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Damm [24]
3 years ago
15

A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.

For a measured manometer deflection of 5 m of water, determine the tank static pressure if the barometric pressure is 101.3 kPa absolute.
(a) 140 kPa absolute

(b) 150 kPa absolute

(c) 160 kPa absolute

(d) 170 kPa absolute
Engineering
1 answer:
ddd [48]3 years ago
4 0

Answer:

P = 150.335\,kPa (Option B)

Explanation:

The absolute pressure of the air-filled tank is:

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 150.335\,kPa

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Tomtit [17]

Answer:

In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .

Explanation:

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3 0
2 years ago
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

4 0
3 years ago
Dean is buying a home for $170,000. The mortgage company he decided to use to finance the home requires a 10% down payment. What
ANTONII [103]
They answer is 3. $17,00
8 0
3 years ago
Read 2 more answers
Derive the following conversion factors: (a) Convert a pressure of 1 psi to kPa (b) Convert a vol ume of 1 liter to gallons (c)
kirill [66]

Answer:

a)6.8 KPa

b)0.264 gallon

c)47.84 Pa.s

Explanation:

We know that

1 lbf=  4.48 N

1 ft =0.30 m

a)

Given that

P= 1 psi

psi is called pound force per square inch.

We know that 1 psi = 6.8 KPa.

b)

Given that

Volume = 1 liter

We know that 1000 liter = 1 cubic meter.

1 liter =0.264 gallon.

c)

1\ \frac{lb.s}{ft^2}=47.84\ \frac{Pa.s}{ft^2}

4 0
3 years ago
Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of
Orlov [11]

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

7 0
3 years ago
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