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Komok [63]
3 years ago
11

Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und

er a certain condition. Each cylinder has one injector and each injector has 6 nozzle holes with a diameter of 200 micrometer. The discharge coefficient of the injector nozzle is 0.75. If the fuel injection pressure difference across the injector nozzle is 100MPa, what will be the fuel injection duration in Crank Angle degrees? Fuel density is 800 kg/m3 .
Engineering
1 answer:
sveticcg [70]3 years ago
4 0

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

combine Mass for 6 cylinder (M) =15 Kg/hr

mass of  each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec

Engine speed (N)= 1500rpm

Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m

Discharge Coefficient (Cd) = 0.75

Pressure difference = 100 MPa

Density of fuel = 800 kg/m^3

velocity of fuel is v  = cd\sqrt{\frac{2*P}{p}}

v = 0.75 \sqrt{\frac{2\times 100\times 10^6}{800}} = 375 m/sec

injected fuel volume  (V) =Area of given  Orifices × Fuel velocity × time of single injection × no of injection/sec

we know that p = m/ V

SoV = \frac{0.000694}{800} =8.68\times10^{-7} m3/sec

putting these value in volume equation and solve for Discharge 8.68\times 10^{-7} = (\frac{(3.14}{4})\times 6\times( .0002\times .0002) \times  375 \times  \frac{(Q}{360}) \times \frac{30}{750} \times \frac{(750}{60)}

Q = 8.845 DEGREE

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kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

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Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0
3 years ago
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