CHE330 F23 HW2 Solved

CHE330, Fall 2023 HW#2 1 Discussion of HW and working together with others is encouraged for homework assignments. Talking it through often helps comprehension. However, this does not include copying of someone else’s work. Please write out the sentence below on your homework assignment followed by your signature to affirm that you did not copy or use someone else’s work. “I affirm that I have upheld my own integrity and that of the university on this assignment” followed by your signature. Show your work for full credit. Include units for all values (if appropriate) and use correct number of significant figures in your answers. Insert graphs into your homework solution and also submit the excel file in which it was generated (or other software program file used). Problem 1: Viscometer data of honey [12] Viscometers are instruments designed to measure the viscosity of a fluid. Imagine a viscometer that applies a shear (tangential force) to a fluid, and measures the force required to apply this shear (the shear stress). Example data from a viscometer is shown below, where the shear stress as a function of the velocity gradient (here, the rate of shear strain) is tabulated. The data below is for different types of honey adapted from: Santos, Francisco Klebson & Filho, Antonio & Leite, R.H.L. & Aroucha, Edna & Santos, Andarair & Oliveira, Thiago. (2014). Rheological and some physicochemical characteristics of selected floral honeys from plants of caatinga. Anais da Academia Brasileira de Ciências. 86. 10.1590/0001-3765201420130064. Shear stress as a function of strain rate for different types of honey shear rate (1/s) shear stress (Pa) croton camprestris cobretum leprosum mimosa tenufiora 0 0 0 0 10 18 48 88 20 40 105 178 30 52 149 249 40 60 199 344 50 91 255 420 60 105 310 520 70 130 351 597 80 145 385 669 90 170 444 771 a. [5] Plot the data on a graph of shear stress as a function of shear rate. Since you have individual data points, plot individual points. You may use any type of software, but Excel is

CHE330, Fall 2023 HW#2 2 recommended. Make sure your plot shows data points, legend, and has labelled axes including units. b. [2] What type of fluid is honey (Newtonian, shear thinning, shear thickening)? Explain your answer using your graph. Newtonian – linear slope (ex. Increasing shear stress with increasing shear rate) c. [5] If a fluid is Newtonian, you can fit a linear trendline through the data from part a, forcing the intercept through the origin. Show the trendline and the equation of the trendline on the plot (note that the intercept of the trendline must be zero- you can force it through zero in the trendline setup in Excel). Determine the viscosity of each type of honey in cP. 0 200 400 600 800 1000 0 20 40 60 80 100 Shear Stress (Pa) Shear Rate (1/s) croton camprestris cobretum leprosum mimosa tenufiora y = 1.8172x R² = 0.9974 y = 4.9716x R² = 0.9994 y = 8.5112x R² = 0.9998 0 100 200 300 400 500 600 700 800 900 0 20 40 60 80 100 Shear Stress (Pa) Shear Rate (1/s) croton camprestris cobretum leprosum mimosa tenufiora Linear (croton camprestris) Linear (cobretum leprosum) Linear (mimosa tenufiora)

CHE330, Fall 2023 HW#2 4 Ketchup – shear thinning, the viscosity decreases with increasing shear rate. b) [5] Determine the shear stress at each shear rate for the given data (Excel recommended). Plot the resulting shear stress versus shear rate data. Make sure your plot shows data points, legend, and has labelled axes including units. Make sure axes numbers do not carry excessive significant figures. How does this plot show the type of fluid for ketchup? shear rate (1/s) viscosity (Pa*s) Shear stress 0.1 100.08 10.01 0.5 34.21 17.11 1 21.54 21.54 5 7.36 36.80 10 4.64 46.40 20 2.92 58.40 30 2.23 66.90 40 1.84 73.60 50 1.59 79.50 60 1.4 84.00 70 1.27 88.90 80 1.16 92.80 90 1.07 96.30 1 10 100 1000 0.1 1 10 100 Viscosity (Pa*s) Shear Rate (1/s)

CHE330, Fall 2023 HW#2 5 Ketchup – shear thinning liquid – increasing shear stress with increasing shear rate c) [2] Ketchup is intentionally engineered to have this behavior. What benefit is there to having such a high viscosity at low shear rates? What benefit is there to having a much lower viscosity at high shear rates? High viscosity at low shear rates – helps keep the ketchup in place and won’t displace too much. Ex. When ketchup is put on a sandwich, it does not just slide right off. A lower viscosity at lower shear rates – helps the liquid move when some energy is applied. Problem 3: Viscosity of air [11] Below is the viscosity of air (in µPa*s) at various temperatures and pressures. a. [5] Plot the viscosity vs. Temperature at P = 0.1 MPa and P = 10 MPa on the same plot. Viscosity is given in µPa*s. Data is taken from CRC Handbook. Carefully label both axes including units. Since this is data, the graph should have points. A straight line segment may be used to connect the points if it helps in visualizing the data. Make sure to include a legend also. 0 20 40 60 80 100 120 0 20 40 60 80 100 Shear Stress (Pa) Shear Rate (1/s)

CHE330, Fall 2023 HW#2 7 Problem 4: Unit conversions [10] a) [5] Show that the gas constant 8.314 Pa m 3 /mol K converts to 0.08206 atm L/mol K and to 10.73 psi ft 3 /lb-mol o R. Note that o R is degree Rankine. 8.314 𝑃𝑃𝑃𝑃 𝑚𝑚 3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐾𝐾 ∗ � 1 𝑃𝑃𝑎𝑎𝑚𝑚 101325 𝑃𝑃𝑃𝑃 � ∗ � 1000 𝐿𝐿 1 𝑚𝑚 3 � = 0.08206 𝑃𝑃𝑎𝑎𝑚𝑚 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚 𝐾𝐾 8.314 𝑃𝑃𝑃𝑃 𝑚𝑚 3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐾𝐾 ∗ � 14 . 7 𝑝𝑝𝑝𝑝𝑝𝑝 101325 𝑃𝑃𝑃𝑃 � ∗ � 35 . 315 𝑓𝑓𝑓𝑓 3 1 𝑚𝑚 3 � ∗ � 1 𝑚𝑚𝑚𝑚𝑚𝑚 2 . 2∗10 −3 𝑚𝑚𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚 � ∗ ( 1 𝐾𝐾 1 . 8 ° 𝑅𝑅 ) = 10.76 𝑝𝑝𝑝𝑝𝑝𝑝 𝑓𝑓𝑓𝑓 3 𝑚𝑚𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚 ° 𝑅𝑅 +_ 0.1 b) [5] Express the mass of a 80.0 kg person as the equivalent lb m . Then convert each to weight on earth in units of N and lb f . (eg. Do the calculation in SI units from kg to N and then do the same calculation in English units from lb m to lb f .) The acceleration due to gravity is 9.81 m/s 2 and 32.2 ft/s 2 . 80 𝑘𝑘𝑘𝑘 ∗ � 2.2 𝑚𝑚𝑙𝑙 𝑚𝑚 1 𝑘𝑘𝑘𝑘 � = 176 𝑚𝑚𝑙𝑙 𝑚𝑚 80 𝑘𝑘𝑘𝑘 ∗ 9.81 𝑚𝑚 𝑠𝑠 2 = 784.8 = 785 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠 2 = 785 𝑁𝑁 176 𝑚𝑚𝑙𝑙 𝑚𝑚 ∗ 32.2 𝑓𝑓𝑎𝑎 𝑠𝑠 2 = 5667.2 𝑚𝑚𝑙𝑙 𝑚𝑚 𝑓𝑓𝑎𝑎 𝑠𝑠 2 ∗ � 𝑚𝑚𝑙𝑙 𝑓𝑓 32.2 𝑚𝑚𝑙𝑙 𝑚𝑚 𝑓𝑓𝑎𝑎 𝑠𝑠 2 � = 176 𝑚𝑚𝑙𝑙 𝑓𝑓

CHE330, Fall 2023 HW#2 8 Problem 5: Shear stress [5] [5] Consider the parallel plate setup shown below. The distance between the plates, h, is 250 µm (micro-meters) and the top plate is moving at V=15 cm/s. The bottom plate is stationary. The fluid is water at 20 o C. a. [1] Determine the fluid velocity at the stationary wall. Velocity = 0 cm/s b. [1] Determine the fluid velocity at the moving wall. Velocity = 15 cm/s c. [1] Determine the fluid velocity at 40 µm from the stationary wall. 𝑣𝑣 𝑥𝑥 ( 𝑦𝑦 ) = 𝑉𝑉 ∗ 𝑦𝑦 𝐻𝐻 𝑣𝑣 𝑥𝑥 (40 𝜇𝜇𝑚𝑚 ) = 15 𝑐𝑐𝑚𝑚 𝑠𝑠 ∗ 40 𝜇𝜇𝑚𝑚 250 𝜇𝜇𝑚𝑚 = 2.4 𝑐𝑐𝑚𝑚 𝑠𝑠 d. [2] Determine the shear stress of this system (in Pa). Note that you may need to look up some parameters for water at 20 o C. 𝜏𝜏 = 𝜇𝜇 ∗ 𝑑𝑑𝑣𝑣 𝑥𝑥 𝑑𝑑𝑦𝑦 , 𝑛𝑛𝑚𝑚𝑎𝑎𝑛𝑛 : 𝑣𝑣𝑣𝑣𝑠𝑠𝑐𝑐𝑚𝑚𝑠𝑠𝑣𝑣𝑎𝑎𝑦𝑦 𝑚𝑚𝑓𝑓 𝑤𝑤𝑃𝑃𝑎𝑎𝑛𝑛𝑤𝑤 𝑃𝑃𝑎𝑎 20℃ = 1 𝑐𝑐𝑃𝑃 𝑑𝑑𝑣𝑣 𝑥𝑥 𝑑𝑑𝑦𝑦 = 𝑐𝑐ℎ𝑃𝑃𝑛𝑛𝑘𝑘𝑛𝑛 𝑚𝑚𝑓𝑓 𝑣𝑣𝑛𝑛𝑚𝑚𝑚𝑚𝑐𝑐𝑣𝑣𝑎𝑎𝑦𝑦 𝑚𝑚𝑣𝑣𝑛𝑛𝑤𝑤 𝑥𝑥 𝑐𝑐ℎ𝑃𝑃𝑛𝑛𝑘𝑘𝑛𝑛 𝑚𝑚𝑓𝑓 𝑦𝑦 = 15 𝑐𝑐𝑚𝑚 / 𝑠𝑠 250 𝜇𝜇𝑚𝑚 𝜏𝜏 = 1 𝑐𝑐𝑃𝑃 ∗ 15 𝑐𝑐𝑚𝑚 𝑠𝑠 250 𝜇𝜇𝑚𝑚 ∗ � 1 𝑚𝑚 100 𝑐𝑐𝑚𝑚 � ∗ � 10 6 𝜇𝜇𝑚𝑚 1 𝑚𝑚 � ∗ � 1 𝑃𝑃𝑃𝑃 ∗ 𝑠𝑠 1000 𝑐𝑐𝑃𝑃 � = 0.6 𝑃𝑃𝑃𝑃