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goldenfox [79]
3 years ago
14

A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr

essure in the tank in atm.
Engineering
1 answer:
navik [9.2K]3 years ago
8 0

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So  atmospheric pressure is equal to  755 mm Hg.

We know that P= ρ g h

Density of Hg=13600 \frac{kg}{m^3}

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30   KPa

So

Absolute pressure=70.72 KPa

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A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum
adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress \sigma = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress \sigma = \dfrac{pd}{2t}

Making thickness t the subject; we have

t = \dfrac{pd}{2* \sigma}

t = \dfrac{560000*3}{2*150000000}

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

\sigma = \dfrac{pd}{4t}

t= \dfrac{pd}{4*\sigma }

t = \dfrac{560000*3}{4*150000000}

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

8 0
3 years ago
1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt
Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = \frac{width}{period}      ...............1

duty cycle = 1.5 × 10^{-3}

and applied voltage will be

applied voltage = duty energy × voltage    ...........2

applied voltage = 1.5 × 10^{-3} × 2

applied voltage = 3 mV

so current will be

current = \frac{applied\ voltage}{resistance}   ................3

current = \frac{3}{150}

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = \frac{1.5}{18.75*10^{-6}}

battery life = 80000 hours

battery life in year = \frac{80000}{8760}

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

4 0
3 years ago
8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)
konstantin123 [22]

Answer:

hi

Explanation:

4 0
3 years ago
What is a Wayfaring graphic?
KonstantinChe [14]

Answer:

ambages. pitiless or without compassion; cruel; merciless. winding, roundabout paths or ways.

Explanation:

6 0
3 years ago
An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa
mote1985 [20]

Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

4 0
3 years ago
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