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IceJOKER [234]
3 years ago
8

A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of

2000 rpm. Calculate the power developed by the engine at this operating condition.
Engineering
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

V_d=2.5\times 10^{-3} m^3

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

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Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam
Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

               Rod 1 : P*cos(Q)  

               Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

               Q = arctan ( y_1 / x_1)

               Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

               Rod 1 : stress_1 = P*cos(Q)  / A_1

               Rod 2: stress_2 = - P*sin(Q)  / A_2

- The cross sectional Area of both rods are A_1 and A_2:

               A_1 = pi*D^2 / 4

               A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

               Rod 1: P_max =  stress_1 * A_1 / cos(Q)

                          P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

                          P_max = 25203.61848 N

               Rod 2: P_max =  stress_2 * A_2 / sin(Q)

                          P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

                          P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

             

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