1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lianna [129]
3 years ago
14

Cold feedwater enters a 200-kPa open feedwater heater of a regenerative Rankine cycle at 70°C with a flow rate of 20 kg/s. Bleed

steam is available from the turbine at 200 kPa and 160°C. At what rate must the bleed steam be supplied to the open feedwater heater so that the feedwater leaves this unit as a saturated liquid?
The mass flow rate of bleed steam to be supplied to the open feedwater heater so that the feedwater leaves this unit as a saturated liquid is ___________ kg/s.

Engineering
1 answer:
Anni [7]3 years ago
8 0

Answer:

The mass flow rate of bleed steam to be supplied is; m'_s = 1.85 Kg/s

Explanation:

From the first table i attached, at temperature of 70°C, the enthalpy of the cold water feed is;

h_c = 293 KJ/Kg

Also from the same table, at a temperature of 160°C and by interpolation, the enthalpy of the steam is;

h_s = 2790 KJ/Kg

Now, the exiting water is leaving the water heater at a combined rate of m'_c + m'_s and we know that it is a saturated liquid at Pressure of 200 KPa.

Where m'_c is mass flow rate of cold water feed while m'_s is mass flow rate of bleed steam.

Thus, from the second table attached and at 200KPa,the enthalpy of the saturated liquid is;

h_e = 504.71 KJ/kg

Now, from energy balance equation, we have;

E'_e = E'_c + E'_s

(m'_c + m'_s)h_e = (m'_c)h_c + (m'_s)h_s

Plugging in the relevant values, we have;

(20 + m'_s)504.71 = (20)293 + (m'_s)2790

10094.2 + 504.71m'_s = 5860 + 2790m'_s

2790m'_s - 504.71m'_s = 10094.2 - 5860

2285.29m'_s = 4234.2

m'_s = 4234.2/2285.29

m'_s = 1.85 Kg/s

You might be interested in
A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
4 years ago
A body of weight 300N is lying rough
kumpel [21]

Answer:

Horizontal force = 89.2 N

Explanation:

The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta

Substituting the given values, we get -

Frictional Force = 0.3*300 * cos 25 = 89.2 N

Horizontal force = 89.2 N

6 0
3 years ago
Que os vizinhos diziam sempre aq lenhador? Responda com elementos do texto.​
Alecsey [184]

Answer:

shsisdnd ajwdhdjeo sisksdne suspended wowodndjd

4 0
3 years ago
Animation can occur before an action
Slav-nsk [51]

Answer:Animation can occur before an action. Prepares the audience for the action. ... A pose or action should clearly communicate to the audience the attitude, mood, reaction or idea of the character. The effective use of long, medium or close up shots, as well as camera angles helps tell the story.

Explanation:May i have brainlist plz only if u wanna give me brainlist though have an nice day and stay safe.

6 0
3 years ago
You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in
shepuryov [24]

Answer:

In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet

Explanation:

The size of the film paper = 3 in. × 5 in.

The focal length of the camera = 5.5 in.

The height and width of the guinea pig = 4 ft.

The height of the aperture above the ground = 2 ft.

Therefore, we have;

Magnification = Height of image/(Height of object)

Withe the 3 in. wide film, we have;

Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625

Magnification = Length of camera/(Distance of object from pin hole)  

∴ Length of camera/(Distance of object from pin hole) = 0.0625

Length of camera = Focal length of the camera = 5.5 in.

Therefore;

5.5 in./(Distance of object from pin hole) = 0.0625

Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft

Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.

3 0
4 years ago
Other questions:
  • A power hacksaw used to cut metal, Link 5 pivots at O5 and its weight forces the saw-blade against the workpiece while the linka
    11·1 answer
  • An_______<br> employee is always prompt and on time.
    11·1 answer
  • the oscillation of rod oa about o is defined by the relation θ= (3/pi)*(sin*pi*t), where theta and t are expressed in radians an
    8·1 answer
  • Put these expressions in a small program that will demonstrate whether they are true or false. Paste the code, and output from t
    10·1 answer
  • g An analog voice signal, sampled at the rate of 8 kHz (8000 samples/second), is to be transmitted by using binary frequency shi
    12·1 answer
  • 1.<br> Defensive driving involves?
    14·1 answer
  • Will mark brainliest!
    13·2 answers
  • A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowr
    15·1 answer
  • Two solid yellow center lines on a two-lane highway indicate:
    13·2 answers
  • In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!