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2 years ago

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A spherical metal ball of radius r_0 is heated in an oven to a temperature of T_1 throughout and is then taken out of the oven a

nd allowed to cool in ambient air at T[infinity] , by convection and radiation. The emissivity of the outer surface of the cylinder is e, and the temperature of the surrounding surfaces is T_{surr}. The average convection heat transfer coefficient is estimated to be h. Assuming variable thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem.

1 answer:

telo118 [61]2 years ago

7 0

Answer:

yes it is u are right

Explanation:

do be secure

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Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago

A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?

stiks02 [169]

Answer:

The tax on this property is $3750$ dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is $\frac{2.5}{100} = 0.025$ dollars

Tax on property of value $150,000 is

$150,000 * 0.025 = 3750$ dollars

The tax on this property is $3750$ dollars

7 0

2 years ago

The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)

xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

$\sigma = n|e|\mu_e+p|e|\mu_h$ (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

$\mu_e=$ mobility of electron

$\mu_h=$ mobility of holes

$\sigma =$ electrical conductivity

Making the substitution in (S)

Mobility of electron

$8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)$

$0.639=\mu_e+\mu_h$

Mobility of hole in (S)

$2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))$

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$

Then, solving the equation:

$0.639=\mu_e+\mu_h$ (1)

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$ (2)

We have,

Mobility of electron $\mu_e = 0.5m^2/V.s$

Mobility of hole is $\mu_h = 0.14m^2/V.s$

6 0

3 years ago

What is the value of the work interaction in this process?

Cloud [144]

Answer:

The answer is " $-121\ \frac{KJ}{Kg}$ ".

Explanation:

Please find the correct question in the attachment file.

using formula:

$\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \ or \ \ \frac{RT_1 -RT_2}{n-1}\\\\$

$W =\frac{R(T_1 -T_2)}{n-1}\\\\$

$=\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\$

7 0

3 years ago

Answer ppeeeeeaaaalll

Bad White [126]

Answer:

what

Explanation:

is this an exam or an test or what is it

3 0

2 years ago