measurements final prac

measurements final prac - Final Practice Problems Peter H...

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Final Practice Problems Peter H. McMurry, Instructor NOTE: These practice problems are taken from previous ME 4031 midterm exams. 1. The Weber number is a dimensionless parameter relating inertial and surface tension forces in a fluid flow. Consider a droplet of water where the Weber number is defined as: We = ρ V 2 d σ The flow conditions are: ρ = 1000 ± 8 kg/m 3 V = 5 ± 0.1 m/s d = 1 ± 0.025 mm σ = 0.072 N/m ± 3% a). What is the Weber number, including uncertainty? Show your answer to an appropriate number of significant figures. (Show your work.) We = ρ V 2 d σ = 1000 5 2 1 × 10 3 0.072 = 347.22 [ ] Δ We We = Δ ρ ρ 2 + 2 ⋅ Δ V V 2 + Δ d d 2 + Δ σ σ 2 = (0.008) 2 + (0.04) 2 + (0.025) 2 + (0.03) 2 = 0.05647 or 5.647% Δ We = 19.6 We = 350 ± 20 [ ] (rounded off) b). If you could choose one of the measurements to have its uncertainty reduced to roughly zero, which one would reduce the overall Weber number uncertainty the most? Reduce Δ V, since it contributes most to Δ We.
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2. a. A pressure gauge is used to measure a system that undergoes an instantaneous step increase in pressure from 10 to 50 Pa. The following data are recorded: time after step, ms P, Pa Γ =(P-Ps)/(Pi-Ps) ln Γ τ =-t/ ln Γ , ms 1 15.06 0.8734 -0.135318 7.390 5 29.67 0.5083 -0.67659 7.390 10 39.66 0.2584 -1.3532 7.390 15 44.75 0.1314 -2.0298 7.390 20 47.33 0.06678 -2.7064 7.390 Determine an approximate value for the time constant of the pressure gauge. Method 1 (Exact. See above table for results): ln Γ = t τ Δ ln Γ Δ t = 1 τ Method 2 (approximate): 1 time constant implies 63.2% of response. ) 10+.632*40=35.28. Using linear interpolation (not the best method, but good enough) this implies: τ =5+(35.28-29.67)/(39.66-29.67)*5=7.81 ms Method 3 (approximate): 2 time constants equals 86.5%, 2.3 time constants = 90%, 5 time constants = 99.3%. Proceed as with method 2.
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2b. The temperature of a water bath is given by the following equation: T [ C ] = 50 + 40sin π t where t is time in seconds. A thermometer with a time constant of 0.5 seconds is used to measure this temperature. Write down the equation that gives the steady-state thermometer output versus time. τ dX dt + X = K F ( t ) = 50 + 40sin π t X ' = X 50 τ dX ' dt + X ' = 40sin π t X ' ( t ) = Ce t τ + A 1 + ( τω ) 2 sin( ω t + ϕ ) where C depends on initial value ϕ ( ω ) = tan 1 ( τω ) "phase error or phase shift angle, in radians" ϕ ( ω ) = tan 1 (0.5 π ) = tan 1 (1.5708) = 57.52 = 1.004 radians Steady state solution: X ' ( t ) = 40 1 + (0.5 ω ) 2 sin( ω t + ϕ ) X ( t ) = 50 + 40 1 + (0.5 π ) 2 sin( π t + 1.004) X ( t ) = 50 + 21.48 sin( π t + 1.004)
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3. You and your friends are developing a system to measure the velocity of a golf cart. Your plan is to paint a white dot on the front tire and then film the wheel with a camera as the golf cart drives down the road. The camera records at a rate of 30 frames per second, and the tire has a diameter of 14 inches. a) Create an equation which relates the angular frequency, f ang [revs/sec], to the velocity, U [ft/sec].
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