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Sauron [17]
3 years ago
12

Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air con

ditioner and returns to the space at 25°C dry bulb and 6.5°C dew point temperature (ϕ2≠100%). If there is any, the condensate leaves the air conditioner at the same temperature of the air leaving the cooling coils (i.e., different from our previous assumption that condensate has same temperature of the cooling coils). The volume flow rate of the air returned to the workspace is 1000 m3 /min (Rair=0.287 kJ/kg∙K). Atmospheric pressure is 101.325 kPa. Compute

Engineering
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

See explaination

Explanation:

The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.

Kindly check attachment for the step by step solution of the given problem.

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A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig
Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0
3 years ago
A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1
Gelneren [198K]

Answer:

final temperature is 424.8 K

so correct option is e 424.8 K

Explanation:

given data

pressure p1 = 1 bar

pressure p2 = 5 bar

index k = 1.3

temperature t1 = 20°C = 293 k

to find out

final temperature  t2

solution

we have given compression is reversible and has an index k

so we can say temperature is

\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }  ...........1

put here all these value and we get t2

\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }

t2 = 424.8

final temperature is 424.8 K

so correct option is e

5 0
3 years ago
Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l
ycow [4]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

3 0
3 years ago
Read 2 more answers
Discuss 7 habits of highly effective people and how important are ethics in today's society​
kompoz [17]

Answer:

Explanation:

The 7 Habits of Highly Effective People, is a book written and first published in 1989. It is a business and self-help book that was written by Stephen Covey. The seven habits include

Being proactive

Starting anything with the end in mind

First things first

Always thinking towards a win-win situation

Seeking initially to understand, then going on to want to be understood

Synergize, and lastly

Growing

7 0
3 years ago
If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement b, no actuator movement c.
Volgvan

If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement.

<h3>What is a pilot-operated check valve (POC)?</h3>

Pilot operated test valves paintings through permitting loose float from the inlet port via the opening port. Supplying a pilot strain to the pilot port permits float withinside the contrary direction. Air strain on pinnacle of the poppet meeting opens the seal permitting air to float freely.

An actuator fault is a form of failure affecting the machine inputs. Due to strange operation or fabric aging, actuator faults might also additionally arise withinside the machine. An actuator may be represented through additive and/or multiplicative fault.

Read more about the pilot-operated check valve:

brainly.com/question/13001928

#SPJ1

7 0
1 year ago
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