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3 years ago

6

Exceeding critical mach may result in the onset of compressibility effects such as:______.

1 answer:

klio [65]3 years ago

4 0

Answer:

Sound barrier.

Explanation:

Sound barrier is a sudden increase in drag and other effects when an aircraft travels faster than the speed of sound. Other undesirable effects are experienced in the transonic stage, such as relative air movement creating disruptive shock waves and turbulence. One of the adverse effect of this sound barrier in early plane designs was that at this speed, the weight of the engine required to power the aircraft would be too large for the aircraft to carry. Modern planes have designs that now combat most of these undesirable effects of the sound barrier.

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A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the a

prisoha [69]

Answer:

W = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)

we know that maximum possible efficiency is given as

$\eta = 1- \frac{T_L}{T_H}$

$\eta = 1 - \frac{ 293}{368}$

$\eta = 0.2038$

rate of heat transfer is given as

$Q_H = \dot m C_p \Delta T$

$Q_H = 0.2 * 4.18 8(95 - 20)$

$Q_H = 62.7 kW$

Maximuim power is given as

$W = \eta Q_H$

W = 0.2038 * 62.7

W = 12.8 KW

3 0

2 years ago

A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo

Natalija [7]

Answer:

For $20^{\circ}$ - 5.556 lb/s

For $15^{\circ}$ - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = $20^{\circ}$

Temperature rise, T' = $15^{\circ}$

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, $C_{p} = 0.24 btu/lb ^{\circ}F$

Now, for a rise of $20^{\circ}$ in temeprature:

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s$

Now, for $15^{\circ}$ :

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s$

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Answer:

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Explanation:

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Answer:

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Explanation:

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