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3 years ago

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A spherical seed of 1 cm diameter is buried at a depth of 1 cm inside soil (thermal conductivity of 1 Wm-1K-1) in a sufficiently

large planter. There is a 1 cm thick layer of mulch (thermal conductivity of 2 Wm-1K-1) on top of the soil. The planter has top surface dimensions of 10 cm by 10 cm and is exposed to 200 Wm-2 of heat. You find the top surface temperature of the mulch (Ts,1) to be uniform at 50˚C. What is the surface temperature of the seed (Ts,2) in ˚C

1 answer:

postnew [5]3 years ago

8 0

Answer:

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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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ivanzaharov [21]

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Read 2 more answers

Who can help me with electric systems for cars?

hoa [83]

Answer: i can see if i can what is the problem

Explanation:

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3 years ago

The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th

Alja [10]

Answer:

Explanation:

$y'''+y=0---(i)$

General solution

$y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)$

$y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1$

in equation (iii)

$c_1=c_2=-1$

Therefore,

$\large\boxed{y=-1-\cos x-2\sin x}$

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Between chain, belt and gear drive, which causes more friction?

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Answer:

All of them cause friction

Explanation:

google

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