HELP PLEASE

Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers

4vir4ik [10]

Answer:

2132hp ed enregia

e

Explanation:

dawdsawdsawdsawdsawdsawdsawdaas

6 0

3 years ago

Suppose you are asked to design an office building. Explain what type of drawing you would use and why.

ohaa [14]

Answer:

birds-eye view perspective

Explanation:

If someone asked me to design an office building, I would draw it from a birds-eye view perspective. I would draw it this way so I could map out where everything in the office would go and make sure I have enough space for everything. I would also draw it this way in order to clearly see where everything would go in the office. For instance, cubicles/desks could go in the bottom left corner, while the boss's office could go in the top right. It would be easier to organize and it would be easier for me to look back on when I need to actually design the office later.

(i'm not sure if this is what your question is asking for so i just made my best guess)

7 0

3 years ago

Read 2 more answers

simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl

Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

a) Air temperature at turbine exit

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) $(\frac{783.05-778.18}{800.13-778.18} )$ = 764.45K

b) The net work output

first we determine the actual work input to compressor

Wc = h2 - h1 ( calculated values )

= 626.57 - 295.17 = 331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4 ( calculated values )

= 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

= 541.88 - 331.4 = 210.48 kJ/kg

c) determine thermal efficiency

лth = Wnet / qin ------ ( 1 )

where ; qin = h3 - h2

equation 1 becomes

лth = Wnet / ( h3 - h2 )

= 210.48 / ( 1324.93 - 626.57 )

= 0.3014 = 30.14%

6 0

3 years ago

Select the correct answer.

Ulleksa [173]

The answer is A. Immediately inform her colleague

4 0

3 years ago

When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs

Reika [66]

Answer: $\mu_{k}$

Explanation:

We use kinetic friction when a body is moving i.e. $\mu_{k}$ for calculations.

Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .

Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

3 0

4 years ago