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3 years ago

Which of the following Identifies the challenges faced by scientists experimenting with using artificial photosynthesis as

Engineering

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

The cost and size of materials needed to produce energy

Explanation:

Artificial photosynthesis is a chemical process that uses solar cells instead of chlorophyll to absorb sunlight and convert it into electricity. This process uses artificial leaves that require man-made catalyst to spilt water present in the air into hydrogen and oxygen. It is clear that the reaction requires heat from the sun for energy production thus the technology is expensive to be applied in most areas of the world. Additionally, results obtained from previous undertaken projects of this type has been ineffective and unsustainable because it involves a lot of trial and error.

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What is the difference between filler and electrode in Welding? Can a filler be an electrode? Can an electrode be a filler? Why?

Vlada [557]

Explanation:

<u>Filler:</u>

Filler is the material rod is used when we are joining two material by using welding process.If thickness of work piece is more so it will become compulsory to provide some filler material for making the welding join to withstand high stresses.

<u>Electrode:</u>

Electrode is the element which is used to complete the electric circuit in welding .Some time electrode is connected with positive terminal and some time with negative terminal ,it depends on the requirement of welding process.In Tungsten inert gas welding electrode is connected negative terminal but on the other hand Metal inert gas welding electrode is connected with positive terminal.Electrode can be consumable non-consumable depends on the condition.

Yes electrode can be work as filler material ,in Metal inert gas welding wire is used as electrode as well as filler material.In Metal inert gas welding consumable electrode is used on the other hand Tungsten inert gas welding non-consumable electrode is used.In Tungsten inert gas welding if thickness of work pieces is less than 5 mm then no need to used any filler material but if thickness is more than 5 mm then we have to use filler material.

8 0

3 years ago

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

3 years ago

Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo

deff fn [24]

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

8 0

3 years ago

Read 2 more answers

A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th

trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

- The mass flow rate of steam through the boiler

- The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

E_6 + E_2 = E_3

flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

y*h_6 + (1-y)*h_3 = h_3

y*2810 + (1-y)*192.8 = 721

Compute y: y = 0.2018

- Heat produced by the boiler q_b:

q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b: q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

q_c = (1-y)*(h_8 - h_1)

q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c: q_c = 1834.26 KJ/ kg

- Net power output w_net:

w_net = q_b - q_c

w_net = 3303.58 - 1834.26

w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

flow(m) = P / w_net

compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

u = 1 - (q_c / q_b)

u = 1 - (1834.26/3303.58)

u = 44.48 %

5 0

3 years ago

What causes a boat to float? My questions are super easy. So you can follow me if you like easy question.

marin [14]

Boats float because the gravity is acting down on it and the buoyant force is acting up on the ship.

6 0

4 years ago

Read 2 more answers