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umka2103 [35]
3 years ago
9

2. In the above figure, what type of cylinder arrangement is shown in the figure above?

Engineering
2 answers:
sergeinik [125]3 years ago
7 0
The answer is c :) ..................................
ivann1987 [24]3 years ago
5 0
Answer: C. Horizontal

Explanation:
Remember engine layouts are usually characterized on their cylinder layout, valves and camshafts.

Types of engines:
•In-line engines have it’s cylinders arranged in a row (ex. In-line 4)

•V-Type engines have it’s cylinder banks are arranged at an angle to each other, so that the banks form a "V" shape when viewed from the front of the engine.

•Horizontal/flat engines have it’s cylinders located on either side of a central crankshaft. (ex. Boxer engine)

•L-Type engines, I’m going to be real I’ve never heard of a L-type layout and I don’t think it’s real.



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

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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.

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

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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}

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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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