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2 years ago

If you could help it would mean alot.

Engineering

2 answers:

scoundrel [369]2 years ago

8 0

Answer:

D is the correct choice.

Explanation:

I'm assuming that this is probably a phase in the textbook or progarm you are studying, and this is just a matter of reading thoroughly.

Engineers usually benefit from catching a mistake, and would also benfit from keeping record of a misstep in order to remain clear of that mistake in the future.

Have a great day, and mark me brainliest if I am most helpful!

gulaghasi [49]2 years ago

3 0

Answer:d

Explanation:

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Solve the inequality. Then graph your solution.<br> -9v – 10 < 7y +6

Cerrena [4.2K]

16

if you add 9+10 you get 18 - 7+6

5 0

2 years ago

An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th

photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

${W_{u}}= \eta {W_{pump}}$ .............1

put here value

${W_{u}}$ = (0.80)(20)

${W_{u}}$ = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

$\Delta{E_{mech}} = {m} \Delta pe$ ..............2

$\Delta {E_{mech}} = {m} g \Delta z$

and that can be express as

$\Delta {E_{mech}} = \rho Q g \Delta z$ ..................3

$\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]$ ......4

solve it we get

$\Delta {E_{mech}} = 13.614$ hp

so here

due to frictional effects, mechanical power lost in piping

we get here

${W_{frict}} = {W_{u}}-\Delta {E_{mech}}$

put here value

${W_{frict}}$ = 16 -13.614

${W_{frict}}$ = 2.37 hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0

2 years ago

A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r

belka [17]

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.