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3 years ago

15

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

ter and is situated in a surrounding ambient temperature of 15°C where the natural convection heat transfer coefficient is 12 W/m^2·K.
If the efficiency of the electrical heater to transfer heat to the plate is 80%, determine the electric power that the heater needs to keep the surface temperature of the plate at 192°C.

1 answer:

MakcuM [25]3 years ago

6 0

Answer:

Heater power = 425 watts

Explanation:

Detailed explanation and calculation is shown in the image below

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3 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

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

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

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4 years ago

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Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to 250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

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3 years ago