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sladkih [1.3K]
2 years ago
8

An electric resistance heater is embedded in a long cylinder of diameter 30 mm. when water with a temperature of 25°c and veloc

ity 1 m/s flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of 80°c is 38 kw/m. when air, also at 25°c, but with a velocity of 10 m/s is flowing, the power per unit length required to maintain the same surface temperature is 400 w/m. assume unit length. calculate and compare the convection coefficients for the flows of water and air. according to your calculations, which is a better convective heat transfer media, air or water?
Physics
1 answer:
victus00 [196]2 years ago
3 0
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows.
 FIND: Convection coefficients for the water and air flow convection processes, hw and ha, respectively.
 ASSUMPTIONS: Flow is cross-wise over cylinder which is very long in the direction normal to flow.
 The convection heat rate from the cylinder per unit length of the cylinder has the form
 q' = h*(pi*D)*(Ts-Tinf)
 and solving for the heat transfer convection coefficient, find
 Water
 hw = q'/((pi*D)*(Ts-Tinf))
 hw = (38*10^3 W/m) / ((pi*(0.030m))*(80-25)C)= 7330.77314  W/m^2K
 Air
 ha = (400W/m) / ((pi*(0.030m))*(80-25)C)=<span> 77.166033 </span> W/m^2K
 COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 95 × ha.
 These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases
 Watter is a better convective heat transfer media than air 

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2 years ago
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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0
2 years ago
A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
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Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how
storchak [24]

Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.69}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}

b) Expression for rate law for first order kinetics is given by:

A=A_0e^{-kt}

where,

k = rate constant  

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

A_0 = initial activity  = 600, 000

A=600000\times e^{-0.346\times 6}

A=75258

Thus there will be 75258 nuclei left at 6 pm.

7 0
2 years ago
A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

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C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

4 0
2 years ago
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