PHY 302 K. Solutions for Problem set # 12. Textbook problem 10.55: Pressure drop due to viscosity in a round pipe of radius R is given by the Poiseuille equation: P L 8η v R 2 8ηF πr 4 (1) where η is viscosity of the fluid, F is the volumetric flow rate, and v F/πR 2 is the average velocity of the flow. Human aorta has approximately round cross-section. In healthy adults, its diameter varies between about 2 and 3.5 cm; for this exercise, we use 2R 2.4 cm (i.e., R 1.2 cm) from the textbook example 10 11 (page 269). The blood flow through aorta depends on what the person is doing (resting or exercising); for this exercise we take v 40 cm/s F πr 2 v 180 cm 3 /s 11 L/min. (2) from the same textbook example 10 11. Finally, the viscosity of whole blood at body temperature is η 4 10 3 Pa s (from table 10 3 on page 275 of the textbook). Plugging all these data into the Poiseuille equation (1), we get P L 8(4 10 3 Pa s)(0.4 m/s) (0.012 m) 2 90 Pa/m 0.9 Pa/cm. (3) Note: over the whole length of the aorta about 50 cm the blood pressure drop due to viscosity is only 45 Pa or 1/3 of a millimeter of mercury. Textbook problem 10.58: Squeezing blood through a thin needle requires a pressure difference: the blood pressure in the tube connected to the needle should be higher than the blood pressure in the patient s vein. Specifically, according to Poiseuille equation (1), we need P P tube P vein 8ηFL πr 4 (4) where R is the inner diameter of the needle, L is its length, η is the viscosity of blood, and F is the desired flow rate. For the needle in question, L 4.0 cm and 2R 0.40 mm, we 1
want F 4.0 cm 3 /min 0.067 cm 3 /s, and the blood viscosity is η 4.0 10 3 Pa s (from textbook table 10 3). Therefore, P 8(4.0 10 3 Pa s)(0.067 cm 3 /2)(4.0 cm) π(0.040 cm/2) 4 17.0 kpa. (5) To create this pressure difference, the bottle of donor s blood is raised above the needle, which creates the hydrostatic pressure P tube near needle P bottle + ρgh (6) where ρ 1050 kg/m 3 is the density of blood (from table 10 1 on page 256 of the textbook). The bottle is made of very soft plastic, so the pressure inside it is equal to the atmospheric pressure, P bottle P atm. Consequently, the gauge pressure in the tube near the needle is P gauge tube P tube P atm ρgh. (7) To be precise, we should correct this formula by the Bernoulli term 1 2 ρv2 (where v is the speed of the blood in the tube) and by the effects of viscosity in the tube, but for a typical tube radius R tube 2 mm, both effects are too small to matter. The gauge pressure of blood in the patient vein is P gauge vein 18 Torr 18 mm.hg 2.4 kpa. (8) hence, to create the pressure difference (5) along the needle, the blood pressure in the tube feeding the needle should be P gauge tube P gauge vein + P 17.0 kpa + 2.4 kpa 19.4 kpa. (9) Comparing this value to the hydrostatic formula (7), we can find the height h of the blood bottle above the needle as h P gauge tube ρg or about 6 feet and 2 inches. 19.4 10 3 Pa 1050 kg/m 3 9.8 N/kg 1.88 m, (10) 2
Non-textbook problem #1: There are three forces acting on a red blood cell: its weight mg, the buoyant force from the plasma F B ρ plasma gv cell ρ plasma g m ρ cell mg ρ plasma ρ cell, (11) and the velocity-dependent viscous drag force F D (v) CLη v (12) where L is the red cell s linear size and C is the viscous drag coefficient for its shape. For a spherical cell of radius R, CL 6πR, i.e. F D (v) 6πRη v. (13) Also, a spherical cell has volume V 4π 3 R3 and hence weight mg 4π 3 R3 ρ cell g (14) and buoyant force F B 4π 3 R3 ρ plasma g. (15) When the cell reaches its terminal velocity v t, it s not accelerating any more, so the net force acting on it should vanish, F net mg F B F D (v t ) 0. (16) In light of eqs. (13), (14), and (15), this gives us 4π 3 R3 ρ cell g 4π 3 R3 ρ plasma g 6πRη v t 0 (17) and hence v t 4π 3 R 3 g(ρ cell ρ plasma ) 6πRη 2R2 g(ρ cell ρ plasma ) 9η plasma. (18) Numerically, for R 6µm, ρ cell 1125 kg/m 3, ρ plasma 1025 kg/m 3, and η plasma 1.5 3
10 3 Pa s, we get v t 5.3 10 6 m/s 19 mm/hr. (19) PS: When typing up this problem, I have confused the radius of a red blood cell with its diameter. In real life, the human red blood cells have diameters between 6 and 8 µm and hence radii between 3 and 4 µm. Plugging R 3.5 µm into eq. (18) would give the terminal velocity v t 6.5 mm/hr. Of course, in real life the red blood cells are disk-shaped rather than spherical, and sometimes a few cells stick together, which lowers the viscous drag. Consequently, the erythrocyte sedimentation rate of a healthy adult varies between 10 and 20 mm/hr, depending on age and sex of the person. Non-textbook problem #2: The forces acting on the Huygens probe about to land on Titan are its weight mg in Titan s gravity and the aerodynamic drag on its parachute, F D (v) C 1 2 ρ aira v 2 (20) where A πr 2 is the parachute s area and C 1.50 is its drag coefficient. Note that unlike the previous problem, the drag on the parachute is due to turbulence rather than viscosity. That s why it depends on the density of the Titan s air rather than its viscosity; also it grows with the parachute s speed as v 2 rather than v. The acceleration the probe depends on its velocity v as ma(v) F net (v) mg F D (v) mg CρA 2 v 2. (21) As the probe s velocity increases, the acceleration slows down. After a while, the probe reaches the terminal velocity v t for which a(v t ) 0 mg CρA 2 v 2 t 0. (22) 4
Solving this equation for the v t, we get v 2 t 2mg Cρ(A πr 2 ) 2(318 kg)(1.35 m/s 2 ) 1.50(5.5 kg/m 3 )π(3.03 m/2) 2 14.4 m2 /s 2 v t 3.8 m/s. Since the Huygens probe had plenty of time to reach the terminal velocity, it landed on Titan s surface with speed v v t 3.8 m/s 8.5 MPH. (23) Textbook problem 13.7: Concrete has thermal expansion coefficient α 12 10 6 ( C) 1. Thus, a concrete slab of length L 0 12 m made at temperature T 0 20 C will expand in hot weather T 1 50 C by L 1 L 0 α(t 1 T 0 ) 4.3 mm (24) and contract in cold weather T 2 30 C by L 2 L 0 α(t 2 T 0 ) 7.2 mm. (25) Consequently, the gaps between adjacent slabs will shrink in hot weather by 4.3 mm and increase in cold weather by 7.2 mm, l gap (50 C) l gap (20 C) 4.3 mm, l gap ( 30 C) l gap (20 C) + 7.2 mm. (26) To avoid horizontal stresses in concrete, the slabs should not interfere with their neighbors thermal expansion. Thus, when the slabs expand in hot weather, they should not bump into each other that s what the gaps are for. The gaps shrink in hot weather, but they should not close until the temperature reaches the maximum expected by this highway, namely T 1 50 C. Thus l gap (50 C) 0 (27) and consequently l gap (20 C) 4.3 mm (28) 5
and l gap ( 30 C) 4.3 mm + 7.2 mm 11.5 mm. (29) In particular, when the concrete slabs are laid down at 20 C, the highway builders should leave gaps of 4.3 mm between each slab. Textbook problem 13.90: As the temperature changes, the length of the platinum bar between two marks changes by L L 0 α T. (30) To keep this change small enough L L0 α T Lmax 1 µm, (31) we need to limit temperature changes to T L max αl 0 1 10 6 m 1 m 9 10 6 (C ) 1 0.11 C. (32) That is, the temperature of the bar should not change by more than ±0.11 C (or ±0.2 F) from the temperature at which the marks on the bar were made. Textbook problem 13.86: In equilibrium, the weight of the fluid displaced by the floating body should equal to the body s own weight. Consequently, the submerged part of the body s volume is related to the total volume of the body as the body s density to the fluid s density, V subm V tot ρ body ρ fluid. (33) When the temperature raises, both mercury and iron expand in volume and their densities 6
decrease, ρ(t ) ρ(t 0 ) 1 + β(t T 0 ) ρ(t 0) (1 β(t T 0 )). (34) But the mercury expands faster than iron β(hg) 180 10 6 / C while β(fe) 35 10 6 / C so the ratio ρ(fe)/ρ(hg) becomes larger: ρ Fe (T ) ρ Hg (T ) ρ Fe(T 0 ) ρ Hg (T 0 ) 1 + β Hg(T T 0 ) 1 + β Fe (T T 0 ) ρ Fe(T 0 ) ( ) ρ Hg (T 0 ) 1 + (β Hg β Fe ) (T T 0 ). (35) According to eq. (33), this means that for an iron cube floating in mercury V subm (T ) V ( ) subm (T 0 ) 1 + (β Hg β Fe ) (T T 0 ) V tot V tot (36) as the temperature increase, the submerged fraction of the cube s volume increases, and the cube floats lower in mercury than at the lower temperature. Specifically, for T increasing from 0 C to 25 C, (β Hg β Fe ) (T T 0 ) (180 35) 10 6 ( C) 1 25 C 3.6 10 3 0.36%, (37) so the submerged fraction of the iron cube s volume increases by 0.36%. Note: this percentage increase is relative to the original fraction V subm V tot (0 C) ρ Fe(0 C) ρ Hg (0 C) 7881 kg/m3 0.5797 57.97% : (38) 13595 kg/m3 As the temperature rises, the submerged fraction increases to 57.97% (1 + 0.0036) 57.97% + 0.21% 58.18%. (39) Thus, relative to the total cube volume, the submerged fraction increases by only 0.21%. 7
Non-textbook problem #3: The volume of alcohol in the thermometer is V V b + πr 2 h (40) where V b is the (inside) volume of the bulb, r is the (inside) radius of the cylindrical tube, and h is the height of alcohol column inside the tube. We assume the glass does not expand, so V b and r are constant and only the height h changes with the temperature-dependent volume of alcohol according to V πr 2 h h V πr 2. (41) As the temperature rises, the volume of alcohol changes by which makes the alcohol height rise by Originally, h 0 3.0 cm and Consequently, which allows us to simplify eq. (43) to V V 0 β T (42) h V 0 β T. (43) πr2 πr 2 h 0 9.4 10 4 cm 3 V b 1.00 cm 3. (44) V 0 V b + πr 2 h 0 V b (45) h V b β T. (46) πr2 For the thermometer in question, V b πr 2 β 1.00 cm 3 π(0.0100 cm) 2 1.09 10 3 ( C) 1 3.47 cm/ C, (47) so when the temperature increases from 10 C to 30 C, the alcohol in the thermometer raises 8
by h (3.47 cm/ C) (20 C) 69.4 cm, (48) form h 0 3.0 cm to h h 0 + h 72.3 cm. Non-textbook problem #4: According to the universal gas law, for any fixed amount of gas P V T nr const. (49) As the helium-filled balloon raises from the ground to the 10,000 ft altitude, helium s pressure and temperature change according to the pressure and temperature of the surrounding air. Consequently, the volume of the balloon changes according to P 2 V 2 P ( ) 1V 1 T2 V 2 V 1 T 2 T 1 T 1 ( P1 P 2 ). (50) Note that the temperatures T 1 and T 2 in this formula are absolute temperatures (counting from absolute zero), so we need to translate them from degrees Fahrenheit to degrees Celsius and hence to Kelvins. Thus, ( 77 32 T 1 77 F 1.8 while and hence At the same time, T 2 23 F ( 23 32 1.8 T 2 T 1 P 1 P 2 ) ( ) 25 C 25 + 273.15 298.15 K (51) ) ( ) 5 C 5 + 273.15 268.15 K (52) 269.15 K 208.15 K 1000 mbar 690 mbar 0.8994. (53) 1.450. (54) Therefore, the volume of the balloon changes from V 1 100 m 3 to ( ) ( ) T2 P1 V 2 V 1 100 m 3 0.8994 1.450 130 m 3. (55) T 1 P 2 9
Non-textbook problem #5: According to the universal gas law P V nrt (56) the amount of gas (in mols) in volume V under pressure P and absolute temperature T is n P V RT. (57) The mass of this gas (in grams!) is m µ n µp V RT (58) where µ is the molecular weight of the gas. Consequently, the density of the gas is ρ m V µp RT. (59) Now let s apply this formula to the atmosphere of Venus. It consists mostly of CO 2 whose molecular weight is µ 44 g/mol and near the surface has pressure P 92 bar 9.2 10 6 Pa, and absolute temperature T 740 K. Consequently, the density is ρ (44 g/mol) (9.2 106 Pa) (8.314 J/K/mol) (740 K) 65, 800 g/m 3 note units! (60) 65.8 kg/m 3, about 60 times denser than the air on Earth. 10
Non-textbook problem #6: In thermal equilibrium, each component of the gas mixture has the same average kinetic energy of a molecule, 12 mv 2 avg 3 2kT (61) where k 1.38 10 23 J/K is the Boltzmann s constant and T is the absolute temperature of the gas. Consequently, for each component (i) of the gas, the mean velocity 2 of a molecule is v(i) 2 avg 3kT m (i) (62) and hence RMS (root-mean-square) molecular velocity is (i) v(i) 2 avg 3kT m (i). (63) units, The molecular weight µ of a gas component is the mass of one molecule in atomic mass m (i) µ i 1 amu, (64) where 1 amu 1 g N A 1.6605 10 27 kg. (65) Consequently, the RMS molecular velocity is (i) 3kT µ i 1 amu. (66) Inside human lungs T 311 K, hence an (i) th component of the gas mixture has (i) 3kT/1 amu µi 2785 m/s µi. (67) Specifically: 11
Helium is a mono-atomic gas of atomic weight µ 4, hence the RMS speed of a helium atom in the mix is He 2785 m/s 4 1393 m/s. (68) Oxygen O 2 has molecular weight µ 2 16 32, hence the RMS speed of an oxygen molecule in the mix is O 2 2785 m/s 32 492 m/s. (69) Water H 2 O has molecular weight µ 2 1 + 16 18, hence the RMS speed of a water molecule in the mix is H 2O 2785 m/s 656 m/s. (70) 18 Carbon dioxide CO 2 has molecular weight µ 12 + 2 16 44, hence the RMS speed of a CO 2 molecule in the mix is CO 2 2785 m/s 44 420 m/s. (71) Textbook problem 13.53: According to the universal gas law P V nrt NkT (72) where n is the amount of gas in mols while N n N A is the net number of molecules, and k R/N A is the Boltzmann s constant. The same constant appears in the formula for the average kinetic energy of a molecule in a gas, 12 mv 2 3 2kT. (73) In terms of the root(mean square) speed of a molecule, 12 mv 2 1 2 m() 2 3 2kT (74) 12
hence ( ) 2 3kT m. (75) Combining this formula with eq. (72), we get ( ) 2 3 ( m kt P V ) 3P V N Nm (76) Note that Nm in the denominator of this formula is the total mass M of the gas and Nm V M V ρ (77) is its density. Plugging this formula into eq. (76), we arrive at ( ) 2 3P ( V Nm 1 ) ρ 3P ρ (78) and hence 3P ρ. (79) Quod erat demonstrandum (which is precisely what had to be proved). 13