1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lorico [155]
3 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]3 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

You might be interested in
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
2 years ago
For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho
ad-work [718]

Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0
3 years ago
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol
pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = \frac{\pi}{4}D^2L

or

V = \frac{\pi}{4}\frac{D^3}{1.1}

on substituting the values, we have

245 =  \frac{\pi}{4}\frac{D^3}{1.1}

or

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

or

L= 6.36 cm

Now,

the compression ratio is given as:

\textup{compression ratio}=\frac{V+v}{v}

on substituting the values, we get

\textup{compression ratio}=\frac{245+27.2}{27.2}

or

Compression ratio = 10.007

4 0
3 years ago
I need help with this question please
solniwko [45]

Answer:

The resultant moment is 477.84 N·m

Explanation:

We note that the resultant moment is given by the moment about a given point

The length of the sides of the formed triangles are;

l = sin(40°) × 4/sin(110°) ≈ 2.736

Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;

The resultant moment, ∑m, is given as follow;

∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m

Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.

6 0
3 years ago
Other questions:
  • A forklift raises a 90.5 kg crate 1.80 m. (a) Showing all your work and using unity conversion ratios, calculate the work done b
    14·2 answers
  • Where Does a Solar Engineer Work? <br> (2 sentences or more please)
    14·2 answers
  • The density of a fluid is given by the empirical equation rho 70:5 exp 8:27 107 P where rho is density (lbm/ft3 ) and P is press
    6·1 answer
  • Who can help me with electric systems for cars?
    14·1 answer
  • A structural component in the shape of a flat plate 29.6 mm thick is to be fabricated from a metal alloy for which the yield str
    11·1 answer
  • What is definition of<br>computational fluid Dynamics <br>unstructured grid <br>domain<br>geometry​
    6·1 answer
  • A step-up transformer has 20 primary turns and 400 secondary turns. If the primary current is 30 A, what is the secondary curren
    15·1 answer
  • Whats the difference between GeForce GTX 1060 and Geforce GTX 3060? Is there any big changes to FPS and other settings?
    7·2 answers
  • Air at 403 K and 1 atm enters a convergent nozzle at a velocity of 150
    9·1 answer
  • Oil, with density of 900 kg/m3 and kinematic viscosity of 0.00001 m2/s, flows at 0.2 m3/s through 500 m of 200-mm-diameter cast-
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!