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3 years ago

A heat pump and a refrigerator are operating between the same two thermal reservoirs. Which one has a higher COP?

Engineering

1 answer:

Mkey [24]3 years ago

4 0

Answer:

Heat pump

Explanation:

COP:

As we know that COP can be defines as the ratio of desired effect to the work in put.

The desired effect can be different for heat pump our desired effect to heat the space and for refrigeration our desired effect is cooled the space.

$COP_{pump}=\dfrac{T_H}{T_H-T_L}$

$COP_{refri}=\dfrac{T_L}{T_H-T_L}$

When heat pump and refrigerator are operating with same temperature limits then from expression we can say that

COP of heat pump = 1 + COP of refrigeration.

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A stream of ethylene gas at 250°C and 3800 kPa expands isentropically in a turbine to 120 kPa. Determine the temperature of the

11Alexandr11 [23.1K]

Answer: for ideal situation

T2 = 16.5158K

Work = - 11.4J

For appropriate generalization correlation

T2 = 308.57K

Work = 177.797MJ

Explanation:detailed calculation and explanation is shown in the image below.

6 0

3 years ago

Find the equivalent impedance Zeq seen by the source when Vs = 4 cos (4t) v, C = 0.5 F, R = 1 Ω and L = 0.5 H. (Give angles in d

lbvjy [14]

Answer:

Zeq = 1.80ohms

Vc = 1.11cos4t volts

Explanation:

Impedance of an AC circuit is defined as the opposition to the flow of current across the resistor, capacitor and inductor of an AC circuit. It is expressed mathematically as;

Zeq = √R²+(Xl-Xc)²

Where R is the resistance

Xl is the inductive reactance

Xc is the capacitive reactance

Xl = 2πfL

Source voltage Vs = Vcos2πft

Given Vs = 4 cos (4t)v

Comparing both source voltage to get the frequency f, we have;

2πft = 4t

2πf = 4...(1)

Given L = 0.5H

C = 0.5F

R = 1.0ohms

Xl = 4×0.5 (note that 2πf = 4 from equation 1)

Xl = 2.0ohms

Xc = 1/2πfC

Xc = 1/4(0.5)

Xc = 1/2

Xc = 0.5ohms

To get the impedance Z

Zeq = √1²+(2-0.5)²

Zeq = √1+1.5²

Zeq = √3.25

Zeq = 1.80ohms

The voltage across the capacitor is expressed as Vc = IXc

Since we know Xc = 0.5ohms

We need current I

To get current we use the relationship Vs = IZ

I = Vs/Z =

4 cos (4t)/1.8

I = 2.22cos4t A

Vc = 2.22cos4t × 0.5

Vc = 1.11cos4t volts

6 0

3 years ago

Water flows through a tee in a horizontal pipe system. The velocity in the stem of the tee is 15 f t/s, and the diameter is 12 i

Romashka [77]

Answer:

The resultant force is 2620.05 lbf acting to the right.

Explanation:

The area for inlet section is:

$A_{1}=\frac{\pi D_{1}^{2} }{4} =\frac{\pi (12/12)^{2} }{4} =0.79 ft^{2}$

The area for oulet section is:

$A_{2} =\frac{\pi D_{2}^{2} }{4} =\frac{\pi (6/12)^{2} }{4} =0.196 ft^{2}$

The volumetric flow rate is:

Q=V1A1=15*0.79=11.85 ft^3/s

The velocities and areas at the exit is the same:

Q=V2A2+V3A3=2V2A2

Clearing V2:

V2=V3=Q/(2*A2)=11.85/(2*0.196)=30 ft/s

The mass flow rate through inlet is:

m1=ρA1V1=1.94*15*0.79=22.99 lbf*s/ft

The mass flow rate through outlet is:

m2=m3=m1/2=22.99/2=11.49 lbf*s/ft

The x-component of force is:

Rx+p1A1=-V1m1

Where p1 is the pressure at inlet

Rx=-(15*22.99)-(2880*0.79)=-2620.05 lbf

Fx=-Rx=2620.05 lbf

The y-component of force is:

Ry+p2A2-p3A3=V2m2-V3m3

Ry+0-0=(30*11.49)-(30*11.49)

Ry=0

Fy=Ry=0

The resultant force is:

$F=\sqrt{Fx^{2}+Fy^{2} } =\sqrt{2620.05^{2}+0 } =2620.05 lbf$

This force is acting to the right.

4 0

3 years ago

According to your equation, what is the functional relationship between the height of the water h_1 and the speed of the flow v_

larisa [96]

This question is incomplete, the complete question is;

Starting with the full Bernoulli’s equation, simplify the equation for the case diagrammed below, where Point 1 is at the top of the water at height h1, with negligible flow speed v1=0, and Point 2 is at height h2=0. Both Point 1 and Point 2 are open to the atmosphere so P1 = P2 = Patm.

According to your equation, what is the functional relationship between the height of the water h1 and the speed of the flow v2 at Point 2?

Answer:

the functional relationship between the height of the water h1 and the speed of the flow v2 at Point 2 is v2 = √2gh1

Explanation:

Gravitational acceleration = g

Density of water = P

Pressure of the water at point 1 = p1 = patm

Speed of the water at point 1 = v1 = 0 m/s

Height of point 1 = h1

Pressure of the water at point 2 = p2 = patm

Speed of the water at point 2 = v2

Height of point 2 = h2 = 0 m

Now, Using Bernoulli's equation at point 1 and 2,

p1 + Pv1²/2 + Pgh1 = p2 + Pv2²/2 + Pgh2

patm + P(0)²/2 + Pgh1 = patm + Pv2²/2 + Pg(0)

Pgh1 = Pv2²/2

gh1 = v2²/2

v2² = 2gh1

v2 = √2gh1

Therefore the functional relationship between the height of the water h1 and the speed of the flow v2 at Point 2 is v2 = √2gh1

7 0

3 years ago

What is the waste water from kitchen sinks called

daser333 [38]

Answer:

grey water??? I think

Explanation:

7 0

3 years ago

Read 2 more answers