can someone help me with this question so :

A four-stroke diesel engine with four cylinders is running at constant speed on a test

bed. An indicator diagram for one cylinder is shown in the figure below and other test

data are given below:

measured output power of engine (brake power) = 55.0 kW

fuel used in 100 seconds = 0.376 litre

calorific value of fuel = 38.6 MJ litre–1

engine speed = 4100 rev min–1

calculate the indicated power?

-so the area of the P-V loop was 470

-so I calculated the indicated engine power to be = 470 x 4 x (4100/60)

-but the markscheme says the indicated power= 470 x 4 x (4100/60) x 0.5

where is the 0.5 from ?

A four-stroke diesel engine with four cylinders is running at constant speed on a test

bed. An indicator diagram for one cylinder is shown in the figure below and other test

data are given below:

measured output power of engine (brake power) = 55.0 kW

fuel used in 100 seconds = 0.376 litre

calorific value of fuel = 38.6 MJ litre–1

engine speed = 4100 rev min–1

calculate the indicated power?

-so the area of the P-V loop was 470

-so I calculated the indicated engine power to be = 470 x 4 x (4100/60)

-but the markscheme says the indicated power= 470 x 4 x (4100/60) x 0.5

where is the 0.5 from ?

Original post by 1234kelly

can someone help me with this question so :

A four-stroke diesel engine with four cylinders is running at constant speed on a test

bed. An indicator diagram for one cylinder is shown in the figure below and other test

data are given below:

measured output power of engine (brake power) = 55.0 kW

fuel used in 100 seconds = 0.376 litre

calorific value of fuel = 38.6 MJ litre–1

engine speed = 4100 rev min–1

calculate the indicated power?

-so the area of the P-V loop was 470

-so I calculated the indicated engine power to be = 470 x 4 x (4100/60)

-but the markscheme says the indicated power= 470 x 4 x (4100/60) x 0.5

where is the 0.5 from ?

A four-stroke diesel engine with four cylinders is running at constant speed on a test

bed. An indicator diagram for one cylinder is shown in the figure below and other test

data are given below:

measured output power of engine (brake power) = 55.0 kW

fuel used in 100 seconds = 0.376 litre

calorific value of fuel = 38.6 MJ litre–1

engine speed = 4100 rev min–1

calculate the indicated power?

-so the area of the P-V loop was 470

-so I calculated the indicated engine power to be = 470 x 4 x (4100/60)

-but the markscheme says the indicated power= 470 x 4 x (4100/60) x 0.5

where is the 0.5 from ?

It is a formula given in AQA Engineering Physics info.

https://filestore.aqa.org.uk/resources/physics/AQA-7407-7408-TG-EP.PDF

You can read more on the following web pages.

https://energyeducation.ca/encyclopedia/Four_stroke_engine

“A four stroke engine delivers one power stroke for every two cycles of the piston (or four piston strokes).”

https://engineering.stackexchange.com/questions/40286/how-many-power-stroke-in-3-cylinders-4-stroke-at-3000-rpm

Original post by Eimmanuel

It is a formula given in AQA Engineering Physics info.

https://filestore.aqa.org.uk/resources/physics/AQA-7407-7408-TG-EP.PDF

You can read more on the following web pages.

https://energyeducation.ca/encyclopedia/Four_stroke_engine

“A four stroke engine delivers one power stroke for every two cycles of the piston (or four piston strokes).”

https://engineering.stackexchange.com/questions/40286/how-many-power-stroke-in-3-cylinders-4-stroke-at-3000-rpm

https://filestore.aqa.org.uk/resources/physics/AQA-7407-7408-TG-EP.PDF

You can read more on the following web pages.

https://energyeducation.ca/encyclopedia/Four_stroke_engine

“A four stroke engine delivers one power stroke for every two cycles of the piston (or four piston strokes).”

https://engineering.stackexchange.com/questions/40286/how-many-power-stroke-in-3-cylinders-4-stroke-at-3000-rpm

Thank you !!! I was so confused for ages about why they’ve halved it

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