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OLga [1]
3 years ago
12

A pressurized 2-m-diameter tank of water has a 10-cm-diameter orifice at the bottom where water discharges to the atmosphere. Th

e water level initially is 3 m above the outlet. The tank air pressure above the water level is maintained at 450 kPa absolute and the atmospheric pressure is 100 kPa Neglecting frictional effects, determine: (a) how long it will take for half of the water in the tank to be discharged; (b) the water level in the tank after 10 s
Engineering
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

A fluid is defined as a material that deforms continuously and permanently under the

application of a shearing stress.

• The pressure at a point in a fluid is independent of the orientation of the surface

passing through the point; the pressure is isotropic.

• The force due to a pressure p acting on one side of a small element of surface dA

defined by a unit normal vector n is given by −pndA.

• Pressure is transmitted through a fluid at the speed of sound.

• The units we use depend on whatever system we have chosen, and they include quantities

like feet, seconds, newtons and pascals. In contrast, a dimension is a more

abstract notion, and it is the term used to describe concepts such as mass, length and

time.

• The specific gravity (SG) of a solid or liquid is the ratio of its density to that of water

at the same temperature.

• A Newtonian fluid is one where the viscous stress is proportional to the rate of strain

(velocity gradient). The constant of proportionality is the viscosity, µ, which is a

property of the fluid, and depends on temperature.

• At the boundary between a solid and a fluid, the fluid and solid velocities are equal;

this is called the “no-slip condition.” As a consequence, for large Reynolds numbers

(>> 1), boundary layers form close to the solid boundary. In the boundary layer,

large velocity gradients are found, and so viscous effects are important.

• At the interface between two fluids, surface tension may become important. Surface

tension leads to the formation of a meniscus, drops and bubbles, and the capillary rise

observed in small tubes, because surface tension can resist pressure differences across

the interface.

You might be interested in
Differences between acidic and basic Bessemer process​
dybincka [34]

Answer:

In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.

5 0
2 years ago
The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0
3 years ago
Repetitive movements at work can lead to injuries. True or False
OverLord2011 [107]
Answer

True

Explanation

RSI can occur when you do repetitive movements. Those movements can cause your muscles and tendons to become damaged over time. Some activities that can increase your risk for RSI are: stressing the same muscles through repetition.
8 0
2 years ago
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws
ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\

E=52000Hp.h

E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\

E=38724920Wh

E=52000Hph(\frac{1977378.4  ft lb}{1Hph}

E=1.028x10^11 ftlb

3 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
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