Fluids Physics Notes PDF

LYCEUM OF ALABANG INC.INC. Fluids GIF Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Fluids Fluids have the ability to flow Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC....

LYCEUM OF ALABANG INC.INC. Fluids GIF Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Fluids Fluids have the ability to flow Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Fluid flow can either be laminar (smooth) or turbulent (erratic). Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. The area of contact between an injection needle and the skin is a very small circle with diameter of 2.54 x 10–3cm. What pressure does the fluid exert if the needle is pushed with a force of 6.9 x 10–9N? Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC. LYCEUM OF ALABANG INC.INC. LYCEUM OF ALABANG INC.INC. LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. What is the total pressure at points A and B? Photo Credit: Richard Wright, Andrews Academy LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. ARCHIMEDES’ PRINCIPLE The buoyant force on a submerged object is equal to the weight of the fluid displaced by the object Photo Credit: Google LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. BERNOULLI’S PRINCIPLE The relationship between velocity of a fluid and the pressure it exerts. Photo Credit: Google LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. The blood flows from the aorta with area of 7.85 x 10–5m2 to the millions of capillaries whose cross section is 0.070m2. The average velocity of flowing blood in the capillaries is 9.00 x 10–4m/s. Determine the speed flowing thru the aorta. Photo Credit: Google LYCEUM OF ALABANG INC.INC. Acap = 0.070m2, vcap = 9.00 x 10–4m/s Aaorta = 7.85 x 10–5m2 vaorta = ? LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. PASCAL’S PRINCIPLE Pascal’s principle states that pressure is transmitted undiminished by a fluid in all directions. Photo Credit: Google  LYCEUM OF ALABANG INC.INC. GIF Credit: GOOGLE LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. Photo Credit: Google LYCEUM OF ALABANG INC.INC. LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. Watch the following videos on YouTube: https://www.youtube.com/watch?v=TdcSZqBqSJ0 https://www.youtube.com/watch?v=1KXmdoUptRQ https://www.youtube.com/watch?v=Yb5wyuWh2zQ https://www.youtube.com/watch?v=YyeX6ArxCYI https://www.youtube.com/watch?v=UJ3-Zm1wbIQ https://www.youtube.com/watch?v=b5SqYuWT4-4 https://www.youtube.com/watch?v=fJefjG3xhW0 https://www.khanacademy.org/science/physics/fluids/density-and- pressure/v/specific-gravity https://www.khanacademy.org/science/physics/fluids/density-and- pressure/v/fluids-part-1 https://www.khanacademy.org/science/physics/fluids/density-and- pressure/v/fluids-part-2 https://www.khanacademy.org/science/physics/fluids/buoyant-force-and- archimedes-principle/v/fluids-part-5  LYCEUM OF ALABANG INC.INC. 1. 2. 3.  LYCEUM OF ALABANG INC.INC. 4.  LYCEUM OF ALABANG INC.INC. STATES of MATTER Photo Credit: Google  LYCEUM OF ALABANG INC.INC. Solid  Atoms are in close contact so they can’t move much  Set volume and shape  Can’t compress Photo Credit: Google  LYCEUM OF ALABANG INC.INC. Liquid  Atoms move past each other  Set volume  Takes shape of container  Hard to compress Photo Credit: Google  LYCEUM OF ALABANG INC.INC. Gas o Atoms far apart o Neither set volume nor shape o Compressible Photo Credit: Google LYCEUM OF ALABANG INC.INC. Density is equal to mass divided by volume Specific gravity is the ratio of a materials’ density to the density of water Cohesion is the force of attraction between similar molecules Adhesion is the force of attraction between molecules of different substances  LYCEUM OF ALABANG INC.INC. Substance Density (kg/m3) SOLIDS Lead Copper Iron and Steel 11.3 x 103 8.9 x 103 7.9 x 103 Densities of Aluminum Granite 2.7 x 103 2.7 x 103 Common Glass 2.6 x 103 Concrete Ice 2.3 x 103 0.917 x 103 Materials Wood (pine) 0.5 x 103 LIQUIDS Mercury 13.6 x 103 Seawater 1.025 x 103 Water (40C) 1.0 x 103 GAS Air 1.29 Carbon dioxide 1.98 Steam 0.60 LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. STRESS Stress describes the magnitude of forces that cause deformation. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Tensile stress - forces pull on an object and cause its elongation, like the stretching of an elastic band Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Compressive stress - forces cause a compression of an object Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Shear stress - forces act tangentially to the object’s surface Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. The heel on a woman’s shoe occupies an area of 2.25 x 10–3m2 on the floor. Determine the compressive stress the woman with a mass of 55kg exert on the floor when standing: a. on one heel only? Photo Credit: GOOGLE b. on both heels? LYCEUM OF ALABANG INC.INC. LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. STRAIN Strain describes the deformation on an object caused by stress. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Elastic modulus is the proportionality constant of the relationship between stress & strain. Young’s modulus – elastic modulus for tensile stress bulk modulus – elastic modulus for compressive stress shear modulus – elastic modulus for shear stress  Approximate Elastic Moduli for Selected Materials LYCEUM OF ALABANG INC.INC. Material Young’s modulus × 1010 Pa Bulk modulus × 1010 Pa Shear modulus × 1010 Pa Aluminum 7.0 7.5 2.5 Bone (tension) 1.6 0.8 8.0 Brass 9.0 6.0 3.5 Brick 1.5 Concrete 2.0 Copper 11.0 14.0 4.4 Crown glass 6.0 5.0 2.5 Granite 4.5 4.5 2.0 Hair (human) 1.0 Hardwood 1.5 1.0 Iron 21.0 16.0 7.7 Lead 1.6 4.1 0.6 Marble 6.0 7.0 2.0 Nickel 21.0 17.0 7.8 Spider thread 3.0 Steel 20.0 16.0 7.5 Acetone 0.07 Ethanol 0.09 Mercury 2.5 Water 0.22  LYCEUM OF ALABANG INC.INC. Stress-Strain graph for a ductile material copper Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. L is the limit of proportionality E is the elastic limit, beyond this point the material is permanently stretched Y is the yield point, beyond this small increases in force result in increases in length B is the breaking point, the material breaks at this point.  LYCEUM OF ALABANG INC.INC. Stress-Strain graph for a brittle material glass Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Watch the following videos on YouTube https://www.youtube.com/watch?v=gGXHdgsFA9s https://www.youtube.com/watch?v=WSRqJdT2COE https://www.youtube.com/watch?v=aQf6Q8t1FQE  LYCEUM OF ALABANG INC.INC. HEAT, TEMPERATURE & THERMAL EXPANSION Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. HEAT Heat is thermal energy which may be absorbed or given off by an object. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. TEMPERATURE Temperature is the measure of the average kinetic energy of the molecules of a substance Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Equation to convert one temperature scale to °𝐂 𝟓 another: = & °𝐂 + 273 = K °𝐅−𝟑𝟐 𝟗 where °𝐂 is temperature in Celsius °𝐅 is temperature in Fahrenheit K is temperature in Kelvin  LYCEUM OF ALABANG INC.INC. A person’s temperature reads 1030F. The normal temperature of a human body is 370C. Does the person have fever? Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC. 0F = 103 °𝐂 𝟓 0F−32 = 𝟗 °𝐂 𝟓 = 103−32 𝟗 °𝐂 𝟓 = 71 𝟗 0C = 71 (5) 9 0C = 39.44  LYCEUM OF ALABANG INC.INC. THERMAL EXPANSION Thermal expansion is the tendency of matter to change its shape, area, volume, and density due to a change in temperature Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. LINEAR EXPANSION GIF Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. LINEAR EXPANSION change in length of an object caused by a change in temperature Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Equation of linear expansion: 𝚫𝐋 = 𝛂𝐋𝐢 𝚫𝐓 where 𝚫𝐋 = 𝐋𝐟 − 𝐋𝐢 in meters (m) α is the coefficient of linear expansion in /0C 𝐋𝐢 is the initial length in m 𝚫𝐓 = 𝑻𝒇 − 𝑻𝒊 in 0C LYCEUM OF ALABANG INC.INC.  LYCEUM OF ALABANG INC.INC. A 1 meter long aluminum rod is heated from 300C to 550C. By how much did it change? Find the final length of the rod. Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC. 𝐋𝐢 = 1m, 𝚫𝐓 = 550C–300C = 250C, 𝛂 = 24x10–6/0C 𝐋𝒇 =?, 𝚫𝐋 = ? 𝚫𝐋 = 𝛂𝐋𝟎 𝚫𝐓 𝚫𝐋 = 24x10–6/0C(1m)(250C) 𝚫𝐋 = 6 x 10–4m Lf − Li = 6 x 10–4m Lf = 6 x 10–4m + 1m 𝐋𝐟 = 1.0006m  LYCEUM OF ALABANG INC.INC. VOLUME EXPANSION GIF Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. VOLUME EXPANSION change in volume of an object caused by a change in temperature Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Equation of volume expansion: 𝚫𝐕 = 𝛃𝐕𝐢 𝚫𝐓 where 𝚫𝐕 = 𝐕𝐟 − 𝐕𝐢 in Liters (L) β is coefficient of volume expansion in /0C 𝐕𝐢 is the initial length in L 𝚫𝐓 = 𝑻𝒇 − 𝑻𝒊 in 0C  LYCEUM OF ALABANG INC.INC. 100mL of ethyl alcohol at 150C is heated to 570C. Calculate its final volume. Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC. 1L 𝐕𝐢 = 100mL 1000mL = 0.1L, 𝚫𝐓 = 570C–150C = 420C, 𝛃 = 1100x10–6/0C 𝚫𝐕 = 𝛃𝐕𝐢 𝚫𝐓 𝚫𝐕 = 1100x10–6/0C (0.1L)(420C) = 4.62 x 10–3L 𝚫𝐕 = 𝐕𝐟 − 𝐕𝐢 4.62 x 10–3L = Vf − 0.1L 4.62 x 10–3L + 0.1L = Vf 0.10462 L = 𝐕𝐟  LYCEUM OF ALABANG INC.INC. HEAT TRANSFER Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. CONDUCTION transfer of heat energy from particle to particle Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. CONVECTION transfer of heat energy by the movement of hot and cold air Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. RADIATION transfer of heat energy by electromagnetic waves Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Heat capacity is the amount of heat required to increase the temperature of a material by 10C Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Specific heat capacity is the amount of heat needed to increase the temperature of 1 gram of the material by 10C Photo Credit: GOOGLE LYCEUM OF ALABANG INC.INC. Substance Specific Heat (J/kg.ºC) water 4184 ethyl alcohol 2430 Specific ice 2090 steam 2010 Heats of aluminum 900 glass 840 Common iron copper 448 390 Materials zinc 386 brass 3800 silver 236 lead 128  LYCEUM OF ALABANG INC.INC. Equation of heat transfer: Q = mcΔT where Q is the amount of heat required in Joules m is mass in kg c is the specific heat in J/kg0C 𝚫𝐓 = 𝐓𝐟 − 𝐓𝐢 (change in temperature)  LYCEUM OF ALABANG INC.INC. How much heat is needed to raise temperature of 2.5kg of water from 200C to its boiling point? Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. m = 2.5kg, c = 4184 J/kg0C, ΔT = 1000C – 200C =800C Q = mcΔT Q = 2.5kg(4184 J/kg0C)(800C) Q = 836800J Q = 8.37 x 105J  LYCEUM OF ALABANG INC.INC. HEAT EXCHANGE the transfer of heat energy from a material with a high temperature to another with a low temperature Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Equation of heat exchange: −𝐐𝐥𝐨𝐬𝐭 = 𝐐𝐠𝐚𝐢𝐧𝐞𝐝 −𝐦𝐜𝚫𝐓𝐥𝐨𝐬𝐭 = 𝐦𝐜𝚫𝐓𝐠𝐚𝐢𝐧𝐞𝐝 where m is mass in kg c is the specific heat in J/kg0C ΔT is the change in temperature in 0C ΔT = 𝐓𝐟 − 𝐓𝐢  LYCEUM OF ALABANG INC.INC. A 300g iron at 3000C is submerged in 200g of water at 200C and quickly cooled. Find the final temperature of the iron and water. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. 1kg mFe = 300g = 0.3kg, cFe = 448J/kg0C, 1000g Ti = 3000C, Tf = ? 1kg mH2O = 200g = 0.2kg, cH2O = 4184J/kg0C, 1000g Ti = 200C, Tf = ?  LYCEUM OF ALABANG INC.INC. −𝐦𝐜𝚫𝐓𝐥𝐨𝐬𝐭 = 𝐦𝐜𝚫𝐓𝐠𝐚𝐢𝐧𝐞𝐝 − 0.3kg(448J/kg0C)(Tf − 3000C) = 0.2kg(4184J/kg0C) (Tf − 200C) −134.4J/0C(Tf ) + 40320J = 836.8J/0C(Tf ) − 16736J 40320J + 16736J = 836.8J/0C(Tf ) + 134.4J/0C(Tf ) 40320J + 16736J = Tf (134.4J/0C + 836.8J/0C) 57056J = 971.2J/0C(Tf ) 𝐓𝐟 = 58.750C  LYCEUM OF ALABANG INC.INC. −𝐦𝐜𝚫𝐓𝐥𝐨𝐬𝐭 = 𝐦𝐜𝚫𝐓𝐠𝐚𝐢𝐧𝐞𝐝 −𝐦𝐜(𝐓𝐟 − 𝐓𝐢 )𝐥𝐨𝐬𝐭 = 𝐦𝐜(𝐓𝐟 − 𝐓𝐢 )𝐠𝐚𝐢𝐧𝐞𝐝 −𝐦𝐜𝐓𝐟 𝐥𝐨𝐬𝐭 + 𝐦𝐜𝐓𝐢 𝐥𝐨𝐬𝐭 = 𝐦𝐜𝐓𝐟 𝐠𝐚𝐢𝐧𝐞𝐝 − 𝐦𝐜𝐓𝐢 𝐠𝐚𝐢𝐧𝐞𝐝 𝐦𝐜𝐓𝐢 𝐥𝐨𝐬𝐭 + 𝐦𝐜𝐓𝐢 𝐠𝐚𝐢𝐧𝐞𝐝 = 𝐦𝐜𝐓𝐟 𝐠𝐚𝐢𝐧𝐞𝐝 + 𝐦𝐜𝐓𝐟 𝐥𝐨𝐬𝐭 𝐦𝐜𝐓𝐢 𝐥𝐨𝐬𝐭 + 𝐦𝐜𝐓𝐢 𝐠𝐚𝐢𝐧𝐞𝐝 = 𝐓𝐟 (𝐦𝐜𝐠𝐚𝐢𝐧𝐞𝐝 + 𝐦𝐜𝐥𝐨𝐬𝐭 ) 40320J + 16736J = Tf (134.4J/0C + 836.8J/0C) 57056J = 971.2J/0C(Tf ) 𝐓𝐟 = 58.750C  LYCEUM OF ALABANG INC.INC. LATENT HEAT The heat energy that must be released or gained by a material per unit mass without an accompanying change in temperature Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Change of phase is the change in the state of a material due to the gain or loss of heat energy Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Solidification is the process of changing a substance from its melted state to its Photo Credit: GOOGLE solid state  LYCEUM OF ALABANG INC.INC. Condensation is the process of changing a gas to its liquid state Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Sublimation is the process of changing a substance from its solid state to its gaseous state. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. The temperature where fusion takes place is the melting point Fusion is the process of changing a substance from its solid state to its liquid state. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. Vaporization is the process of changing a liquid to vapor The temperature where vaporization takes place is the boiling point. Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. PROCESS ENDOTHERMIC EXOTHERMIC  Solidification/Freezing  Condensation  Vaporization/Boiling  Fusion/Melting  Sublimation  LYCEUM OF ALABANG INC.INC. Heat of Transformations Substance Melting Heat of Fusion, Hf Boiling Heat of Vaporization, Hv Point Point (J/kg) (cal/g) (J/kg) (cal/g) (K) (K) water 273 3.34 x 105 79.7 373 2.26 x 105 539 iron 1808 2.89 x 105 69.2 3023 63.4 x 105 1520 lead 600 0.232 x 105 5.54 2023 5.69 x 105 205 copper 1356 2.05 x 105 48.9 2840 48.0 x 105 1150 Ethyl alcohol 159 1.04 x 105 25 78 8.5 x 105 204  LYCEUM OF ALABANG INC.INC. Equation of heat of fusion: 𝐐 = 𝐦𝐇𝐟 where 𝐐 is the heat gained/lost by the material in Joules, m is mass in kg; and Hf if heat of fusion in J/kg  LYCEUM OF ALABANG INC.INC. Twenty grams of ice at –100C melts completely. How much heat is absorbed? GIF Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. 1g mice = 20g = 0.02kg, Ti = –100C, 1000g Tf = 00C (melting point of ice) cice = 2090 J/kg.0C Hf = 3.34 x 105J/kg 𝐐 = 𝐦𝐇𝐟 & Q = mcΔT LYCEUM OF ALABANG INC.INC. Q = mcΔT Q = 0.02kg(2090 J/kg.0C)(00C–(–100C)) Q = 0.02kg(2090 J/kg.0C)(100C) Q = 418J (heat needed to start the ice melting) 𝐐 = 𝐦𝐇𝐟 𝐐 = 0.02kg(3.34 x 105J/kg) Q = 6680J (heat needed to completely melt the ice) LYCEUM OF ALABANG INC.INC. Qtotal = mcΔT + 𝐦𝐇𝐟 Qtotal = 418J + 6680J Qtotal = 7098J  LYCEUM OF ALABANG INC.INC. Equation of heat of vaporization: 𝐐 = 𝐦𝐇𝐯 where 𝐐 is the heat gained/lost by the material in Joules, m is mass in kg; and Hv if heat of vaporization in J/kg  LYCEUM OF ALABANG INC.INC. Fifty grams of water at 250C is heated until it evaporates completely. How much heat did it absorb? Photo Credit: GOOGLE  LYCEUM OF ALABANG INC.INC. 1g mwater = 50g = 0.05kg, Ti = 250C, 1000g Tf = 1000C (boiling point of water) cwater = 4184 J/kg.0C, Hv = 2.26 x 105J/kg 𝐐 = 𝐦𝐇𝐯 & Q = mcΔT LYCEUM OF ALABANG INC.INC. Q = mcΔT Q = 0.05kg(4184 J/kg.0C)(1000C–250C) Q = 0.05kg(4184 J/kg.0C)(750C) Q = 15690J (heat needed to start the water boiling) 𝐐 = 𝐦𝐇𝐟 𝐐 = 0.05kg(2.26 x 105J/kg) Q = 11300J (heat needed to completely boil the water) LYCEUM OF ALABANG INC.INC. Qtotal = mcΔT + 𝐦𝐇𝐯 Qtotal = 15690J + 11300J Qtotal = 26990J