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Helga [31]
3 years ago
12

The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)

at room temperature: σ (????∙m) –1 n (m–3 ) p (m–3 ) Intrinsic 8.9 × 10^4 8.7 × 10^23 8.7 × 10^23Extrinsic 2.3 × 10^5 7.6 × 10^22 1.0 × 10^25 (p-type)Calculate electron and hole mobilities.

Engineering
1 answer:
xxTIMURxx [149]3 years ago
6 0

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

\sigma = n|e|\mu_e+p|e|\mu_h (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

\mu_e= mobility of electron

\mu_h=mobility of holes

\sigma = electrical conductivity

Making the substitution in (S)

Mobility of electron

8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)

0.639=\mu_e+\mu_h

Mobility of hole in (S)

2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))

0.1436 = 7.6*10^{-3}\mu_e+\mu_h

Then, solving the equation:

0.639=\mu_e+\mu_h (1)

0.1436 = 7.6*10^{-3}\mu_e+\mu_h (2)

We have,

Mobility of electron \mu_e = 0.5m^2/V.s

Mobility of hole is \mu_h = 0.14m^2/V.s

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A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
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2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

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==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

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h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
3 years ago
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Answer:

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Explanation:

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The exergy balance equation  is given as follows;

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(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively

Explanation:

Find attachment for explanation

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