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allsm [11]
2 years ago
11

Air at 40°C flows over a long, 25-mm-diameter cylinder with an embedded electrical heater. In a series of tests, measurements we

re made of the power per unit length, P’, required to maintain the cylinder surface temperature at 300°C for different free stream velocities u of the air. The results are as follows:
Air velocity, u (m/s) 1 2 4 8 12
Power, P’ (W/m) 450 658 983 1507 1963

(a) Determine the convection coefficient for each velocity, and display your results graphically.
(b) Assuming the dependence of the convection coefficient on the velocity to be of the form h = Cu n , determine the parameters C and n from the results of part (a)

Engineering
1 answer:
Murljashka [212]2 years ago
7 0

Answer:

a) See attachment

b) C = 21.626 W / m^2 .K , n = 0.593

Explanation:

a)

dT = 300 - 40 = 160 K

A = pi*D*l where D = 0.025 m

Power = P' * l = h * dT * A

P' l = h * pi * D * l* dT

Hence,

h = P' / (pi*D*dT)

We will use the above equation to compute for respective values of P'

Note: The results are tabulated and attached

b)

Assuming the dependence of the convection coefficient on the velocity to be of the form h=CV^n, determine the parameters C and n from the results of part (a).

Taking logarithm on both sides:

Ln (h) = Ln(C) + n*Ln(V)

n = (Ln(h2) - Ln(h1)) / (Ln(V2) - Ln(V1))

n = (Ln (32.223/22.037)) / Ln(2))

n = 0.593

Using regression we can compare:

C = 21.626 W/m^2 . K

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Identify an object in your house that contains a physical system and list three questions you could use to define the system
jonny [76]

Answer:

ALL CAREFULLY ANSWERED CORRECTLY

Explanation:

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✓ How can the environment affect the edibility of the bread

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✓ What process is involved in these constituents mixing to form the loaf.

2) The law of thermodynamics makes us to understand that when heat/energy passes through a system, the systems internal energy changes with respect to the conservation of energy law. That is energy lost = energy gained. Typically, ice would melt in a cup of hot tea because of the thermal energy in the molecules of the hot tea. When you heat a material, you are adding thermal kinetic energy to its molecules and usually raising its temperature. The temperature of the ice raises due to the kinetic energy added to it and it melts to water.

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4 0
2 years ago
Tech A says that as moisture levels increase in brake fluid, the boiling point of the brake fluid decreases. Tech B says that fi
dezoksy [38]

Answer: Both Technician A and B are correct.

Explanation:

Technicians A and B are both right about their diagnosis. The Society of Automotive Engineers performed extensive research on vehicle brake fluids and found that there is typically a 2% moisture content in the brake fluid after a year of operating a vehicle. And as the moisture content of the brake fluid rises, the boiling point of the brake fluid decreases as well.

4 0
2 years ago
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses
yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature T_1 = 25° C

T{\infty} =600^0C

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}

Bi = \dfrac{2.5}{231}

Bi = 0.0108

The time constant value \tau_t is :

\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}

\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}

\tau_t = \dfrac{2702* 0.025*1033}{100}

\tau_t = 697.79

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]

where;

Q = -\Delta E _{st} which correlates with the change in the internal energy of the solid.

So;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}

The maximum value for the change in the internal energy of the solid  is :

(pVc)\theta_1 = -\Delta E _{st}max

By equating the two previous equation together ; we have:

\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{  (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

0.75=  [1-e^{\dfrac {-t}{ \tau_1}}]}

So;

0.75=  [1-e^{\dfrac {-t}{ 697.79}}]}

1-0.75=  [e^{\dfrac {-t}{ 697.79}}]}

0.25 =  e^{\dfrac {-t}{ 697.79}}

In(0.25) =  {\dfrac {-t}{ 697.79}}

-1.386294361= \dfrac{-t}{697.79}

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}

\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}

\dfrac{T - 600}{25-600}= 0.25

\dfrac{T - 600}{-575}= 0.25

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0
2 years ago
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