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koban [17]
3 years ago
9

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25 °C,

into the flow, which is at 75 °C, and measuring its surface temperature at some time during the transient heating process. If the sphere has a diameter of 0.1 m, a thermal conductivity of 22 W/(m·K), and a thermal diffusivity of 0.40 ×10-5 m2/s, at what time will a surface temperature of 60 ºC be recorded if the convection coefficient is 300 W/(m2·K)?
Engineering
1 answer:
sashaice [31]3 years ago
7 0

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity = \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}

theraml conductivity = k = 22 W/m.K

h = 300 W/ m^2.K

T_i = 25 degree C = 298 k

T_o = 60 degree C = 333 k

T_{\infty}= 75 degree C =  348 L

diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226

\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}

\tau = 0.036 t

\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}

at Bi = 0.226

Ai = 0.982

\lambda = 0.876

\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}

0.3 = 0.982 e^{-0.2t}

0.305 = e^{-0.2t}

-1.187 = - 0.02t

t = 59.37 s

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3 years ago
Define waves as it applies to electromagnetic fields
julsineya [31]

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5 0
1 year ago
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

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The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
3 years ago
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com
Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For  vapor-compression refrigeration cycle

P1 = P4  ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

<u>From pressure table of R 134a refrigerant</u>

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

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also h4 = 95.47 kJ/kg

To determine the coefficient of performance  

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

3 0
3 years ago
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