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Svetach [21]
3 years ago
10

Design a filter that has infinite DC gain, a gain of one from 1Hz to 100 Hz and filters (1storder) any signals above 100 Hz.a) S

ketch the bode plotb) Sketch the s-planec) write the transfer function of the filterd) write the differential equatione) write out the unforced transient responsef) write out the frequency response
Engineering
1 answer:
EastWind [94]3 years ago
4 0

Answer:

Attached below are the  sketches

answer :

c) G(s) = 100 / ( s + 100 )

d) y'(t)  + 100Y(s) = 100 X(s)

e) g(t) = e^-100t  u(t)

Explanation:

a) Sketch the bode plot

The filter here is a low pass filter

b) Sketch the s-plane

attached below.     pole ( s ) is at 100

c) write the transfer function of the filter

Transfer function ; G(s) = 100 / ( s + 100 )

d) write the differential equation

Y(s) / X(s) = 100 / s + 100

Y(s) [ s + 100 ] = 100 X(s)

= sY(s) + 100Y = 100 X(s)

∴ differential equation = y'(t)  + 100Y(s) = 100 X(s)

e) write out the unforced transient response

g(t) = e^-100t  u(t)

f) write out the frequency response

attached below

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Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
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Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

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\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

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The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

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\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

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3 years ago
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