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2 years ago

12

The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar rad

iation is incident on the spacecraft at a rate of 1200 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed. The surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed is K.

1 answer:

kondaur [170]2 years ago

8 0

Answer:

T surface = 3.9°C

Explanation:

given data

emissivity 0.6

absorptivity = 0.2

solar radiation is incident rate = 1200 W/m²

solution

we get here surface temperature by equality of emitted and absorbed heat rate that is

Q (absorbed) = Q (heat ) .................1

α Qinc = $\epsilon * \sigma *A*T^4(surface)$

T surface = $\sqrt[4]{\frac{\alpha Qinc}{\epsilon *\sigma * A} }$ ..........................2

put here value and we get

T surface = $\sqrt[4]{\frac{0.2*1000}{0.6*5.67**10^{-8}} }$

T surface = 276.9 K

T surface = 3.9°C

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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

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2 years ago

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp

cricket20 [7]

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

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2 years ago

A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can

Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a) ln(T2/T1)=μβ β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb T1=80 lb μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

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2 years ago

An insulated, vertical piston-cylinder device initially contains 10kg of water, 6kg of which is in the vapor phase. The mass of

Alexeev081 [22]

Answer:

a)120C

b)29kg

Explanation:

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To solve this exercise follow the steps below

1. we will call 1 the initial state, 2 the steam that enters and 3 the final state

2. We find the quality of the initial state, dividing the mass of steam by the total mass.

$q1=\frac{6kg}{10kg} =0.6$

3 Find the internal energy in the three states using thermodynamic tables

note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg

u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg

u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg

4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables

T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C

5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final

m1u1+m2u2=(m1+m2)u3

where

m1=inital mass=10kg

m2=the mass of the steam that has entered.

solve for m2

(m1)(u1-u3)=(m2)(u3)-(m2)(u2)

$m2=m1\frac{u1-u3}{u3-u2} =10\frac{1719-2529}{2529-2808} =29kg$

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2 years ago

Imagine you compare the effectiveness of four different types of stimulant to keep you awake while revising statistics using a o

Arlecino [84]

Answer:

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