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2 years ago

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A 6cm OD, 2cm thick copper hollow sphere [k=386W/m.C] is uniformly heated at the inner surface at a rate of 150W/m2. The outside

surface is cooled with air at 20C with a heat transfer coefficient of 10W/m2.oC. Calculate the temperature of the outer surface

1 answer:

Misha Larkins [42]2 years ago

6 0

Answer:

The outside surface temperature=21.66 C.

Explanation:

Given that

Outer diameter=6 cm

Inner diameter=2 cm

$Heat\ flux\ at\ inner\ surface\q=150\frac{W}{m^2}$

Outside temperature=20 C

$Outside\ heat\ transfer\ coefficient=10\frac{W}{m^2-K}$

To find the outside temperature

Heat out from sphere=Heat absorb by surrounding

$q\times A_i=hA_o\Delta T$

$q\times d_i^2=hd_o^2(T_o-20)$

Now by putting the values

$150\times 2_i^2=10\times 6_o^2(T_o-20)$

$So\ T_o=21.66C$

So the outside surface temperature=21.66 C.

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Explanation:

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2 years ago

three balls each have a mass m if a has a speed v just before a direct collision with B determine the speed of C after collision

ratelena [41]

Answer:

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Explanation:

Conservation momentum, when ball A strikes Ball B

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MV + 0= MVg2

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2 years ago

Read 2 more answers

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor ack

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Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

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2 years ago

The force exerted on a bridge pier in a river is to be tested in a 1:10 scale model using water as the working fluid. In the pro

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Answer:

Force on the prototype is 5000 N

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As per the question:

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Now,

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where

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$\frac{\frac{F'}{A'}}{\frac{F}{A}} = \frac{1}{10}$

where

F' = Force on model

F = Force on prototype

A' = Area of model

A = Area of prototype

Now:

$\frac{F'}{F}.\frac{A}{A'} = \frac{1}{10}$

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2 years ago

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which

Westkost [7]

Answer:

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$COP_{H, rev} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

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$\dot W = \frac{300\,kW}{17.749}$

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By the First Law of Thermodynamics, heat pump works at steady state and likewise, the heat released from cold reservoir is now computed:

$-\dot Q_{H} + \dot W + \dot Q_{L} = 0$

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = 300\,kW - 16.902\,kW$

$\dot Q_{L} = 283.098\,kW$

According to the Second Law of Thermodynamics, a reversible heat pump should have an entropy generation rate equal to zero. The Second-Law model for the system is:

$\dot S_{in} - \dot S_{out} - \dot S_{gen} = 0$

$\dot S_{gen} = \dot S_{in} - \dot S_{out}$

$\dot S_{gen} = \frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}}$

$\dot S_{gen} = \frac{283.098\,kW}{280.15\,K} - \frac{300\,kW}{297.15\,K}$

$\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$

Albeit entropy generation rate is positive, it is also really insignificant and therefore means that such heat pump satisfies the Second Law of Thermodynamics. Furthermore, $\dot S_{in} = \dot S_{out}$ .

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2 years ago