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3 years ago

How do batteries and other types of power sources make physical computing systems more mobile?

Engineering

2 answers:

Roman55 [17]3 years ago

6 0

Answer:

Batteries store energy reasonably well and for a long time. Primary batteries (non-rechargeable) hold more energy than secondary (rechargeable) and the self-discharge is lower. Lead-, nickel- and lithium-based batteries need periodic recharges to compensate for lost energy. Hope this helps!

Explanation:

Sonbull [250]3 years ago

6 0

Because batteries store a lot of energy

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Elodia [21]

Answer:

if ur mad you may drive faster if ur sad u may drive slower due to the amount of adrenaline and dopamine levels in your body in that given moment

Explanation:

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3 years ago

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natali 33 [55]

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Explanation:

80 grit is coarsest grit that may be used on aluminum

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3 years ago

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3 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

4 years ago

In an air compressor the compression takes place at a constant internal energy and 50KJ of heat are rejected to the cooling wate

lozanna [386]

Answer:

work is 50 kj

Explanation:

Given data

heat (Q) = 50 kj

To find out

work input for the compression stroke per kilogram of air

Solution

we will apply here "first law of thermodynamics" i.e.

The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.

ΔU = Q – W ................1

here ΔU is change in internal energy, Q is heat and W is work done

here U = 0 because air compressor the compression takes place at a constant internal energy in question

so that by equation 1

Q = W

and Q = 50

so work will be 50 kj

8 0

3 years ago