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3 years ago

13

For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

te with a temperature of 377 K.

2 answers:

Burka [1]3 years ago

6 0

Answer:

Sound will travel with a speed of 302.9 m/sec

Explanation:

We have given $c_p=840.4j/kg-K$

And $c_v=651.5j/kg-K$

Temperature T = 377 K

Gas constant $R=c_p-c_v=840.4-651.5=188.9j/kg-K$

And $\gamma =\frac{c_p}{c_v}=\frac{840.4}{651.5}=1.289$

Speed is given by $v=\sqrt{\gamma RT}=\sqrt{1.289\times 188.9\times 377}=302.9794m/sec$

So sound will travel with a speed of 302.9 m/sec

Crank3 years ago

3 0

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

$c=\sqrt(\gamma*R*T)$

Where R is the gas's particular constant defined in terms of Cp and Cv:

$R=Cp-Cv$

For particular values given:

$R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}$

$R=188.9 \frac{J}{Kg-K}$

The gamma undimensional constant is also expressed as a function of Cv and Cp:

$\gamma=Cp/Cv$

$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

$c=313 m/s$

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Explanation:

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Concerning the storage battery, what category of the primary sources is voltage produced?

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2 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Design a 10-to-4 encoder with inputs in the l-out-of-10 code and outputs in a code like normal BCD except that input lines 8 and

zvonat [6]

Answer:

See image attached.

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