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Vilka [71]
3 years ago
15

Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin

e is 630 kJ per kg of steam flowing through the turbine. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, deter- mine (a) the isentropic turbine efficiency, (b) the rate of entropy gen- eration, in kJ/K per kg of steam flowing. Air at 40°F, 1 atm enters a compressor operating at steady state and exits at 620°F, 8.6 atm. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, determine (a) the isentro- pic compressor efficiency and (b) the rate of entropy generation, in Btu/ºr per lb of air flowing. 2.7

Engineering
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

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The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long,
Alja [10]

Answer:

F=6200\ \text{N}\\

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

\sigma=\cfrac{3FL}{2bd^2}

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}

3 0
3 years ago
6. What types of injuries can occur in an electronics lab and how can they be prevented?
marysya [2.9K]

Answer:

The most common injuries in a chemistry lab is making a fire, heat burns, chemical burns, cuts and scrapes, contamination, inhalation, and spills and breaks.

1.) You can prevent making a fire by making sure you close and seal flammable materials.

2.) You can prevent heat burns by teaching the students how to properly use tongs,water baths, and other cooling equipment. 

3.) You can prevent chemical burns by treating the chemicals with caution, measure carefully, and use the approved containers.

4.) You can prevent cuts and scrapes by telling the students how to use the blades safely, and also when they are disposing broken or sharp items they should know how to wrap them up so no one else will get hurt. 

5.) You can prevent contamination by washing your hands, protect their clothing and skin with a lab coat or a lab apron, gloves and glasses, and cleaning your area where the germs of the chemicals were so no one will become.

6.) You can prevent inhalation by opening up windows, using ventilation fans, and using an equipment that measures the amount of gas emission in a room.

7.) Finally, you can prevent spills and breaks by telling the students what will happen if anything spills, and tell them to clean up.  

8 0
3 years ago
A mountain bike has a sprocket and chain drive system designed to adjust the force needed by an operator. The system consists of
Tatiana [17]

Answer:

B. 26 rpm

Explanation:

The sprocket has a diameter of 10 in

The back wheel has a diameter of 6.5 in

One complete revolution formula is : 2πr -------where r is radius

For the sprocket , one revolution = π * D where D=2r

π * 10 = 31.4 in

For the back wheel, one revolution = π* 6.5 = 20.42 in

The pedaling rate is : 40 rpm

Finding the ratio of revolutions between the sprocket and the back tire.

In one revolution; the sprocket covers 31.4 in while the back tire covers 20.42 in so the ratio is;

20.42/ 31.4 = 0.65

So if the speed in the sprocket is 40 rpm then that in the back tire will be;

40 * 0.65 = 26 rpm

8 0
3 years ago
When the electrical connection to the alternator from the battery is not tight, it can cause?
Oksi-84 [34.3K]

Answer:

affects the flow of electricity

Explanation:

A loose battery terminal affects the flow of electricity. There is less power going to the electrical systems and the vehicle will not start or start sluggishly. Also, a loose battery terminal causes the car's electrical components like navigation, car lights, and audio among others to dim or fail completely.

3 0
3 years ago
Technician A uses three prong electrical cords when possible.
Zolol [24]
A

Correct me if I’m wrong tnx
8 0
3 years ago
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