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notka56 [123]
2 years ago
15

Mercury at 25°c flows over a 3-m-long and 2-m-wide flat plate maintained at 75°c with a velocity of 0. 01 m/s. Determine the rat

e of heat transfer from the entire plate?
Physics
1 answer:
Colt1911 [192]2 years ago
8 0

The rate of heat transfer from the entire plate is 710.8 kW

The properties of mercury at the film temperature of (75+25)/2= 50°C

k = 8.83632 W/m °C

v = 1.056 × 10^-^7 m²/s

Pr = 0.0223

The local Nusselt number relation for liquid metals is given by

 N_u =\frac{h_sx}{k} = 0.565 9Re_x Pr)^\frac{1}{2}

The average heat transfer coefficient for the entire surface can be determined from

h = \frac{1}{L} \int\limits^1_0 {h_x} \, dx

Substituting the local Nusselt number relation into the above equation and performing the integration we obtain

N_u = 1.13 (Re_s Pr)^\frac{1}{2}

The Reynolds number is

Re_L = \frac{VL}{v} = \frac{(0.01)(3m)}{1.056 * 10^-^7 m^2/s} = 0.028 * 10^7

Using the relation for the Nusselt number, the average heat transfer coefficient and the heat transfer rate are determined to be

                       Q = 710.8 kW

Therefore, the rate of heat transfer from the entire plate is 710.8 kW

Learn more about heat transfer here:

brainly.com/question/16055406

#SPJ4

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Covalent and ionic bonds
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8 0
3 years ago
What is an analogy of two different roads or rivers to compare a series and parallel circuit?
faltersainse [42]

Answer:

In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.

5 0
2 years ago
Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Vedmedyk [2.9K]

Answer:

(a) V (A) =  0.7 m/s,

(b) V (A) =  0.7 m/s,

(c) V (B) =  0.7 m/s

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a)  law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V      (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V =  0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) =  0.7 m/s

b)  V(A) =  0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) =  0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s =  3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1)  ( dividing both side by 3.50 Kg)

For elastic collision  

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u  ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0
4 years ago
A space traveller leaves Earth for 10 years at .85c. According to an observer on Earth, how much time has passed?
eduard
First of all, you didn't tell us WHO measured the "10 years".

If it was the people on Earth, then 10 years passed according to them.

If it was 10 years on the space traveler's clock,  then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.

If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by

       √(1 - v²/c²) .

You said that  v/c  = 0.85 .

v²/c² = (0.85)² = 0.7225

1 - v²/c² =  1 - 0.7225 = 0.2775

√(1 - v²/c²)  =  √0.2775 = 0.5268

If one clock counts up 10 years, then the other one counts up

(10years) / 0.5268 =  <em>18.983 years </em>


I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.

8 0
4 years ago
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