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klasskru [66]
3 years ago
7

The performance of a heat pump degrades (i.e., its COP decreases) as the temperature of the heat source decreases. This makes us

ing heat pumps at locations with severe weather conditions unattractive. Consider a house that is heated and maintained at 26°C by a heat pump during the winter. What is the maximum COP for the heat pump if heat is extracted the out door air at 10 C?
Engineering
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

COP_max = 18.69

Explanation:

We are given;

Heated space temperature; T_H = 26°C = 273K + 26 = 299K

Temperature at which heat is extracted; T_L = 10°C = 273 + 10 = 283K

Now the Coefficient of performance (COP) of a heat pump will be a maximum when the heat pump operates in a reversible manner. The COP of a reversible heat pump depends on the temperature limits in the cycle only and is determined by the formula;

COP_max = 1/(1 - (T_L/T_H))

Thus,

COP_max = 1/(1 - (283/299))

COP_max = 1/(1 - 0.9465)

COP_max = 1/0.0535 = 18.69

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Juli2301 [7.4K]

Answer:

jjj jk to the day of the day of the day of the day of the day

Explanation:

hey hi I am a br of August and I am not able to join the meeting and will

7 0
2 years ago
A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [1 1 1] direction. Wha
Ivenika [448]

Answer: required tensile stress is 0.889 MPa

Explanation:

Given that;

tensile load is oriented along the [1 1 1] direction

shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane

first we determine

λ which is Angle between  [1 1 1]  and  [1 0 1]

so

cosλ = [ 1(1) +  1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]

= 2 / √3√2 =  2/√6

Next, we determine ∅ which is angle between [1 1 1]  and  [1 1 -1]

so,

cos∅ = [ 1(1) +  1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]

cos∅ = [ 2-1] / [√3√3 ]

cos∅ = 1/3

Now, we know that;

σ = T_stress/cosλcosθ

so we substitute

σ = 0.242 / ( 2/√6 × 1/3 )

σ = 0.242 / 0.2721

σ = 0.889 MPa

Therefore, required tensile stress is 0.889 MPa

3 0
3 years ago
Are the communication skills required for a team leader in a manufacturing plant different from those of a customer service exec
Svetllana [295]

Answer:

Explanation: Excellent customer service isn’t just down to your frontline staff, but a customer might be won or lost there. That’s why it’s so important every employee works together to resolve issues and create memorable moments – and empower your frontline agents to be as valuable to the customer as possible. It could be the difference between the customer coming back again, or going elsewhere.

The following tips are designed to help both customer service representatives, customer service management, and operations staff to work together to make experiences that matter. Developing customer service skills is important for the whole team to thrive – and to ensure customers keep coming back.

Customer service representative tips

Practice active listening

Learn to empathize with your customers

Use positive language

Improve your technical skills

Know your products and services

Look for common ground

Communicate clearly

Be solutions-focused

Admit mistakes

Be willing to learn

Customer service team management and operations tips

Provide first-class training

Set your standards high

Have a clear escalation pathway

Align your touchpoints

Create a culture of excellence

Be smart about automation

Use tools that boost speed and efficiency

Measure and analyze customer feedback

Use closed-loop feedback

Listen, understand and take action

5 0
2 years ago
Question
creativ13 [48]

Answer:

its d

Explanation:

4 0
2 years ago
A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap
adell [148]

Answer:

<em>a) 4.51 lbf-s^2/ft</em>

<em>b) 65.8 kg</em>

<em>c) 645 N</em>

<em>d) 23.8 lb</em>

<em>e) 65.8 kg</em>

<em></em>

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

145 pound = x slug

x = 145/32.174 = <em>4.51 lbf-s^2/ft</em>

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = x kg

x = 145/2.205 = <em>65.8 kg</em>

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = <em>645 N</em>

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

x m/s^2 = 5.30 ft/s^2

x = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = <em>23.8 lb</em>

e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth

mass on the moon = <em>65.8 kg</em>

3 0
3 years ago
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