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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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Answer:

Tension can be easily explained in the case of bodies hung from chain, cable, string

Explanation

uniform speed, tension; T = W.

T=m(g±a)

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2 years ago
Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV  = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

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3 years ago
What is the biggest expectation when engineers test out designs?
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The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.
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Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,
iren [92.7K]

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA,  IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters          IE       IC           IB            VE       VB         Vc

Unit                     mA     mA          μA            V           V          V

Value                  4.65    4.6039   46.039    10.7      10     4.6039

4 0
3 years ago
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