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MrMuchimi
3 years ago
10

An ideal Diesel cycle has a compression ratio of 17 and a cutoff ratio of 1.3. Determine the maximum temperature of the air and

the rate of heat addition to this cycle when it produces 140 kW of power and the state of the air at the beginning of the compression is 90 kPa and 578C. Use constant specific heats at room temperature.

Engineering
1 answer:
Radda [10]3 years ago
6 0

Answer:

maximum temperature = 1322 k

rate of heat addition = 212 kw

Explanation:

compression ratio = 17

cut off ratio = 1.3

power produced = 140 Kw

state of air at the beginning of the compression = 90 kPa and 578 c

Determine the maximum temperature of air

attached below is the detailed solution

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faust18 [17]

Answer:

#WeirdestQuestionOfAllTime

Explanation:

8 0
3 years ago
Ronny wants to calculate the mechanical advantage. He needs to determine the length of the effort arm and the length of the load
kakasveta [241]

Answer:

I hope it's helpful.

Explanation:

Simple Machines

Experiments focus on addressing areas pertaining to the relationships between effort force, load force, work, and mechanical advantage, such as: how simple machines change the force needed to lift a load; mechanical advantages relation to effort and load forces; how the relationship between the fulcrum, effort and load affect the force needed to lift a load; how mechanical advantage relates to effort and load forces and the length of effort and load arms.

Through investigations and models created with pulleys and levers, students find that work in physical terms is a force applied over a distance. Students also discover that while a simple machine may make work seem easier, in reality the amount of work does not decrease. Instead, machines make work seem easier by changing the direction of a force or by providing mechanical advantage as a ratio of load force to effort force.

Students examine how pulleys can be used alone or in combination affect the amount of force needed to lift a load in a bucket. Students find that a single pulley does not improve mechanical advantage, yet makes the effort applied to the load seem less because the pulley allows the effort to be applied in the direction of the force of gravity rather than against it. Students also discover that using two pulleys provides a mechanical advantage of 2, but that the effort must be applied over twice the distance in order to gain this mechanical advantage Thus the amount of work done on the load force remains the same.

Students conduct a series of experiments comparing the effects of changing load and effort force distances for the three classes of levers. Students discover that when the fulcrum is between the load and the effort (first class lever), moving the fulcrum closer to the load increases the length of the effort arm and decreases the length of the load arm. This change in fulcrum position results in an increase in mechanical advantage by decreasing the amount of effort force needed to lift the load. Thus, students will discover that mechanical advantage in levers can be determined either as the ratio of load force to effort force, or as the ratio of effort arm length to load arm length. Students then predict and test the effect of moving the fulcrum closer to the effort force. Students find that as the length of the effort arm decreases the amount of effort force required to lift the load increases.

Students explore how the position of the fulcrum and the length of the effort and load arms in a second-class lever affect mechanical advantage. A second-class lever is one in which the load is located between the fulcrum and the effort. In a second-class lever, moving the load changes the length of the load arm but has no effect on the length of the effort arm. As the effort arm is always longer than the load arm in this type of lever, mechanical advantage decreases as the length of the load arm approaches the length of the effort arm, yet will always be greater than 1 because the load must be located between the fulcrum and the effort.

Students then discover that the reverse is true when they create a third-class lever by placing the effort between the load and the fulcrum. Students discover that in the case of a third-class lever the effort arm is always shorter than the load arm, and thus the mechanical advantage will always be less than 1. Students also create a model of a third-class lever that is part of their daily life by modeling a human arm.

The CELL culminates with a performance assessment that asks students to apply their knowledge of simple machine design and mechanical advantage to create two machines, each with a mechanical advantage greater than 1.3. In doing so, students will demonstrate their understanding of the relationships between effort force, load force, pulleys, levers, mechanical advantage and work. The performance assessment will also provide students with an opportunity to hone their problem-solving skills as they test their knowledge.

Through this series of investigations students will come to understand that simple machines make work seem easier by changing the direction of an applied force as well as altering the mechanical advantage by afforded by using the machine.

Investigation focus:

Discover that simple machines make work seem easier by changing the force needed to lift a load.

Learn how effort and load forces affect the mechanical advantage of pulleys and levers.

8 0
3 years ago
At an auction of antiques, a bidder for a particular porcelain statue would be trying to ________(a) Maximize the difference bet
alex41 [277]

Answer: MINIMIZE INPUT

Explanation: AUCTION this is a process of selling a product,an Art work other tradeable assets like stocks, bonds based on the person with the highest BIDDING( Higher amount). In most cases the person buying will try to control his bidding to the MINIMUM AMOUNT in order for him to avoid spending higher than expected. Auction sale is sometimes used when trying to sell off old products which has been held for a long time, sometimes Auctions are used to raise funds for a particular Reason like the sale of ARTIFACTS.

5 0
4 years ago
Briefly discuss if it would be better to operate with pumps in parallel or series and how your answer would change as the steepn
Aleksandr [31]

Answer:

1) In series, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) In parallel, the combined head and volume flow will move along the system curve from point 1 to point 3.

Explanation:

1) Pump in series:

When two or more pumps are connected in series, their resulting pump performance curve will be obtained by adding their respective heads at the same flow rate as shown in the first diagram attached.

In the first diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3.

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the head of two identical pumps connected in series is twice the head of a single pump flowing at the same rate.

- Point 3 is the point where the system is operating when both pumps are running.

Now, since the flowrate is constant, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) Pump in parallel:

When two or more pumps are connected in parallel, their resulting pump performance curve will be obtained by adding their respective flow rates at same head as shown in the second diagram attached.

In the second diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the flow rate of two identical pumps connected in series is twice the flow rate of a single pump.

- Point 3 is the point where the system is operating when both pumps are running.

In this case, the combined head and volume flow will move along the system curve from point 1 to point 3.

5 0
3 years ago
120 litres of water is discharge from container in 25 seconds. Find the rate of discharge in cumecs.if the discharge took place
notsponge [240]
<h2>Answer:</h2>

Rate of discharge in cumecs: <u>0.0048m³/s</u>.

Velocity flow: <u>24m/s</u>.

<h2>Explanation:</h2>

<h3>1. Find the rate of discharge in cumecs.</h3>

a. Convert from litres to m³.
120L*1000= 120000mL

120000mL=120000cm³

120000cm³/100³=0.12 m³.

b. Rate of discharge.

<em>If  0.12 m³ where discharged in 25 seconds, the rate of discharge is:</em>

0.12m³/25s = 0.0048m³/s.

<em />

<em />

<h3>2. Find the velocity flow.</h3>

Let's refer to the fluid mechanics equation that relates volume flow, area and velocity. This is the formula:

\frac{dV}{dt}=Av; where the expression \frac{dV}{dt} is the volume flow rate (in m³/s); A is the cross-sectional area of the pipe (in m²), and v is the velocity flow (in m/s).

a. Solve the equation for v.

\frac{dV}{dt}=Av\\ \\(\frac{dV}{dt})/A=v\\ \\v=(\frac{dV}{dt})/A

b. Calculate the cross-sectional area of the pipe.

<em>The cross-sectional area of the pipe is a circle. Hence, the formula of this area is:</em>

A=\pi r^{2}

<em>We'll have to convert the diameter to meters, because the formula for flow velocity needs the area in m². Let's go ahead and do that.</em>

<em />50mm/1000=0.05m.

<em>We were given the diameter, and the formula uses the radius, but the radius is just half of the diameter, therefore, we can substitute in toe formula like this:</em>

A=\pi (\frac{0.05}{2} )^{2}=0.0020m^{2}

c. Substitute in the new expression for velocity flow and calculate.

v=(\frac{dV}{dt})/A\\ \\v=(\frac{0.048m^{3} }{1s})/(0.0020m^{2} )\\\\ v= 24m/s

8 0
2 years ago
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