Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider isentropic compressor of a vaporcompression refrigeration cycle. What are the isentropic efficiency and second-law efficiency of this compressor? Justify your answers. Is the second-law efficiency of a compressor necessarily equal to its isentropic efficiency? Explain.

Short Answer

Expert verified
Explain. Answer: Yes, in the case of an isentropic compressor, the isentropic efficiency and the second-law efficiency are equal to each other, and both are 100%. In an isentropic compressor, the actual work input is equal to the isentropic work input as there are no losses or irreversibilities. Keep in mind, this equality is valid for the theoretical case of an isentropic compressor. However, in real-world compressors, the isentropic and second-law efficiencies are generally not equal, as they involve losses and irreversibilities.

Step by step solution

01

Define isentropic efficiency

The isentropic efficiency of a compressor is defined as the ratio of the actual work input required to compress the refrigerant to the ideal (isentropic) work input. This efficiency takes into account the losses due to irreversibilities in the compression process. Isentropic efficiency (η_isentropic) = (Isentropic work input) / (Actual work input)
02

Define second-law efficiency

The second-law efficiency of a compressor is defined as the ratio of the minimum theoretical work input required to achieve the desired effect (increase in pressure and temperature of the refrigerant) to the actual work input. Second-law efficiency (η_second-law) = (Minimum theoretical work input) / (Actual work input)
03

The isentropic efficiency

Since we are considering an isentropic compressor, which is a theoretical case without any friction or other losses, the actual work input will be equal to the isentropic work input. Consequently, the isentropic efficiency will be equal to 1 (or 100%). η_isentropic = (Isentropic work input) / (Actual work input) = (Isentropic work input) / (Isentropic work input) = 1
04

The second-law efficiency

In an isentropic compressor, the actual work input is equal to the isentropic work input. Therefore, the second-law efficiency of the compressor will also be equal to 1 (or 100%), because the minimum theoretical work input is equal to the actual work input. η_second-law = (Minimum theoretical work input) / (Actual work input) = (Isentropic work input) / (Isentropic work input) = 1
05

Comparing isentropic and second-law efficiencies

In the case of an isentropic compressor, the isentropic efficiency and the second-law efficiency are equal to each other, and both are 100%. This equality is valid for the isentropic compressor, where the compression process ideally doesn't involve any losses or irreversibilities. However, in real-world compressors, there are always some losses and irreversibilities, leading to a decrease in both isentropic and second-law efficiencies. Therefore, for a real compressor, the second-law efficiency is generally not equal to its isentropic efficiency. The second-law efficiency is often higher than the isentropic efficiency because the actual work input includes losses that are not considered in the minimum theoretical work input.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Thermoelectric coolers that plug into the cigarette lighter of a car are commonly available. One such cooler is claimed to cool a \(12-0 z(0.771-1 b m)\) drink from 78 to \(38^{\circ} \mathrm{F}\) or to heat a cup of coffee from 75 to \(130^{\circ} \mathrm{F}\) in about \(15 \mathrm{min}\) in a well-insulated cup holder. Assuming an average COP of 0.2 in the cooling mode, determine ( \(a\) ) the average rate of heat removal from the drink, \((b)\) the average rate of heat supply to the coffee, and ( \(c\) ) the electric power drawn from the battery of the car, all in \(\mathrm{W}\).

An ammonia-water absorption refrigeration cycle is used to keep a space at \(25^{\circ} \mathrm{F}\) when the ambient temperature is \(70^{\circ} \mathrm{F}\). Pure ammonia enters the condenser at 300 psia and \(140^{\circ} \mathrm{F}\) at a rate of \(0.04 \mathrm{lbm} / \mathrm{s}\). Ammonia leaves the condenser as a saturated liquid and is expanded to 30 psia. Ammonia leaves the evaporator as a saturated vapor. Heat is supplied to the generator by geothermal liquid water that enters at \(240^{\circ} \mathrm{F}\) at a rate of \(0.55 \mathrm{lbm} / \mathrm{s}\) and leaves at \(200^{\circ} \mathrm{F}\). Determine ( \(a\) ) the rate of cooling provided by the system, in \(\mathrm{Btu} / \mathrm{h}\), the \(\mathrm{COP}\), and (b) the second-law efficiency of the system. The enthalpies of ammonia at various states of the system are: condenser inlet \(h_{2}=665.7 \mathrm{Btu} / \mathrm{lbm},\) evaporator inlet \(h_{4}=190.9 \mathrm{Btu} / \mathrm{lbm}\) evaporator exit \(h_{1}=619.2 \mathrm{Btu} / \mathrm{lbm} .\) Also, take the specific heat of geothermal water to be \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\).

How is the coefficient of performance of an absorption refrigeration system defined?

A refrigerator operates on the ideal vapor compression refrigeration cycle with \(\mathrm{R}-134 \mathrm{a}\) as the working fluid between the pressure limits of 120 and 800 kPa. If the rate of heat removal from the refrigerated space is \(32 \mathrm{kJ} / \mathrm{s}\), the mass flow rate of the refrigerant is \((a) 0.19 \mathrm{kg} / \mathrm{s}\) \((b) 0.15 \mathrm{kg} / \mathrm{s}\) \((c) 0.23 \mathrm{kg} / \mathrm{s}\) \((d) 0.28 \mathrm{kg} / \mathrm{s}\) \((e) 0.81 \mathrm{kg} / \mathrm{s}\)

Consider a \(300 \mathrm{kJ} / \mathrm{min}\) refrigeration system that operates on an ideal vapor-compression refrigeration cycle with refrigerant- 134 a as the working fluid. The refrigerant enters the compressor as saturated vapor at \(140 \mathrm{kPa}\) and is compressed to \(800 \mathrm{kPa}\). Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine ( \(a\) ) the quality of the refrigerant at the end of the throttling process, ( \(b\) ) the coefficient of performance, and ( \(c\) ) the power input to the compressor.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free