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Lorico [155]
4 years ago
10

A compression ignition engine when tested gave an indicator card having area 3250mm^2 and length 73mm. The calibration factor wa

s 0.2 bar/mm. The mechanical efficiency of the engine is 80%. Calculate the IMEP and the BMEP
Engineering
1 answer:
atroni [7]4 years ago
7 0

Answer:

BMEP = 8.904 bar

IMEP =  11.13 bar

Explanation:

given data

area A = 3250 mm²

length L = 73 mm

calibration factor = 0.2 bar/mm

mechanical efficiency = 80 %

to find out

IMEP and the BMEP

solution

first we calculate the BMEP break mean efficiency pressure tat is express as

BMEP = \frac{A}{L}

here A is area and L is length

so BMEP is

BMEP = \frac{3250}{73} = 44.52 mm

BMEP = 44.52 × 0.2 bar = 8.904 bar

and

we know mechanical efficiency is

mechanical efficiency = \frac{BMEP}{IMEP}

so put here value we get IMEP

IMEP = \frac{8.904}{0.80}

IMEP =  11.13 bar

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Gnoma [55]
what is it ill know your question??!!
5 0
3 years ago
Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state
lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

P_1=1\ bar,V_1=4.25 {m^3}

P_2=6.8\ bar

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

Now by putting the values                (1.4 bar =140 KPa)

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=140\times 4.25 \ln \dfrac{1.4}{6.8}

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

7 0
4 years ago
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to su
Andrews [41]

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

6 0
4 years ago
Read 2 more answers
Teachers
IRINA_888 [86]
Uh I’m just gonna say yes because I think this is just something random
5 0
3 years ago
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
Stella [2.4K]

Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}

\dfrac{V_3}{V_2}=\dfrac{3200}{520}

So cut-off ratio\dfrac{V_3}{V_2}=6.15

\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}

Now putting the values in above equation

\dfrac20=\dfrac{V_4}{V_3}\times 6.15

\dfrac{V_4}{V_3}=3.25

So expansion ratio\dfrac{V_4}{V_3}=3.25.

\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}

\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

3 0
4 years ago
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