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Ilya [14]
4 years ago
3

Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 0.5 kg/s and leaves at 600 kPa and 450 K. Neglec

ting kinetic energy changes, determine (a) the volume flow rate of the carbon dioxide at the compressor inlet (b) the power input to the compressor.
Physics
1 answer:
ra1l [238]4 years ago
0 0

Answer:

a)\dot V = 0.28335 m3/s

b)\dot W_{in} = 259.63 kW

Explanation:

part a)

inlet volume of air V1 is given as = \frac{RT_1}{P_1}

putting all value in the above formula

V_1 = = \frac{0.1889 kpa - m3/kg .K * 300 K}{100kPa}

V_1 = 0.5667 m3/kg

we know that  volume of flow rate

\dot V = \dot mV_1

\dot V = 0.5 kg/s *0.5667 m3/kg = 0.28335 m3/s

\dot V = 0.28335 m3/s

part b

we know that total energy remain constant, so we have

E_in - E_out = Change in energy in system

E_in = E_out

therefore

\dot W_{in} + \dot m h_1 =\dot m h_2

\dot W_{in} = \dot m h_2- \dot m h_1

the power input to the compressor is

\dot W_{in} = \dot m c_p (h_2- h_1 )

\dot W_{in} = 0.5*5.1926*(450-350 )

\dot W_{in} = 259.63 kW

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So then if we replace we got:

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Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

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c = 3x10^8 m/s

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So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

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E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

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