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OleMash [197]
3 years ago
12

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

ty of 600 m/s. Determine the final temperature if the steam is superheated in the final state, and the quality if it is saturate
Physics
1 answer:
Rudiy273 years ago
6 0

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

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D

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As the mass of an object increases so does its gravitational force. <br> a. True <br> b. False
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A ray of light has a wavelength of
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Answer:

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Explanation:

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We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

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So, the wavelength in vacuum is equal to 428.8 nm.

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Cell phones do not work everywhere. There are places where your cell phone cannot get a signal
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A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s
3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer.  The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\  340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

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