Concept explainers
OBD II systems use several modes of operation. List three of them.
OBD 2 system and modes of operation.
Explanation of Solution
The OBD system is an On-Board Diagnosis system. It can be installed in all cars and some light duty trucks. It helps in controlling the engine parts and it has proved its importance by keeping an eye on the other parts of a vehicle also. For example, chassis, accessories and other components of cars.
It helps in a vehicle’s maintenance and diagnoses the parts of a vehicle. It helps in producing less emission thus protecting the environment from harmful emission caused by a vehicle’s emission. The diagnosis or maintenance engine light is present at the dashboard. With these light signals, we can check the service needed and the problems caused to a vehicle. It is used for an overall scanning of the vehicle. If there is any problem in the vehicle, the light indicator will indicate so. It provides a list of codes which helps in running on-board diagnosis.
These diagnoses are carried out by collecting the information from the vehicle. This information is analyzed to come up with a suitable solution.
Modes of operation of this system are listed below:
- Live date track
- Freeze frames
- Diagnosis trouble codes.
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Chapter 25 Solutions
AUTMOTIVE TECH, TECH MANUAL & MIND TAP
- Example(2): Double effect evaporator is used for concentrating a certain caustic soda solution 10000kg/hr from 9wt% to 47wt%. The feed at 30°C enters the first evaporator. Backward arrangement evaporators are used. steam is available at 167.7°C and the vapor space in the second effect is 14.6Kpa. The overall heat transfer coefficients of the two effects are 8380 and 6285kcal/ W.CH ork -conce -SOLFFF and-ans.. 112.1 а DiD 3 respectively and the specific heat capacity of all caustic soda solution 3.771 KJ/Kg. °C, determine the heat transfer area of each effect معدلة 5:48 م Oarrow_forwardgive me solution math not explinarrow_forwardgive me solution math not explinarrow_forward
- use Q Strips of material 10 mm thick are dried under constant drying conditions from 28 to 13 per cent moisture in 25 ks. The equilibrium moisture content is 7 per cent. The relation between E, the ratio of the final free moisture content at time t to the initial free moisture content, and the parameter J is given by: E 1 0.64 0.49 0.38 0.295 0.22 0.14 J 0 0.1 0.2 0.3 0.5 0.6 العنوان 0.7 It may be noted that J = kt/12, where, k = constant, t = time (ks) 1 = thickness/2 of the sheet of material (mm) a. Based on the given data, plot a graph of E against J b. Determine the time taken to dry 60 mm planks from 22 to 10 per cent moisture under the same conditions assuming no loss from the edges? ina östler ۲/۱arrow_forward14.25.2.5 kg/s of a solution at 288 K containing 10 per cent of dissolved solids is fed to a forward-feed double-effect evaporator, operating at 14 kN/m² in the last effect. If the product is to consist of a liquid containing 50 per cent by mass of dissolved solids and dry saturated steam is fed to the steam coils, what PROBLEMS 1179 should be the pressure of the steam? The surface in each effect is 50 m² and the coefficients for heat transfer in the first and second effects are 2.8 and 1.7 kW/ m² K, respectively. It may be assumed that the concentrated solution exhibits a boiling-point rise of 5 deg K, that the latent heat has a constant value of 2260 kJ/kg and that the specific heat capacity of the liquid stream is constant at 3.75 kJ/kg K Oarrow_forward: +0 العنوان use only 5) A 100 kg batch of granular solids containing 30% moisture is to be dried in a tray drier to 15.5% by passing a current of air at 350 K tangentially across its surface at a velocity of 1.8 m/s. If the constant rate of drying under these conditions is 0.7 g/s m2 and the critical moisture content is 15%, calculate the approximate drying time. Assume the drying surface to be 0.03 m2 /kg dry mass. мониarrow_forward
- give me solution math not explinarrow_forward۲/۱ : +0 العنوان seoni 4) 1 Mg (dry weight) of a non-porous solid is dried under constant drying conditions with an air velocity of 0.75 m/s parallel to the drying surface. The area of drying surface is 55 m2 If initial rate of drying is 0.3 g/m2 s, how long it will take to dry a material from 0.15 to 0.025 kg water/kg dry solid? The critical moisture content is 0.125 and the equilibrium moisture is negligible. The falling rate of drying is linear in moisture content. If air velocity increases to 4 m/s, what will be the anticipated saving in drying time? 0 ostherarrow_forward14.23. A double-effect forward-feed evaporator is required to give a product consisting of 30 per cent crystals and a mother liquor containing 40 per cent by mass of dissolved solids. Heat transfer coefficients are 2.8 and 1.7 kW/m² K in the first and second effects respectively. Dry saturated steam is supplied at 375 kN/m² and the condenser operates at 13.5 kN/ m². (a) What area of heating surface is required in each effect assuming the effects are identical, if the feed rate is 0.6 kg/s of liquor, containing 20 per cent by mass of dissolved solids, and the feed temperature is 313 K? (b) What is the pressure above the boiling liquid in the first effect? The specific heat capacity may be taken as constant at 4.18 kJ/kg K. and the effects of boiling-point rise and of hydrostatic head may be neglected. O Oarrow_forward
- 5) A 100 kg batch of granular solids containing 30% moisture is to be dried in a tray drier to 15.5% by passing a current of air at 350 K tangentially across its surface at a velocity of 1.8 m/s. If the constant rate of drying under these conditions is 0.7 g/s m2 and the critical moisture content is 15%, calculate the approximate drying time. Assume the drying surface to be 0.03 m2 /kg dry mass. Oarrow_forwardSolve for v and Iarrow_forwardG = 0.350MPa, P = 900N, a=20mm, b=50mm, c=80mmarrow_forward
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