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timama [110]
4 years ago
14

Air, at a pressure of 700 kPa and a temperature of 80°C, flows through a convergent– divergent nozzle. The inlet area is 0.005 m

2 and the pressure on the exit plane is 40 kPa. If the mass flow rate through the nozzle is 1 kg/s, find, assuming onedimensional isentropic flow,

Engineering
1 answer:
vodomira [7]4 years ago
7 0

Answer: The complete part of the question is to find the exit velocity

Explanation:

Given the following parameters

Inlet pressure = 700kpa

outlet pressure = 40kpa

Temperature = 80°C = 353k

mass flow rate =  1 kg/s

The application of the continuity and the bernoulli's equation is employed to solve the problem.

The detailed steps and the appropriate formula is as shown in the attached file.

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Answer:

Condition A

Heat flux is 1400 W/M^2

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Explanation:

Given that:

T_s is given as  30 degree celcius

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heat\ flux = \frac{Q}{a}= h\Delta = 40{30 - (-5)} = 1400 w/m^2

condition A

water temperature  = 10 degree c

convection coefficient = 800 w/m^2.k

heat\ flux = \frac{Q}{A} = H(\Delta} = 800\times (30-14) = 12800w/m^2

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3 years ago
What are the benefits of using the engineering design process
Shalnov [3]

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Some of the benefits are tangible for they are visible in the design and production process, while the other benefits are intangible which may not be visible directly but result in improvement in the quality of product, better control over designing and production process, reduction of stress on the designers etc.

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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

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L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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