Updated: Aug 12, 2020
(Rgpv online exam important questions & answers 2020 )|| Chp-1 || RAC )
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Air refrigeration important question, chapter-1:-
Important Questions From Chapter-1
1. Working principle of the vortex tube?
2. Thermoelectric refrigeration system
Thermo-electric cooling is described as a solid-state method of heat transfer generated primarily through the use of dissimilar semiconductor materials.To understand the cooling method, it is first necessary to know how thermoelectric cooling systems differ from their conventional refrigeration counterparts. Like conventional refrigeration, it obeys the basic laws of thermodynamics. Only the actual system for cooling is different. In a conventional refrigeration system, the main working parts are the evaporator, condenser, and compressor. The evaporator surface is where the liquid refrigerant boils, changes to vapor and absorbs heat energy. The compressor circulates the refrigerant and applies enough pressure to increase the temperature above ambient level. The condenser helps discharge the absorbed heat into the ambient air.
BASIC PRINCIPLES:- Peltier Effect- when a voltage or DC current is applied to two dissimilar conductors, a circuit can be created that allows for continuous heat transport between the conductor’s junctions.Seebeck Effect- is the reverse of the Peltier Effect. By applying heat to two different conductors a current can be generated.The current is transported through charge carriers (opposite the hole flow or with electron flow).Heat transfer occurs in the direction of charge carrier movement.Applying a current (e- carriers) transports heat from the warmer junction to the cooler junction
Refrigeration based on the Peltier effect: It is obtained by arranging a series of thermo-electric cells in a horizontal array which is then encased in plates made of an electrical insulator. Each thermo-electric cell consists of a pair of dissimilar semi-conductors ( called modules) which are connected by electrical conductors at either end.. The passage of electric current through them causes one of the plates to become hot and the other to become cold. When there is adequate cooling to the heated plate, the opposing plate can reach a low temperature or extra heat on a continuous basis. The Coefficient of Performance (COP) of a Peltier module is defined in the same way as for a conventional refrigeration system :Coefficient of Performance = Rate of heat extraction divided by Electrical Power input. Critical materials parameters to ensure a high COP are a high thermo-electric coefficient to generate the cooling effect, a high electrical conductivity to suppress Ohmic heating and a low thermal conductivity to prevent much heat being conducted from the hot side of the module to the cold side of the module.
3. Explain the reverse Carnot cycle of refrigeration. why it is not used in practice?
reversed Carnot CycleReversed Carnot Cycle
Reversing the Carnot cycle does reverse the directions of heat and work interactions.
A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump.
The reversed Carnot cycle is the most efficient refrigeration cycle operating between two specified temperature levels.
It sets the highest theoretical COP.
Reversed Carnot CycleReversed Carnot Cycle
Consists of 4 processes
1. Adiabatic Compression
2. Isothermal Compression
3. Adiabatic Expansion
4. Isothermal Expansion
Point to remember ….! Point to remember ….!
Practically, the reversed Carnot cycle cannot be used for refrigeration purpose as the adiabatic process requires very high speed operation, whereas the isothermal process requires very low speed operation.
4. 1 ton of refrigeration
5. Cop of refrigeration
The coefficient of performance (COP) of refrigeration is always more than 1
6. Merit & Demerit of Air refrigeration system?
7. Bootstrap air evaporative cooling system Without evaporating & with evaporating?
8. Regenerative air refrigeration system?
9. How throttling can produce cooling?
10. Working of a simple air refrigeration system?
Numerical based on Simple refrigeration system
Numerical based on Bootstrap refrigeration
Numerical based on reversed Carnot cycle
MCQ Questions OF RAC || RGPV online exams 2020
One ton refrigeratiqn corresponds to
(a) 50 kcal/min
(b) 50 kcal/kr
(c) 80 kcal/min
(d) 80 kcal/hr
(e) 1000 kcal/day.
In S.J. unit, one ton of refrigeration is equal to
(a) 210 kJ/min