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nadezda [96]
2 years ago
9

What type of test can show a chemical engineer if a material remains in a system and accumulates or if it moves right through?

Engineering
1 answer:
UkoKoshka [18]2 years ago
7 0

A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

<h3>What is a chemical engineer?</h3>
  • Processes for manufacturing chemicals are created and designed by chemical engineers.
  • To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
  • A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
  • Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.

To learn more about chemical engineer, refer to:

brainly.com/question/23542721

#SPJ4

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Heater power = 425 watts

Explanation:

Detailed explanation and calculation is shown in the image below

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A gas metal arc welder is also known as a _____ welder.<br> A) TIGB) GTAWC) GMAWD) Resistance spot
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GMAW

Explanation:

It's literally the initials of that type of welding

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90% of traffic crashes are due to driver error.<br> True<br> False
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90% of traffic crashes are due to driver error.

True

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Jordan's dad made a new recipe for dinner. Jordan looked at the food and saw that it was white, yellow, and purple in color. She
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2 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

4 0
3 years ago
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