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maw [93]
3 years ago
15

A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of

0.3 kg/s. The heat pump is used to heat a home that loses heat to the environment at a rate of 180,000 Btu/hr. Refrigerant leaves the condenser as a subcooled liquid.
Determine the mass flow rate of refrigerant.

Engineering
1 answer:
solmaris [256]3 years ago
5 0

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

<u>At State A</u>:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

<u>At State B</u>:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg                          (By interpolation)

<u>At State C</u>:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

<u>m = 0.352 kg/s</u>

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3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat
Yuki888 [10]

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

4 0
3 years ago
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com
Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For  vapor-compression refrigeration cycle

P1 = P4  ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

<u>From pressure table of R 134a refrigerant</u>

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance  

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

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3 years ago
Sketch T-s and p-v diagrams for the Diesel cycle.
labwork [276]

Answer:

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1-2.Reversible adiabatic compression

2-3.Heat addition at constant pressure

3-4.Reversible adiabatic expansion

4-1.Heat addition at constant volume

When air inters in the piston cylinder after that it compresses and gets heated due to compression after that heat addition take place at constant pressure after that power is produces when piston moves to bottom dead center.

From the diagram of P-v And T-s we can understand so easily.

3 0
3 years ago
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