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miv72 [106K]
2 years ago
10

Calculate the mean piston speed, bmep, and specific power of: (a) the spark-ignition engines in Figs. 1.11 and 1.13 at their max

imum rated power and torque; and (b) the mean piston speed, bmep, and specific power of the diesel engines in Figs. 1.31 and 1.32 at their maximum rated power and torque. Briefly explain any significant differences between these two types of engine

Engineering
1 answer:
Kisachek [45]2 years ago
5 0

Answer:

Explanation:

See the attached picture for detailed answer.

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Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t
alukav5142 [94]

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0
2 years ago
A liquid with a specific gravity of 2.6 and a viscosity of 2.0 cP flows through a smooth pipe of unknown diameter, resulting in
sasho [114]
.......,.,.,.,.,.,,.,.,,.,.,.,.,.,.,.,,.,.,,.,,,.,,.,,.,.,.,.,,............,,,,,,’mmdjidvdhxxkf jkkk he in d
6 0
2 years ago
Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No
Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is 1.557\times10^{-7}m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

Re=\frac{vd}{\nu}

Here, v is velocity, \nu is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

Re=\frac{vd}{\nu}

Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

5 0
2 years ago
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0
1 year ago
A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75 . The nozzle is supplied from an air reser
Flura [38]

Answer:

a. 4.279 MPa

b. 3.198 MPa to 4.279 MPa

c. 0.939 MPa

d. Below 3.198 MPa

Explanation:

From the given parameters

M_{exit} = 1.75 MPa  

M at 1.6 MPa gives A_{exit}/A* = 1.2502

M at 1.8 MPa gives  A_{exit}/A* = 1.4390

Therefore, by interpolation, we have M_{exit} = 1.75 MPa  gives A

However, we shall use M_{exit} = 1.75 MPa and A

Similarly,

P_{exit}/P₀ = 0.1878

a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have

A_{exit}/A* = 1.387. by interpolation M

Therefore P_{exit} = P₀ × P

Which shows that the nozzle is choked for back pressures lower than 4.279 MPa

b) Where there is a normal shock at the exit of the nozzle, we have;

M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa

Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406

Where the normal shock occurs at the nozzle exit, we have

P_b = 3.406\times 0.939 = 3.198 MPa

Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as P_b = 4.279 MPa

Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa

c) At M_{exit} = 1.75 MPa  and P

d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane    

8 0
2 years ago
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