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Mama L [17]
3 years ago
9

A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat

urally underground- as the heat source.If a supply of hot water at 140oC is discovered at a location where the environmental temperature is 20oC, determine the maximum thermal efficiency a geothermal power plant built at that location can have.
Engineering
1 answer:
jeka943 years ago
3 0

Answer:

the maximum thermal efficiency is 29%

Explanation:

the maximum efficiency for a thermal engine that works between a cold source and a hot source is the one of a Carnot engine. Its efficiency is given by

Maximum efficiency= 1 - T2/T1

where

T2= absolute temperature of the cold sink (environment)= 20°C + 273 = 293

T2= absolute temperature of the hot source (hot water supply) = 140°C + 273 = 413

therefore

Maximum efficiency= 1 - T2/T1 = 1 - 293/413 = 0,29 =29%

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Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
When a rubber is stretched during a tensile test, its elongation is initially proportional to the applied force, but as it reach
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A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 84998 N , and experienc
german

Answer:

Elastic Modulus = 227 GPa

Explanation:

Given,

Load = 84998 N

Length of bar = 300 mm = 0.3 m

Elongation = 0.18 mm = 0.00018 m

Cross sectional Area of the bar = (25mm × 25mm) = 0.025 × 0.025 = 0.000625 m²

From Hooke's law, the stress experienced by a material is proportional to the strain experienced by the same body, as long as the elastic limit isn't exceeded.

Stress ∝ strain

The coefficient of proportionality is the elastic modulus, E.

Stress = E × (Strain)

Stress = (Load)/(Cross sectional Area)

Stress = (84998 ÷ 0.000625) = 135,996,800 N/m²

Strain = (Change in length)/(Original length)

Strain = (ΔL/L) = 0.00018 ÷ 0.3 = 0.0006

E = (Stress/Strain)

E = 135,996,800 ÷ 0.0006 = 226,661,333,333.3 Pa = (2.267 × 10¹¹) Pa

1 GPa = 10⁹ Pa

(2.267 × 10¹¹) = 2.267 × 10² × 10⁹ = 226.7 GPa = 227 GPa to the nearest GPa. (No decimal place)

Hope this Helps!!!

4 0
3 years ago
Read 2 more answers
Determine the resultant horizontal and vertical force component that the water exerts on the side of the dam. The dam is 25 ft l
Leya [2.2K]

Answer:

Your question is not complete. However, i have provided the missing part both the diagram and remaining data as stated in attached full.

Resultant vertical force component is 260005.2 Ibs

Resultant horizontal force component is 487500 Ibs

Explanation:

From P = Specific Weight * height

P = 62.4 * 25 = 1560Ib/ft^2

Then, W = P * Z

W = 1560 * 25 = 39000Ib/ft

Resultant forces

Let the base length be 10ft

A = 2/3 (a*b)

A = 2/3 (10* 25) = 166.67ft^2

Resolving vertical component

F vertical = Specific weight * Volume

F vertical = 62.4 (166.67 * 25)

F vertical = 260005.2 Ibs

Resolving horizontal component

F horizontal = 1/2 (W * height )

F horizontal = 1/2 ( 39000 *25)

F horizontal = 487500 Ibs

3 0
3 years ago
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