All categories

3 years ago

Derive the COP of Carnot Refrigerator and Carnot heat Pump.

Engineering

1 answer:

deff fn [24]3 years ago

8 0

Answer:

Answered

Explanation:

COP is the coefficient of performance of refrigerator and Heat pump

COP of refrigerator is given by

cooling effect/work input = $\frac{Q_L}{W_{in}}$

And for ideal refrigeration cycle

COP= $\frac{T_L}{T_H-T_L}$

COP of heat pump is given by

Heating effect/ work input = $\frac{Q_H}{W_{in}}$

And for ideal Heat pump cycle

COP= $\frac{T_H}{T_H-T_L}$

Now, we can conclude that

$COP_{HP}= COP_{R}+1$

You might be interested in

Disconnecting a circuit while in operation can create a voltage blank

zlopas [31]

Answer:

what is the question

Explanation:

confused

5 0

3 years ago

A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.

AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass $m_{in}+m_{out}=\bigtriangleup m_{system}$

so, $m_e=m_1-m_2$

$Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1$

#Also, given the properties of water as;

$(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg$

#We assume constant properties for the steam at average temperatures: $h_e=\approx(h_1+h_2)/2$

#Replace known values in the equation above; $h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg$

#Using the mass and energy balance relations;

$m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg$

#We have $Q_i_n+m_eh_e=m_2u_2-m_1u_1$ : we replace the known values in the equation as;

$Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ$

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0

3 years ago

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p

Natali [406]

Answer:

Mechanical Efficiency = 83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = $V=8ft^3/s\\$

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

$P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW$

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

$=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\$

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0

3 years ago

A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so

irina1246 [14]

Answer: idk sorry we both on the same thing

7 0

3 years ago

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000 $\frac{ft lbm}{s^{2} }$

We know that moment of inertia is given as

$I = \frac{bt^{3} }{12}$

$I = \frac{5 (1.5^{3} )}{12}$

I = 1.40625 $in^{4}$

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = $\frac{PL^{3} }{48EI}$

$0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}$

E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0

3 years ago