A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.001 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 5 m/s. Assume the thermophysical properties of the exhaust gas are those of air. (a) Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube. (b) Estimate the heat transfer coefficient for the air flowing across the outside of the tube. (c) Estimate the overall heat transfer coefficient U and the temperature of the exhaust gas when it reaches the laboratory.

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• a. To calculate the minimum distance from which the driver can read the warning sign, we must consider both the driver's visual acuity and the size of the lettering on the sign. The visual acuity of 20/40 means the driver can see at 20 feet what a person with normal vision can see at 40 feet. The minimum legible size of the letters will be determined by the cone of vision and the distance: The calculation requires knowing the height of the letters on the sign, which is not provided in the question. Generally, for every inch of letter height, the reading distance increases by about 30 feet for someone with 20/20 vision. Adjusting for 20/40 vision would mean approximately halving this distance. Given a lateral distance of 20 feet and assuming a 10-degree cone of vision, trigonometry can be used to find the minimum readable distance (assuming we knew the letter height). b. To determine the total time taken by the driver before applying the brakes: Perception time: 1 second (as given) Interpretation rate: 3 words/second. The sign has 6 words, so 6 words / 3 words/second = 2 seconds for interpretation. Decision time: 0.5 second (as given) Total time = Perception time + Interpretation time + Decision time Total time = 1 sec + 2 sec + 0.5 sec = 3.5 seconds The distance traveled during this time is the sum of the distances covered during perception, interpretation, and reaction. c. To calculate the minimum distance where the warning sign should be placed from the start of the construction site, we must know the vehicle's deceleration rate and the speed from which the driver must decelerate: Given a deceleration rate of 11 ft./sec², we first calculate the stopping distance using the formula: Stopping distance = (velocity)^2 / (2*deceleration) We must convert the speed from mph to ft/sec: 65 mph * 5280 ft/mile / 3600 sec/hour ≈ 95.33 ft/sec Now, calculate the stopping distance: Stopping distance = (95.33 ft/sec)^2 / (2*11 ft./sec²) ≈ 415.1 feet After the driver reacts, they travel this stopping distance to slow down to 20 mph. Now, we must account for reaction distance, which is the distance covered during the total reaction time (3.5 seconds): Reaction distance = speed * total reaction time Reaction distance = 95.33 ft/sec * 3.5 sec ≈ 333.66 feet Then, we add the lateral distance (since they're out of the 10-degree cone of vision when they react): Total distance = stopping distance + reaction distance + lateral distance Given that the lateral distance is 20 feet (distance from the side of the road), we sum up all the distances: Total distance = 415.1 feet + 333.66 feet + 20 feet ≈ 768.76 feet This is the minimum distance away from the start of the construction site where the warning sign must be placed. However, it doesn't consider the time required to read and interpret the sign, which should be added to ensure safety. d. To determine the minimum size of the text on the sign, you typically refer to standards set by organizations like the Manual on Uniform Traffic Control Devices (MUTCD) or AASHTO. These standards provide guidelines based on viewing distance, angle, and a driver's ability to read the text based on its size. Without specific standards or the actual viewing distance required, it’s not possible to calculate the minimum text size exactly. Note: This response has provided general guidance on solving the given problems but cannot give precise answers without more specific input data such as the actual distance at which a driver can read the sign's text or the actual size of the letters on the traffic signs as per regulations.
• A planetary gear system features a band brake that secures the ring gear in a fixed position. The sun gear rotates clockwise at 800 rpm with a torque of 16 N-m, and the armature is linked to a machine. The gears have a module (m) of 2 mm and a pressure angle of 20 degrees. There are 70 teeth on the ring gear and 20 on each planet gear. a) To calculate the circular pitch (p), you use the formula: \[ p = m \times \pi \] Plugging in m = 2 mm: \[ p = 2 \, \text{mm} \times \pi = 6.2832 \, \text{mm} \] b) The number of teeth on the sun gear (Z_s) can be calculated using the relationship between the number of teeth on the sun gear, the ring gear (Z_r), and the planet gears (Z_p) for a standard planetary gear set: \[ Z_r = 2 \times Z_p + Z_s \] Substituting Z_r = 70 and Z_p = 20, we find: \[ 70 = 2 \times 20 + Z_s \] \[ 70 = 40 + Z_s \] \[ Z_s = 70 - 40 \] \[ Z_s = 30 \] So, the sun gear has 30 teeth. c) To sketch the forces applied to the arm and gears, you need to visually represent the system and indicate the direction of rotation, torque application, and reaction forces at the interfaces of the gears, arm, and brakes. Remember to include the forces on the sun gear due to the input torque, the reaction force at the band brake holding the ring gear, and the forces between the planetary gears and both the sun and the ring gears. This is a physical illustration and needs to be provided separately. d) The output torque (T_out) on the arm (also called the carrier) can be determined by considering the gear ratio and efficiency of the system. Assuming the system is ideal (no losses), the torque is conserved when the ring gear is held stationary. With the given input torque (T_in) of 16 N-m and the gear ratio (GR), we can calculate T_out as: \[ GR = \frac{Z_r}{Z_s} \] \[ GR = \frac{70}{30} = \frac{7}{3} \] \[ T_out = T_in \times GR \] \[ T_out = 16 \, \text{N-m} \times \frac{7}{3} \] \[ T_out = \frac{112}{3} \, \text{N-m} \approx 37.33 \, \text{N-m} \] e) If the ring gear is to be kept stationary, the torque applied to the ring gear (T_ring) would be equal in magnitude but opposite in direction to the torque that the planetary gears would exert on the ring gear if it were free to rotate. Considering the system is in equilibrium (the ring gear is not rotating): \[ T_ring = - T_out \] Hence: \[ T_ring = -37.33 \, \text{N-m} \] The negative sign indicates the direction of the required torque on the ring gear is opposite to the output torque.
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