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abruzzese [7]
3 years ago
9

You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 ch

annels). How much space will you save by recording this song with 16-bit, 44.1 kHz sampling for a stereo + subwoofer system (3 channels)? Note that you are making these recordings without using any compression. Please give you answer in terms of Gibibits and round to the nearest hundredth.
Engineering
1 answer:
djyliett [7]3 years ago
8 0

Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample =  24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at  96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × 10^{9}  bits

as 1 Gib bit = 2^{30} bits

so required space = 2.48832 × 10^{9}  bits ÷  2^{30} bits

required space = 2.3174 Gibibits   ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × 10^{9}  bits

space required = 0.381024 × 10^{9}  bits  ÷  2^{30} bits  

space required = 0.3548 Gibibits    ...........2

so

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

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A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl
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  • let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
  • let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
  • let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

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<u>Using effectiveness-NUT method</u>

  1. <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

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C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

          .<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

          .<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

            .<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

           .<em>5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = \\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )

NTU = \frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582   -1} )

NTU = 0.661

          <em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = \frac{NTU C{min} }{U}

As = \frac{0.661 x 5016 W. °c }{640 W/m²}

As = 5.18 m²

            <em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = \frac{As}{\pi D}

L = \frac{5.18 m² }{\pi (0.015 m)}

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0
3 years ago
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