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2 years ago

A Hartnell governor has two rotating balls, of mass 2.7 kg each. The ball radius is 125 mm in the mean position when the

Engineering

1 answer:

docker41 [41]2 years ago

4 0

Answer:

A Hartnell governor has two rotating balls, of mass 2.7 kg each. The ball radius is 125 mm in the mean position when the

ball arms are vertical and the speed is 150 r.p.m. with the sleeve rising. The length of the ball arms is 140 mm and the

length of the sleeve arms 90 mm. The stiffness of the spring is 7 kN/m and the total sleeve movement is 12 mm from the

mean position. Allowing for a constant friction force of 14 N acting at the sleeve, determine the speed range of the

governor in the lowest and highest sleeve positions. Neglect the obliquity of the ball arms

Explanation:

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If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas

jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ( $\eta$ ), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$ , then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.

8 0

3 years ago

A sensor produces a signal with amplitude 15 mV. A voltage amplifier must amplify the signal such that the amplitude of the outp

Nady [450]

Answer:

42.50 dB

Explanation:

Determine the minimum voltage gain

amplitude of input signal ( Vi ) = 15 mV

amplitude of output signal ( Vo) = 2 V

Vo = 2 v

therefore ; minimum gain = Vo / Vi = 2 / ( 15 * 10^-3 )

= 133.33

Minimum gain in DB = 20 log ( 133.33 )

= 42.498 ≈ 42.50 dB

8 0

3 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$