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2 years ago

How does the two-stroke Otto cycle differ from the four-stroke Otto cycle?

Engineering

1 answer:

Digiron [165]2 years ago

3 0

Answer:

Two stroke cycle Four stroke cycle

1.Have on power stroke in one revolution. 1.have one power

stroke in two revolution

2.Complete the cycle in 2 stroke 2.Complete the cycle in 4 stroke

3.It have ports 3.It have vales

4.Greater requirement of cooling 4.Lesser requirement of cooling

5.Less thermal efficiency 5.High thermal efficiency

6.Less volumetric efficiency 6.High volumetric efficiency

7.Size of flywheel is less. 7.Size of flywheel is more.

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1 year ago

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2 years ago

The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a

Arturiano [62]

Answer:

Explanation:

Given that:

Torque T = 2300 lb - ft

Bending moment M = 1500 lb - ft

axial thrust P = 2500 lb

yield points for tension σY= 100 ksi

yield points for shear τY = 50 ksi

Using maximum-shear-stress theory

$\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}$

where;

$A = \pi c^2$

$I = \dfrac{\pi}{4}c^4$

$\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}$

$\tau_A = \dfrac{T_c}{\tau}$

where;

$\tau = \dfrac{\pi c^4}{2}$

$\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{55200 }{\pi c^3}}$

$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)$

Let say :

$|\sigma_1 - \sigma_2| = \sigma_y$

Then :

$2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)$

$(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6$

$6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0$

According to trial and error;

c = 0.75057 in

Replacing c into equation (1)

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$