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bekas [8.4K]
3 years ago
8

A square aluminum plate 5 mm thick and 200 mm on a side is heated while vertically suspended in quiescent air at 40C. (A) Determ

ine the average heat transfer coefficient for the plate when its temperature is 15C by two methods: using results from the similarity solution to the boundary layer equations, and using results from an empirical correlation. (B) Calculate the convective heat transfer loss to the air from both surfaces (C) Calculate the net radiation heat loss to a very large environment at 35C. Assume it is a two surface enclosure and all surfaces are gray and diffuse. e=1.0. Do you think you can neglect the radiation heat loss?
Engineering
1 answer:
Kamila [148]3 years ago
4 0

brainly.com/question/14000001

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The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin
vfiekz [6]

Answer:

Ф = 0.02838 ft

F  = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝  × ΔT  × L  ... where Ф= Gap Delta in ft

                                          ΔT= Temperature rise in F

                                               = 90- (-20)

                                               =  110 F

Ф =  6.45 ×10^(-6) × 110 × 40

Ф =  28,380 × 10^(-6) ft

Ф = 0.02838 ft     .... total gape required for expansion of steel rails

Stress induced in rails is given by,

   σ     =  ∝  × ΔT  × E

          =  6.45 ×10^(-6)   × 110  × 200

  σ      =  1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider  ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

8 0
3 years ago
Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M
Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

                            C = (20*10^6)*14.6096

                           C = 292 Mbps

3 0
2 years ago
Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i
stealth61 [152]

Answer:

\dot m = 2.74 kg/s

Explanation:

given data:

pressure 1 MPa

diameter of pipe  =  30 cm

average velocity = 10 m/s

area of pipe= \frac[\pi}{4}d^2

                 = \frac{\pi}{4} 0.3^2

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

\dot m = \rho A v

for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

\dot m = 4.068 * 0.070* 10

\dot m = 2.74 kg/s

7 0
3 years ago
Turn on your____
storchak [24]

Answer:

b

Explanation:

5 0
2 years ago
Read 2 more answers
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