All categories

3 years ago

Which option is a potential environmental risk of adopting a new technology?

Engineering

1 answer:

kvv77 [185]3 years ago

7 0

Answer:

its b

Explanation:

You might be interested in

A 200‑m rigid vessel contains a saturated liquid‑vapor mixture with a vapor quality of 75%. The temperature of the vessel is mai

DerKrebs [107]

Answer:

Given,

Temperature;

T = 393;;K

Convert to Celcius;

T = (393-273) degrees

T = 120°C

Using Table A-4 (Saturated water - Temperature table), at T = 120 C;

vf = 0.001060 m³/kg

vg = 0.89133 m³/kg

Quality is given as;

75% = 0.75

Specific volume is given as;

v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)

v= 0.66876 m³/kg

We know;

v = V/m

0.66876 = 100/m

m = 149.53 kg

6 0

3 years ago

A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an

Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = $1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}$

Cyclone is to 80 % efficient

So particulate remaining = $0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136$

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0

2 years ago

Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro

sasho [114]

Answer:

Just answered this to confirm my profile.

Explanation:

I dont have a clue, this is just to confirm my profile.

8 0

3 years ago

During his military campaign in what is now Germany, Julius Caesar lead his army of 40,000 soldiers to the western bank of the R

CaHeK987 [17]

Answer:

identifying a problem

Explanation:

its right

5 0

2 years ago

An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductan

german

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

$X_L=0.2\times 12=2.4$ ohm

$X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm$

$Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}$

$Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}$

=5.021 ohm

so amplitude of current = $\frac{v}{z}=\frac{120}{5.021}=23.89$

4 0

3 years ago