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3 years ago

How do information systems support the activities in a supply chain?

Engineering

1 answer:

Degger [83]3 years ago

6 0

Answer:

the no. of activities supply in a cahin like in the figuration wise they supply the chain

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Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow r

Georgia [21]

Answer:

a) $Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}$ , b) $Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}$ , c) 1600

Explanation:

a) The Reynolds Number is modelled after the following formula:

$Re = \frac{\rho \cdot v \cdot D}{\mu}$

Where:

$\rho$ - Fluid density.

$\mu$ - Dynamics viscosity.

$D$ - Diameter of the tube.

$v$ - Fluid speed.

The formula can be expanded as follows:

$Re = \frac{\rho \cdot \frac{4Q}{\pi\cdot D^{2}}\cdot D }{\mu}$

$Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}$

b) The Reynolds Number has this alternative form:

$Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}$

c) Since the diameter is the same than original tube, the Reynolds number is 1600.

8 0

3 years ago

Functional and nonfunctional requirements documents are used to _____.

Fiesta28 [93]

Answer:

Non-functional requirements when defined and executed well will help to make the system easy to use and enhance the performance

Explanation:

7 0

3 years ago

Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i

Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

$T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k$

Find the work done by the compressor

$\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg$

Find the actual workdone by the compressor

$\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg$

Find the temperature at exit of the turbine

$T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k$

Find the actual workdone by the turbine

$1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg$

Find the temperature of the regeneration

$\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k$

Find the heat supplied

$Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg$

Find the thermal efficiency

$n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4$

60.4%

Find the mass flow rate

$m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42$

Find the actual workdone by the compressor

$\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg$

Find the actual workdone by the turbine

$\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg$

Find the temperature of the compressor exit

$\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k$

Find the temperature at the turbine exit

$4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k$

Find the temperature of regeneration

$\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k$

6 0

3 years ago

Read 2 more answers

In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at

Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

$V = 690V$

$P_{real} = 2.3MW$

Real power in 3 phase

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Now the Phase Voltage is,

$V_{ph} = \frac{V}{\sqrt{3}}$

$V_{ph} = \frac{690}{\sqrt{3}}$

$V_{ph} = 398.37V$

The current phase would be,

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Rearranging,

$I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}$

Replacing,

$I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}$

$I_{ph}= 2.26kA/phase$

Therefore the current per phase is 2.26kA

6 0

3 years ago

Explain why the scenario below fails to illustrate an understanding of the importance of metrology. Situation: Natalie is a cali

VMariaS [17]

Answer:

Explanation:

The situation being described completely fails in regard to the importance of metrology. This is because the main importance of metrology is making sure that all of the measurements in a process are as accurate as possible. This accuracy allows an entire process to function efficiently and without errors. In a food production plant, each individual department of the plant relies on the previous function to have completed their job with the correct and accurate instructions so that they can fulfill their functions correctly and end up with a perfect product. If the oven (like in this scenario) is a couple of degrees off it can cause the product to come out burned or undercooked, which will then get transferred to the next part of production which will also fail due to the failed input (burned or undercooked product). This will ultimately lead to an unusable product at the end of the process and money wasted. Which in a large production plant means thousands of products in a single batch are thrown away.

8 0

3 years ago