HEAT CAPACITIES OF AN IDEAL GAS Example 19.6. A typical dorm room or bedroom contains about 2500 moles of air. Find the change in the internal energy of this much air when it is cooled from 35.0 C to 20.0 C at constant pressure of 1.00 atm.

SECTION 19.7: INTERNAL ENERGY OF IDEAL GAS The internal energy of an ideal gas depends only on its temperature, not on its pressure of volume. For example, when partition is broken to start a free expansion of gas into the vacuum region, both the volume and pressure are changed in such a way that temperature remains constant. It is easy to see by considering isochoric process that du = dq dw = dq = nc V dt (50)

SECTION 19.8: ADIABATIC PROCESS OF IDEAL GAS During adiabatic process dq = 0 du = dw = pdv. (51)

SECTION 19.8: ADIABATIC PROCESS OF IDEAL GAS For an ideal gas the equation of state is pv = nrt (52) and which implies du = nc V dt (53) TV γ 1 = constant (54) By substituting back into equation of state we get It is also straightforward to calculate work pv T = pvγ = constant. (55) W = 1 γ 1 (p 1V 1 p 2 V 2 ). (56)

SECTION 19.8: ADIABATIC PROCESS OF IDEAL GAS Example 19.7. The compression ratio of a diesel engine is 15.0 to 1; that is, air in a cylinder is compressed to 1 1.50 of its initial volume. (a) If the initial pressure is 1.01 10 5 Pa and the initial temperature is 27 C (300 K), find the final pressure and the temperature after adiabatic compression. (b) How much work does the gas do during the compression if the initial volume of the cylinder is 1.00 L = 1.00 10 3 m 3? (Use the value C V = 20.8 J/mol K and γ = 1.400 for air.)

CHAPTER 12: FLUID MECHANICS Density and pressure Buoyancy Fluid flow Average density and pressure Specific density (relative to water) Gauge pressure (compared to air) Calculating pressure with depth Archimedes s principle Calculating buoyancy force Reaction force to buoyancy force Understanding incompressible flow Applying continuity equation Applying Bernoulli s equation

CHAPTER 19: THE FIRST LAW OF THERMODYNAMICS Thermodynamics Processes Ideal gas Thermodynamic states Equation of state of ideal gas Work done by a system The first law of thermodynamics Heat capacities at const. pressure and const. volume Adiabatic (heat and first law of thermodynamics) Isochoric (work and first law of thermodynamics) Isobaric (work and first law of thermodynamics) Isothermal (int. energy and first law of thermodynamics) PV diagrams of these processes Relations between Cp and C V capacities Internal energy of ideal gas Solving for adiabatic process of ideal gas