All categories

3 years ago

What is compression ratio of an Otto cycle? How does it affect the thermal efficiency of the cycle?

Engineering

1 answer:

GuDViN [60]3 years ago

6 0

Answer:

$r= \dfrac{V_1}{V_2}$

$\eta =1-\dfrac{1}{r^{\gamma-1}}$

Explanation:

Compression ratio:

Compression ratio is the ratio of volume in the cylinder.Generally it is denoted by r.

The compression ratio for Otto cycle is about 8 to 12.But the compression ratio of diesel engine is about 14 to 25.

So from the Otto cycle compression ratio can be define as

$r= \dfrac{V_1}{V_2}$

As we know that efficiency defines as the ratio of work out to the heat input.In the Otto cycle the efficiency is given as in terms of compression ratio r as

$\eta =1-\dfrac{1}{r^{\gamma-1}}$

Where γ is the heat capacity ratio.

Read more about career advancement

<em>brainly.com/question/7053706</em>

7 0

2 years ago

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = $\frac{MC}{I}$ = $\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}$ = 159.07 MPa

B) Determine max transverse shear stress

attached below

ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam = 159.07 / 2 = 79.535 MPa

attached below is the remaining solution

6 0

3 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

6 0

2 years ago

A drilling operation is performed on a steel part using a 12.7 mm diameter twist drill with a point angle of 118 degrees. The ho

Masteriza [31]

Answer:

a. Rotational speed of the drill = 375.96 rev/min

b. Feed rate = 75 mm/min

c. Approach allowance = 3.815 mm

d. Cutting time = 0.67 minutes

e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min

Explanation:

Here we have

a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min

b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min

c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm

d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes

e. R = 0.25πD²fr = 9525 mm³/min.

7 0

3 years ago

Please define the coefficient of thermal expansion?

Vikki [24]

Answer:

The coefficient of thermal expansion tells us how much a material can expand due to heat.

Explanation:

Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.

Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).

7 0

3 years ago