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3 years ago

The two stroke engines has greater cooling and lubrication requirements than four stroke engine. Explain why?

Engineering

1 answer:

AlexFokin [52]3 years ago

8 0

Answer:

A two stroke engine produces twice the power compared to a four stroke engine of same weight and size.

Explanation:

In a two stroke engine, all the four processes namely, intake stroke, compression stroke, power stroke and exhaust stroke takes place in one revolution of crankshaft or two strokes of the piston. While in a four stroke engine, all the four processes namely intake stroke, compression stroke, power stroke and exhaust stroke take place in two revolution of crankshaft or four strokes of the piston.

Therefore, there is one power stroke in one revolutions of the crankshaft in case of a two stroke engine as compared to the four stoke engine where there is one power stroke for two revolutions of the crank shaft.

So the power developed in a two stroke engine is more ( nearly twice ) as compared to a four stroke engine of the same capacity. When power produced is more, the heat dissipation is also more in case of a two stroke engine. So greater cooling is required to dissipate heat from a two stroke engine as compared to a four stroke engine.

Also in a two stroke engine, the lubricating oil is used with the oil whereas a four stroke engine has a separate tank for lubricating oil. So the lubricating oil gets burnt quickly in a two stroke engine.

Thus, to dissipate more heat, a two stroke engines has greater cooling and lubrication requirements than a four stroke engines as power produce in a two stroke engine is more than a four stoke engine with same weight or size.

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Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press

ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1$ (wind direction factor)

$k _{zt} = 1$ = topographical factor (for flat terrain)

$q_n$ = velocity pressure at height h

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

$= 36.86 k_z$

But for height h = 30 ft, $k_z$ = 0.98 (from table)

$\therefore q_n = 36.86 \times 0.98$

= 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ , so $C_p=-0.5$ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

$GC_{pi} = \pm 0.18$ (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

4 0

4 years ago

What is the function of the rotor and the stator in an AC motor

sweet-ann [11.9K]

The rotor is situated inside the stator and is mounted on the AC motor's shaft. It is the rotating part of the AC motor. And while we know this, the major function of the rotor and the stator is helping the motor shaft rotate.

7 0

4 years ago

The electron concentration in silicon at T = 300 K is given by

puteri [66]

Answer:

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

Explanation:

From the question we are told that:

Temperature of silicon $T=300k$

Electron concentration $n(x)=10^{16}\exp (\frac{-x}{18})$

$\frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})$

Electron diffusion coefficient is $Dn = 25cm^2/s \approx 2.5*10^{-3}$

Electron mobility is $\mu n = 960 cm^2/V-s \approx0.096m/V$

Electron current density $Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2$

Generally the equation for the semiconductor is mathematically given by

$Jn=qb_n\frac{dn}{dx}+nq \mu E$

Therefore

$-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E$

$E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}$

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

7 0

3 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

Isormophous phase diagram

shusha [124]

Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

6 0

3 years ago