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3 years ago

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Air at 25°C and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligible

, determine the exit temperature of air. Solve using appropriate software.

1 answer:

Vsevolod [243]3 years ago

4 0

Answer:

$T_{2} = 25^{\circ}C$

Solution:

As per the question:

Temperature, $T_{1} = 25^{\circ}C$

Pressure, $P_{1} = 5 atm[\tex]Now,We know that the process of throttling occurs in adiabatic conditions where no heat is transferred in between the system and the surrounding, i.e., Q = 0Thus no work is done in this process, i.e., W = 0Enthalpy also remains same from one state to the other state, i.e., [tex]\Delta h = 0$

Therefore, from the eqn:

$Q - W = \Delta h + \Delta KE$

We can write:

$\Delta h = h_{1} - h_{2} = 0$

$\Delta h = C_{p}\Delta T$

$C_{p}(T_{2} - T_{1}) = 0$

Thus

$T_{1} = T_{2} = 25^{\circ}C$

where

$T_{2}$ = Exit temperature

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An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

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Which is not expecting acceleration?

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Non-contact forces act between two bodies even when they are at a distance apart. For example: gravity, electric force, magnetic force etc.

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Inertia is a property of matter by virtue of which it tends to remain in its state of motion or rest. It does depend on mass of the object, more the mass, more is inertia. For example, cycle can be easily moved but we need real push hard for a car to move.

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3 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

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Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

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After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

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4 years ago

An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat

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Answer:

The maximum speed is 21.39 m/s.

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3 years ago

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Answer: Option (A) and (D) are the correct statements.

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