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strojnjashka [21]

3 years ago

14

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

meter of the fin is 3 cm. The thermal conductivity of the fin is 150 W/m·K. The heat transfer coefficient is 123 W/m^2·K. Estimate the fin temperature in °C at a distance of 10 cm from the base.

1 answer:

Mrrafil [7]3 years ago

8 0

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

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Answer:

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3 years ago

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

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$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

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$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

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Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.

<h3>What is wasted energy?</h3>

Wasted energy is energy that is not useful when the transformation in the system occurs.

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The efficiency of the system = 50 %

Output work (OW) is calculated as:

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OW = 375 J

Wasted energy = Total energy - output work

= 750 - 375

= 375 J

Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.

Learn more about wasted energy here:

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Explanation:

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3 years ago

A thick steel slab ( 7800 kg/m3 , c 480 J/kg K, k 50 W/m K) is initially at 300 C and is cooled by water jets impinging on one o

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Answer:

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