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densk [106]
3 years ago
10

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

Engineering
1 answer:
motikmotik3 years ago
8 0

Answer:

The answer is "112.97 \ \frac{ft}{s}"

Explanation:

Air flowing into thep_1 = 20 \ \frac{lbf}{in^2}

Flow rate of the mass m  = 230.556 \frac{lbm}{s}

inlet temperature T_1 = 700^{\circ} F

PipelineA= 5 \times 4 \ ft

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}

      = \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}

V= \frac{mv}{A}

   = \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}

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Answer:

True

Explanation:

Dual home host - it is referred to as the firewall that is incorporated with two or more networks. out of these two networks, one is assigned to the internal network and the other is for the network. The main purpose of the dual-homed host is to ensure that no Internet protocol traffic is induced between both the network.

The most simple example of a dual-homed host is a computing motherboard that is provided with two network interfaces.

7 0
3 years ago
For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.
Elza [17]

Assumptions:

  • Steady state.
  • Air as working fluid.
  • Ideal gas.
  • Reversible process.
  • Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

  • Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
  • Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

        r =\frac{V_1}{V_2}

  • Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
  • Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

         r = \frac{V_4}{V_3} = \frac{V_1}{V_2}

  • Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
  • Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

           \eta = 1-(\frac{1}{r^{\gamma - 1} } )

where:

           \gamma = \frac{C_{p} }{C_{v}} : specific heat ratio

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

           \gamma = 1.4

Answer:

See image attached.

5 0
3 years ago
Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M
Naddika [18.5K]

Answer:

a. 318.2k

b. 45.2kj

Explanation:

Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

7 0
3 years ago
Jodie bought some shirts for 6$ each marge brought some shirts for 8$ each
Alex_Xolod [135]

Answer:

you need more details but if you have to find the difference, its $2.00

Explanation:

8-6=2

3 0
3 years ago
A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show
aleksandrvk [35]

Answer:

GM

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter d=2m

Length l=2.5m

Weight W=22kN

Specific weight of sea water \mu= 10.25kN/m^3

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

W=(pwg)Vd

Where

V_d=\pi/4(d)^2y

Therefore

22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m

Therefore

Center of Bouyance B

B=\frac{y}{2}=0.26m\\\\B=0.75

Center of Gravity

G=\frac{I.B}{2}=2.6m

Generally the equation for\BM is mathematically given by

BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\

Therefore

BG=2.6-0.476\\\\BG=0.64m

Therefore

GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\

Therefore

GM

So the bouy does not float with its axis vertical

4 0
3 years ago
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