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VladimirAG [237]
3 years ago
12

Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa, 518 oC, and mass flow rate of 5,322 kg/h and exits the system at 96

kPa and 421 m/s. Considering the nozzle having an inlet area of 37 cm2, calculate the inlet velocity in m/s
Physics
1 answer:
Orlov [11]3 years ago
3 0

Answer:

The velocity at the nozzle at inlet V_{1} = 3584 \frac{m}{sec}

Explanation:

Pressure at inlet P_{1} = 1 × 10^{6} Pa

Temperature at inlet T_{1} = 518 ° c = 791 K

Mass flow rate = \frac{5322}{60} \frac{kg}{sec} = 88.7

Gas constant for carbon die oxide is R = 189 \frac{J}{kg k}

Mass flow rate inside the nozzle is given by the formula = \frac{P_{1} }{R T_{1} } × A_{1} × V_{1}

⇒ P_{1} = = 1 × 10^{6} Pa

⇒ RT_{1} = 791 × 189 = 149499 \frac{J}{kg}

⇒ A_{1} = 0.0037 m^{2}

Put all the above values in above formula we get,

⇒ 88.7 = \frac{10^{6} }{149499} × 0.0037 × V_{1}

⇒ V_{1} = 3584 \frac{m}{sec}

This is the velocity at the nozzle at inlet.

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4 0
3 years ago
What happens when you decrease the thrust on your scooter? A. You stop B. Nothing happens C. You fall over D. You speed up Reset
mars1129 [50]

Answer:

D. You speed up

Explanation:

hope it helps

4 0
3 years ago
A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef
AveGali [126]
The thermal efficiency of an engine is
\eta= \frac{W}{Q}
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%
8 0
3 years ago
49. A block is pushed across a horizontal surface with a
nata0808 [166]

Answer:

(a) 37.5 kg

(b) 4

Explanation:

Force, F = 150 N

kinetic friction coefficient = 0.15

(a) acceleration, a = 2.53 m/s^2

According to the newton's second law

Net force = mass x acceleration

F - friction force = m a

150 - 0.15 x m g = m a

150 = m (2.53 + 0.15 x 9.8)

m = 37.5 kg

(b) As the block moves with the constant speed so the applied force becomes the friction force.

F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4

8 0
3 years ago
A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has
Pani-rosa [81]

Answer:

Part a)

EMF = 0.38 V

Part b)

\frac{dA}{dt} = 0.43 m^2/s

Explanation:

Part a)

Initial value of magnetic flux is given as

\phi_1 = BAcos\theta

\phi_1 = (2.1)(0.35 \times 0.55) cos65

so we have

\phi_1 = 0.17 Wb

Final flux through the loop is given as

\phi_2 = 0

now EMF is given as

EMF = \frac{\phi_1 - \phi_2}{\Delta t}

EMF = \frac{0.17 - 0}{0.45}

EMF = 0.38 V

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

E = Bcos65 \times \frac{dA}{dt}

0.38 = 2.1 cos65(\frac{dA}{dt})

\frac{dA}{dt} = 0.43 m^2/s

5 0
3 years ago
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