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Stels [109]
3 years ago
11

Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

Engineering
1 answer:
IceJOKER [234]3 years ago
6 0

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

q=-\kappa \bigtriangledown T

Where,

q is the heat flux

\kappa is the thermal conductivity

\bigtriangledown T is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

C=\frac{E}{\Delta T}

Where,

C is the Heat capacity

E is the energy absorbed/released

\Delta T is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

\alpha=\frac{\kappa}{C_p \times \rho}

Where,

\alpha is thermal diffusivity

\kappa is the thermal conductivity

C_p is specific heat capacity at constant pressure

\rho is density

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1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is expo
Alexeev081 [22]

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

                     q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]      

Where,

          σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

          Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

          T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

                    q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]    

                    q' = (1.4251*10^-8)* [ 24645536240 ]    

                    q' = 351.22 W / m   ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

                  Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

                  Thermal conductivity ( k ) = 0.025596

                  Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

                 Re = U∞*d / v

                      = ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

                      = 52941.56574   ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

           Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3}  }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4}  }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5}    \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3}  }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4}  }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5}  \\\\

           Nu_D = 0.3 + \frac{127.59828 }{ 1.13824  }*1.27251  = 142.95013

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

            h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K

- The heat loss per unit length due to convection heat transfer is given by:

           q'_convec = h*( π*d )* [ Ts - T∞ ]

           q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

           q'_convec = 11.49493* 130

           q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

           q'_total = q_a + q_convec

           q'_total = 351.22 + 1494.34009

           q'_total = 1845.56 W / m  ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

             q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}  

             q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

             

Where,

             T_o = T∞ = 20°C

            T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

             r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

             r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

            q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

            q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m

Therefore,

             q'_loss = 351.22 - 97.10

            q'_loss = 254.12 W / m   .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.

6 0
3 years ago
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