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Stels [109]
2 years ago
11

Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

Engineering
1 answer:
IceJOKER [234]2 years ago
6 0

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

q=-\kappa \bigtriangledown T

Where,

q is the heat flux

\kappa is the thermal conductivity

\bigtriangledown T is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

C=\frac{E}{\Delta T}

Where,

C is the Heat capacity

E is the energy absorbed/released

\Delta T is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

\alpha=\frac{\kappa}{C_p \times \rho}

Where,

\alpha is thermal diffusivity

\kappa is the thermal conductivity

C_p is specific heat capacity at constant pressure

\rho is density

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2 years ago
Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe
NemiM [27]

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

6 0
3 years ago
How can statistical analysis of a dataset inform a design process
Shtirlitz [24]

Answer:

Explanation:.

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2 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

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3 years ago
In order to avoid a rollover, what is the highest degree incline one should mow on? 10-degree incline 5-degree incline 30-degree
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Answer: B: 20-degree incline

Explanation:

A tractor user should avoid slopes of more than 20 degrees in order to avoid rollovers

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