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likoan [24]
3 years ago
13

The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the

pump to an equipment room on the twenty-seventhfloor is 40ftof water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
Physics
1 answer:
Artemon [7]3 years ago
5 0

This question is incomplete, the complete question is;

The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventh floor is 40ft of water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor

Assume 12ft of elevation per floor

Answer: 48.68 psig

Explanation:

First  we calculate the elevation of the building

hb = 27 story * 12ft per floor/story

hb =  324 ft

given that the head lost in the vertical riser hL = 40 ft

now the delivery head required in the riser on he 27th floor;

hd = 8 psig *  (2.31 ft / 1 psig)

hd = 18.46 ft

Now calculate the suction head required by balancing the energy per unit weight of water, considering pump as the control volume

hp = (hb + hL + hd) - hs

hs = hb + hL + hd - hp

where hp is the head developed by the pump (270 ft)

hb is the elevation of the 27th floor of the building ( 324 ft)

hL is the head lost in the vertical riser ( 40 ft)

hd is the head required to exist in the riser on the 27th floor (18.46 ft)

so we substitute

hs = 324 ft + 40 ft + 18.46 ft - 270 ft

hs = 112.46

so 112.46ft * (1 psig / 2.31 ft)

= 48.68 psig

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Answer:

B) 27.3 m

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Where:  

x: horizontal position in meters (m)

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vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

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v₀x = vx = 30*cos40° = 22.98 m/s

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Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

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vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

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We replace t = 1.19 s , in the equation (1)

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3 0
3 years ago
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
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Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

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Given that,

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Using equation of ideal gas

PV=nRT

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3 years ago
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tino4ka555 [31]

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