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4 years ago

Guys do u know who is

2 answers:

6 0

I think so, I’ve seen her on here a couple of times. Though I’ve never spoken with said person

Hope this helps you (also why?)

beks73 [17]4 years ago

6 0

Answer:

i found out a way to go on her profile and it said she helps remove reported answers

Explanation:

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What are the different types of documents used to communicate engineering designs?

Ipatiy [6.2K]

Answer:

COMMON ENGINEERING DOCUMENTS

Inspection or trip reports.

Research, laboratory, and field reports.

Specifications.

Proposals.

Progress reports.

ect...

Explanation:

7 0

3 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

max2010maxim [7]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

7 0

3 years ago

Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui

Burka [1]

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

6 0

3 years ago

Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th

Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0

4 years ago

An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

Darina [25.2K]

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

8 0

3 years ago