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Slav-nsk [51]
3 years ago
14

A large truck drives down the highway at 10 m/s hauling a rectangular trailer that is 6 m long, 2 m wide, and 2 m tall. The trai

ler contains frozen food and is therefore temperature-controlled. Its external surface can be approximated to be a consistent 15°C, while the outside air is at 20°C. Assume the heat transfer on the front, back, and bottom of the trailer is negligible.
a) How much cooling power must be provided to maintain the temperature controlled trailer? (i.e. What is the total heat transfer rate from the air to the trailer)?b) What is the minimum local heat transfer coefficient on the surface of the trailer? Where does the minimum occur?c) What percentage of the total heat transfer is occuring over a laminar boundary layer?d) If the cooling system can only provide 5 KW of cooling, what is the fastest speed that this truck can drive while still adequately maintaining the temperature within the trailer?

Engineering
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

3w/m²k

Explanation:

Base on the scenario been described in the question, the solution to the given problem solve in the file attached below

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What were some challenges engineers faced in designing aqueducts.
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What is manufacturing machining ?
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Read 2 more answers
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold
PIT_PIT [208]

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle (\eta_{r}), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\% (1)

Where:

T_{C} - Temperature of the cold reservoir, in Kelvin.

T_{H} - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: (T_{C} = 400\,K, T_{H} = 1200\,K)

\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%

\eta_{r} = 66.667\,\%

In exchange, real efficiency for a power cycle (\eta), no unit, is defined by this expression:

\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\% (2)

Where:

Q_{C} - Heat released to cold reservoir, in kilojoules.

Q_{H} - Heat gained from hot reservoir, in kilojoules.

W_{C} - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for \eta < \eta_{r}, reversibily for \eta = \eta_{r} and it is impossible for \eta > \eta_{r}.

Now we proceed to solve for each case:

a) Q_{H} = 900\,kJ, W_{C} = 450\,kJ

\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 50\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

b) Q_{H} = 900\,kJ, Q_{C} = 300\,kJ

\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 66.667\,\%

Since \eta = \eta_{r}, the power cycle operates reversibly.

c) W_{C} = 600\,kJ, Q_{C} = 400\,kJ

\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%

\eta = 60\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

d) Since \eta >  \eta_{r}, the power cycle is impossible.

8 0
3 years ago
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<u>Explanation:</u>

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The Universal Design process involves building products that can be used by a wide range of users at ease. For example, you may ask yourself: Is my product/service easily accesible to those with disabilities?

Other processes include;

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  • Follow the existing standards of product design
  • Evaluate and review your universal design methods
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