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4 years ago

A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.

Engineering

2 answers:

Anna007 [38]4 years ago

4 0

Answer:

COP of the heat pump is 3.013

OP of the cycle is 1.124

Explanation:

W = Q₂ - Q₁

Given

Q₂ = Q₁ + W

= 15 + 7.45

= 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

GuDViN [60]4 years ago

3 0

Answer:

3.013
1.124

Explanation:

W = Q2 - Q1 ----- equation 1

W = work done

Q2 = final energy

Q1 = initial energy

A) calculate the COP of the heat pump

COP = $\frac{Q2}{W}$

from equation 1

Q2 = Q1 + W = 15 + 7.45 = 22.45 KW

therefore COP = $\frac{22.45}{7.45}$ = 3.013

B) COP when cycle is reversed

COP = $\frac{Q1}{W}$

from equation 1

Q1 + W = Q2 ------ equation 2

Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2

Q1 = 8.375 KW

COP = $\frac{8.375}{7.45}$ = 1.124

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3 years ago

Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur

Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

BP = 11995.2 Pa = 11.99 KPa

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

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7 0

3 years ago

A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well

ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

= 14.9 * 20.1 = 299.5 m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T* t )

= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

= 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = $\frac{Q}{4\pi T} * W (u)$

where : Q = 2725.5

T = 299.5

W(u) = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0

3 years ago

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = $\frac{MC}{I}$ = $\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}$ = 159.07 MPa

B) Determine max transverse shear stress

attached below

ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam = 159.07 / 2 = 79.535 MPa

attached below is the remaining solution

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3 years ago

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Fantom [35]

Summary

Students learn about the variety of materials used by engineers in the design and construction of modern bridges. They also find out about the material properties important to bridge construction and consider the advantages and disadvantages of steel and concrete as common bridge-building materials to handle compressive and tensile forces.

This engineering curriculum aligns to Next Generation Science Standards (NGSS).

Engineering Connection

When designing structures such as bridges, engineers carefully choose the materials by anticipating the forces the materials (the structural components) are expected to experience during their lifetimes. Usually, ductile materials such as steel, aluminum and other metals are used for components that experience tensile loads. Brittle materials such as concrete, ceramics and glass are used for components that experience compressive loads.

Learning Objectives

After this lesson, students should be able to:

List several common materials used the design and construction of structures.

Describe several factors that engineers consider when selecting materials for the design of a bridge.

Explain the advantages and disadvantages of common materials used in engineering structures (steel and concrete).

Educational Standards

NGSS: Next Generation Science Standards - Science

Common Core State Standards - Math

International Technology and Engineering Educators Association - Technology

State Standards

Suggest an alignment not listed above

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Worksheets and Attachments

Strength of Materials Worksheet (doc)

Strength of Materials Worksheet (pdf)

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Strength of Materials Worksheet Answers (pdf)

Strength of Materials Math Worksheet (doc)

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Strength of Materials Math Worksheet Answers (doc)

Strength of Materials Math Worksheet Answers (pdf)

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