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3 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:

Oxana [17]3 years ago

6 0

Answer:

COP of heat pump=3.013

COP of cycle=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 =Q2/W

from equation 1

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

therefore COP =22.45/7.45 = 3.013

B) COP when cycle is reversed

COP = 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 =8.375/7.45 = 1.124

Verizon [17]3 years ago

4 0

Answer:

a) COP = 3.01

b) COP = 1.12

Explanation:

a) The heat is:

Q₂ = Q₁ + W = 15 + 7.45 = 22.45 kW

the COP of the heat pump is

COP = Q₂/W = 22.45/7.45 = 3.01

b) Operating as an air conditioner, the heat is

Q₁ = Q₂ - W = 8.375 kW

The COP is:

COP = Q₁/W = 8.375/7.45 = 1.12

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The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

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Answer:

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3 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

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