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mash [69]
2 years ago
12

Some of our modern kitchen cookware is made of ceramic materials. (a) List at least three important characteristics required of

a material to be used for this application. (b) Make a comparison of three ceramic materials as to their relative properties and, in addition, to cost. (c) On the basis of this comparison, select the material most suitable for the cookware.
Engineering
1 answer:
Alisiya [41]2 years ago
8 0

Answer:

A)

It should be Non- toxic

It should possess high Thermal conductivity

It should have the Required Thermal diffusivity

B)

  • stoneware : This material has good thermal diffusivity and it  is quite affordable and it is used in making pizza stones
  • porcelain: mostly used for mugs and it is non-toxic
  • Pyrex : posses good thermal conductivity used in oven

C)   All the materials are suitable because they serve different purposes when making modern kitchen cookware

Explanation:

A) characteristics required of a ceramic material to be used as a kitchen cookware

  • It should be Non- toxic
  • It should possess high Thermal conductivity
  • It should have the Required Thermal diffusivity

B) comparison of three ceramic materials as to their relative properties

  • stoneware : This material has good thermal diffusivity and it  is quite affordable and it is used in making pizza stones
  • porcelain: mostly used for mugs and it is non-toxic
  • Pyrex : posses good thermal conductivity used in ovens

C) material most suitable for the cookware.

 All the materials are suitable because they serve different purposes when making modern kitchen cookware

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In a paragraph of 125 words, explain at least three ways that engineers explore possible solutions in their
joja [24]

Answer:

Three ways that engineers explore possible solutions in their projects are;

1) Prototyping

2) Simulation

3) Calculations

Explanation:

1) Prototyping is the process of experimental testing of samples of design, or model of a product with the possibility of the inclusion of control of parameters in order to determine the workability of a solution.

2) Simulation is the process of creating an imitation of a situation, operation or process which can be used to determine if a particular solution will be able to work as required in the simulated environment of a problem.

3) Calculations are used to find preliminary results of particular situations, their cause and effects based on scientific laws, theories and hypothesis such that the factor of the problem is equated with the available ideas to find the best possible solution.

3 0
2 years ago
A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo
Natalija [7]

Answer:

For 20^{\circ} - 5.556 lb/s

For 15^{\circ} - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = 20^{\circ}

Temperature rise, T' = 15^{\circ}

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, C_{p} = 0.24 btu/lb ^{\circ}F

Now, for a rise of 20^{\circ} in temeprature:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s

Now, for 15^{\circ}:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s

4 0
2 years ago
The heat transfer coefficient decreases with increasing x for both the laminar and turbulent regions a. True b. False
REY [17]

Answer:

A) True  

Explanation:

Yes this is true when length is creases the heat transfer coefficient decease with length.

The heat transfer(h) coefficient is varying with x by given expression

For Laminar flow

h \alpha \dfrac{1}{x^{\frac{1}{2}}}

For turbulent flow

h \alpha \dfrac{1}{x^{\frac{1}{5}}}

But when flow is in transitional state the heat heat transfer(h) coefficient is increases with x.But for laminar as well as turbulent flow h is decrease when x increases.

3 0
2 years ago
Which statement about cathode ray tubes (CRTs) is true?
yuradex [85]

Answer:

CRTs consume less power than LCDs.

hope it helps (^^)

# Cary on learning

6 0
2 years ago
Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

8 0
3 years ago
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