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2 years ago

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A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055

(1 + 2.8 × 10–3T) W/mK where T is in °C. The outside surface temperature is 40°C. Determine the heat flow/m length and the temperature at the mid radius

1 answer:

Alekssandra [29.7K]2 years ago

4 0

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

$K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}$

So the mean conductivity

$K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )$

$T_m=\dfrac{160+273+40+273}{2}$

$T_m=373 K$

$K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )$

$K_m=0.112 \frac{W}{mK}$

So heat conduction through cylinder

$Q=kA\dfrac{\Delta T}{L}$

$Q=0.112\times \pi \times 0.15^2\times 120$

Q=0.95 W/m

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g The function below takes a single string parameter: sentence. Complete the function to return everything but the middle 10 cha

ale4655 [162]

Answer:

def processString(sentence):
middle = len(sentence) // 2
output = sentence[0: middle - 5] + sentence[middle+5:]
return output
print(processString("I have a dream"))

Explanation:

Create a function processString that take sentence as input paratemer (Line 1).

Next create a variable middle to hold the value of middle index of sentence string (Line 2)

Build the output string by slicing the input sentence from first character to character middle - 5th and from middle + 5th till the end of the string (Line 3).

Test the function using a sample sentence and we shall get the output "I am"

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2 years ago

A long, horizontal, pressurized hot water pipe of 15cm diameter passes through a room where the air temperature is 24degree C. T

solmaris [256]

Answer:

Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m

Explanation:

Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.

We know that heat transfer by convection is given by

$\dot{Q}=hA(T-T_{\infty })$

where,

h = heat transfer coefficient = 10.45 $W/m^{2}K$ (free convection in air)

A = Surface Area of the pipe

Applying the given values in the above formula we get

$\dot{Q}=10.45\times \pi DL\times (130+273-(24+273))\\\\\frac{\dot{Q}}{L}=10.45\times 0.15\times \pi \times (130-24)\\\\\frac{\dot{Q}}{L}=521.99W/m$

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2 years ago

Compute the mass fractions of α ferrite and cementite in pearlite. Assume T=726⁢ C∘.

kotykmax [81]

Answer:

The mass composition of ∝ ferrite is 88.94%.

The mass composition of cementite is 11.05%.

Explanation:

Given that

T=726 °C

We have to find the mass fractions of ferrite nad cementite in pearlite.

Lets take data from ideal Iron -carbon diagram at 726 °C

Composition of ∝ ferrite=0.022 5

Composition of pearlite =0.76 %

Composition of cementite =7.6%

We know that if we want to find the mass fraction the we use Lever rule .So now by using lever rule

$The\ mass\ composition\ of\alpha \ ferrite =\dfrac{6.7-0.76}{6.7-0.022}$

So the mass composition of ∝ ferrite is 88.94%.

$The\ mass\ composition\ of\ cementite =\dfrac{0.76-0.022}{6.7-0.022}$

So the mass composition of cementite is 11.05%.

4 0

2 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

3 years ago

An unconstrained 10mm thick plate of steel 100mm on a side with a 25mm diameter hole in the center is heated from 20 degrees C t

prisoha [69]

Answer:

The correct answer is "25.03 mm".

Explanation:

Given:

Thickness of plate,

= 10 mm

On a side,

= 100 mm

Diameter hole,

= 25 mm

Coefficient of thermal expansion,

CTE = $12\times 10^{-6} /^{\circ} C$

Now,

⇒ $D_i\times (12\times 10^{-6}) \Delta \theta = \Delta D$

= $25\times 12\times 10^{-6} \Delta \theta$

= $3\times 10^{-4} \Delta \theta$

= $3\times 10^{-2}$

hence,

The final diameter of hole will be:

$D_f=25.03 \ mm$

8 0

2 years ago