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3 years ago

/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a

Engineering

1 answer:

Digiron [165]3 years ago

4 0

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 * 7 = 0.24 kg/s

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A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel

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Answer:

The load reflection coefficient, $\Gamma =0.62\angle 82.875^{\circ} \Omega$

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Characteristic impedance of the co-axial cable, $Z_{c} = 75 \Omega$

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In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance $Z_{i}$ is given by:

$Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}$ (1)

Now, we calculate the value of $\beta$ :

$\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}$

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$\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53$

Now, Substituting the value in eqn (1):

$Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega$

Now, the load reflection coefficient is given by:

$\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}$

Thus

$\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega$

Similarly,

Reflection coefficient at input:

$\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}$

$\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega$

Now, the SWR is given by:

SWR, Standing Wave Ratio = $\frac{1 +|\Gamma|}{1 - |\Gamma|}$

SWR = $\frac{1 +|0.62|}{1 - |0.62|} = 4.26$

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