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USPshnik [31]
3 years ago
15

In which forms do MOST of the Sun's energy reach Earth's surface?

Engineering
1 answer:
natta225 [31]3 years ago
4 0

Answer:

I believe the answer is B, visible and infrared light.

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A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 l
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modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048

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Modulus is the ratio of tensile stress to tensile strain

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And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in

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The two windings of transformer is: a)- Conductively linked. b)- Not linked at all. c)- Inductively linked d)- Electrically link
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The two windings of transformer is c)- Inductively linked

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An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75
Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process            T_{3} = 2042.56 K

(b). The value of pressure at the end of heat addition process                    P_{3} = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle   E_{otto} = 0.4478

(d). The value of mean effective pressure of the cycle P_{m} = 1506.41 \frac{k pa}{kg}

Explanation:

Compression ratio r_{p} = 8

Initial pressure P_{1} = 95 k pa

Initial temperature T_{1} = 278 °c = 551 K

Final pressure P_{2} = 8 × P_{1} = 8 × 95 = 760 k pa

Final temperature T_{2} = T_{1} × r_{p} ^{\frac{\gamma - 1}{\gamma} }

Final temperature T_{2} = 551 × 8 ^{\frac{1.4 - 1}{1.4} }

Final temperature T_{2} = 998 K

Heat transferred at constant volume Q = 750 \frac{KJ}{kg}

(a). We know that Heat transferred at constant volume Q_{S} = m C_{v} ( T_{3} - T_{2}  )

⇒ 1 × 0.718 × ( T_{3} - 998 ) = 750

⇒ T_{3} = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ \frac{P_{3} }{P_{2} } = \frac{T_{3} }{T_{2} }

⇒ P_{3} = \frac{2042.56}{998} × 760

⇒ P_{3} = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle Q_{R} = m C_{v} ( T_{4} - T_{1}  )

For the compression and expansion process,

⇒ \frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }

⇒ \frac{2042.56}{998} = \frac{T_{4} }{551}

⇒ T_{4} = 1127.7 K

Heat rejected Q_{R} = 1 × 0.718 × ( 1127.7 - 551)

⇒ Q_{R} = 414.07 \frac{KJ}{kg}

Net heat interaction from the cycle Q_{net} = Q_{S} - Q_{R}

Put the values of Q_{S} & Q_{R}  we get,

⇒ Q_{net} = 750 - 414.07

⇒ Q_{net} = 335.93 \frac{KJ}{kg}

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ Q_{net} = W_{net}

⇒ W_{net} = 335.93 \frac{KJ}{kg}

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

E_{otto} = 1- \frac{T_{1} }{T_{2} }

Put the values of T_{1} & T_{2} in the above formula we get,

E_{otto} = 1- \frac{551 }{998 }

⇒ E_{otto} = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure P_{m} :-

We know that mean effective pressure of  the Otto cycle is  given by

P_{m} = \frac{W_{net} }{V_{s} } ---------- (1)

where V_{s} is the swept volume.

V_{s} = V_{1}  - V_{2} ---------- ( 2 )

From ideal gas equation P_{1} V_{1} = m × R × T_{1}

Put all the values in above formula we get,

⇒ 95 × V_{1} = 1 × 0.287 × 551

⇒ V_{1} = 0.6 m^{3}

From the same ideal gas equation

P_{2} V_{2} = m × R × T_{2}

⇒ 760 × V_{2} = 1 × 0.287 × 998

⇒ V_{2} = 0.377 m^{3}

Thus swept volume V_{s} = 0.6 - 0.377

⇒ V_{s} = 0.223 m^{3}

Thus from equation 1 the mean effective pressure

⇒ P_{m} = \frac{335.93}{0.223}

⇒ P_{m} = 1506.41 \frac{k pa}{kg}

This is the value of mean effective pressure of the cycle.

4 0
3 years ago
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