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nirvana33 [79]
3 years ago
15

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:

<u><em>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. </em></u> <em>Hope this helps!</em>

Explanation:

Sonbull [250]3 years ago
6 0
Because batteries store a lot of energy
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The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0
3 years ago
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati
Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0
3 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
3 years ago
2.1 What is the minimum number of pins required for a so-called dual-op-amp IC package, one containing two op amps? What is the
cupoosta [38]

Answer:

8 for dual-op-amp package, and 14 for quad-op-amp

Explanation;

This is because every op-amp has 2 input terminal 4 pns

So one output terminal that is 2 pins which are required for power

and the same for a minumum number of pins required by quad op amp which is 14

5 0
3 years ago
State two faults that are common in a simple cell​
Step2247 [10]

Answer:

the two defects of a simple cell are:

1. Polarization

2. Local action

4 0
2 years ago
Read 2 more answers
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