6-) The packaging unit of a cement factory is worked in the following way: The pakaging is done in sets, each set is consisted of 10 cement bags. If the number of cement bags are more or less than 10 then the warning arises and the system stops. The standard weight of the cement bags is 50 kg. Tolerance is specified as +-1%. The cement bags with more than 1% margin of error are shifted to the conveyor belt, called as faulty conveyor belt. The cement bags without any error are loaded to trucks, by passing through three different coveyor belts. When the 'Start Loading' button is pressed, the third conveyor belt starts immediately, whereas the second conveyor belts starts 20 seconds after the third belt and the first belt starts 20 seconds after the second belt. Once the "Stop Loading" button is pressed, the first belt stops immediately, the second belt stops 10 seconds after the first belt and the third belt stops 10 seconds after the second belt. If there is a fault in third conveyor belt, the first and second conveyor belts stop immediately. If there is a fault in the second conveyor belt, the first conveyor belt stops immediately. In case of any fault in the first conveyor belt, only the first belt stops. When the Emergency Stop button is pressed, all the conveyor belts are stoped. Draw the ladder diagram of this system. Symbol Address Description Emergency Stop 10.0 Emergency stop button Counter Sensor M10.0 Bag counter sensor Weight Sensor IW64 Bag weight sensor (analog) Loading Stop 10.1 Loading stop button Loading Start 10.2 Loading start button Belt 1 Fault M4.1 Belt 1 fault signal Belt 2 Fault M4.2 Belt 2 fault signal Beltt 3 Fault M4.3 Bant 3 fault signal Faulty Belt Q0.0 Faulty belt motor First Belt Q0.1 Belt 1 motor Second Belt Q0.2 Belt 2 motor Third Belt Q0.3 Belt 3 motor
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Problem 1: (50 points) Consider the lab jack as shown. A pair of scissor mechanisms are used to lift parts in experiments, therefore the load is split into 4 levers dividing up the forces as shown. The top and bottom plates are connected to the levers with a pin and roller connection respectively. In order to lift a weight, W, the jack must apply forces, F, as shown. As the angle of the scissor changes the force, F, that must be provided changes. As the weight is lifted the pins and levers are under a load which is a function of the angle θ. Assuming symmetry of the jack stand we can break the jack into parts and solve for the internal loads and select pins and levers that are sufficiently sized for the design. Each lever is 5in long and is supported by pins at all 3 joints. 1. Determine the force, F, required to hold the weight, W, as a function of θ. 2. If the weight, W, is 40lb determine the maximum required force, F, over the range of (15° < θ < 65°). 3. Determine the maximum shear force on the pins, over the range of (15° < θ < 65°). Note the shear force on the pins will include the normal force and the shear force on the levers. 4. If the pins have an allowable shear stress of 20kpsi, determine the required diameter for the pins. Note the pins are in single shear. 5. Determine the maximum bending moment on the levers, over the range of (15° < θ < 65°). Note that you may need a shear force and bending moment diagram to determine this. 6. If the levers have an allowable normal stress of 36kpsi and the width of the cross section is 1/16in wide, determine the minimum height for a factor of safety of 2 due to normal stress due to bending.
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7.2 Consider the following power system network. Both generators are at 30 MW output and both loads are consuming 30 MW. Bus 1 is the reference bus. The data for this network is as follows: From Bus To Bus x MW Limit 1 2 0.20 35.0 1 3 0.30 35.0 1 4 0.30 25.0 2 3 0.30 40.0 3 4 0.40 25.0 The initial power flows are calculated with a DC power flow and result in the following base flows. The line flow limits are also shown with the percent loading: BASE TRANSMISSION LOADING Path From To Low Flow High Percent Loading 1 1 2 -35.0 -8.87 35.0 25.4 2 1 3 -35.0 15.21 35.0 43.5 3 1 4 -25.0 23.66 25.0 94.6 4 2 3 -40.0 21.13 40.0 52.8 5 3 4 -25.0 6.34 25.0 25.4 The PTDF factors and the LODF factors for this system are as follows: POWER TRANSFER DISTRIBUTION FACTOR (PTDF) MATRIX Monitored Transaction Line From(Sell) - To(Buy) 1 to 3 1 to 4 2 to 3 2 to 4 1 to 2 0.2958 0.1268 -0.4225 -0.5915 1 to 3 0.4930 0.2113 0.2958 0.0141 1 to 4 0.2113 0.6620 0.1268 0.5775 2 to 3 0.2958 0.1268 0.5775 0.4085 3 to 4 -0.2113 0.3380 -0.1268 0.4225 LINE OUTAGE DISTRIBUTION FACTOR (LODF) MATRIX Monitored Line Outage of one circuit From - To 1 to 2 1 to 3 1 to 4 2 to 3 3 to 4 1 to 2 0.0000 0.5833 0.3750 -1.0000 -0.3750 1 to 3 0.7000 0.0000 0.6250 0.7000 -0.6250 1 to 4 0.3000 0.4167 0.0000 0.3000 1.0000 2 to 3 -1.0000 0.5833 0.3750 0.0000 -0.3750 3 to 4 -0.3000 -0.4167 1.0000 -0.3000 0.0000 a. In this problem we are only concerned with outages on lines 1-2, 1-3, and 1-4. Do any of these outages, taken one outage at a time, result in overloads? If so, how much and what lines are overloaded? b. The generator at bus 1 is going to reduce its output and at the same time the load at bus 4 is going to reduce its load until there are no overloads due to the lines listed in part (a). How much should the load on bus 4 and the generation on bus 1 be reduced to eliminate all overloads?
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