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Phantasy [73]
3 years ago
9

A plane wall of thickness 2L = 60 mm and thermal conductivity k= 5W/m.K experiences uniform volumetric heat generation at a rate

qdot , while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T[infinity] = 30 oC . Under steady-state conditions, the temperature distribution in the wall is of the form T(x) = a + bx+ cx2 where a = 860C, b=-2000C /m, c=-2× 104 0C /m2, and x is in meters. The origin of the x-coordinate is at the midplane of the wall.
a) Sketch the temperature distribution and identify significant physical features.

b) What is the volumetric rate of heat generation in the wall?

c) Determine the surface heat fluxes at x=-L, +L, and How are these fluxes related to the heat generation rate?

d) What are the convection coefficients for the surfaces at x=-L and x=+L?
Engineering
1 answer:
aniked [119]3 years ago
6 0

Answer:

Explanation:

A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq  

˙

q

q

˙

​  

, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT  

∞

​  

=20  

∘

C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx  

2

 where a=82.0∘C,b=−210∘C/m,c=−2×104C/m2a=82.0  

∘

C,b=−210  

∘

C/m,c=−2×10  

4

C/m  

2

, and x is in meters. The origin of the x-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation q in the wall? (c) Determine the surface heat fluxes, q

′′

x

(−L)q  

x

′′

​  

(−L) and q

′′

x

(+L)q  

x

′′

​  

(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q

′′

x

(x)q  

x

′′

​  

(x). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated (q=0), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with q=0? How much energy must be removed by the fluid per unit area of the wall (J/m2)(J/m  

2

) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m  

3

 and 800J/kg⋅K800J/kg⋅K, respectively.

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Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o
jekas [21]

Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

 H_m=(\frac{200*10^3}{1000*9.8}-10)

 H_m=10.39m

Generally the equation for Power is mathematically given by

 P=\rho gQH_m

Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

 A=\pi r^2\\A=3.142 (0.05)^2

 A=7.85*10^{-3}

Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

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5 0
3 years ago
Describe a project in which you would use a pleater, ruffling foot, or gathering foot. Explain each of these tools and choose th
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All of the tools can be used in the creation of a dress, for example, a pleater can be used in the top section of the dress to give it a nice texture and make it different from the skirt. In the same way, others such as the ruffling foot or the gathering foot can be used in the sleeves of the dress.

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3 years ago
I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0
3 years ago
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch
Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 \frac{ft^{3}}{sec}

         D_{1}= 6 inch=0.5 ft

        D_{2}=2 inch=0.1667 ft

As we know that Q=AV

A_{1}\times V_{1}=A_{2}\times V_{2}

So V_{2}=\frac{Q}{A_2}

     V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}

     V_{2=0.687 ft/sec

We know that Head loss due to sudden contraction

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If nothing is given then take K=0.5

So head lossh_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}

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So head loss=0.00366 ft

4 0
3 years ago
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