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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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4 0
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1. 6.1 PSPICEMULTISIM The current in a 50μH inductor is known to be iL=18te−10tAfor t≥0. 1. Find the voltage across the inductor
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Answer:

a. Voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Power = -59.3μW

c. Inductor is delivering power.

d. Energy = 5934.3nJ

e. Time = 100ms; Energy = 1095941.025nJ

Explanation:

Given

Current; iL=18te^(−10t)A or t≥0.

L.= inductor = 50μH

a. The voltage, V across the inductor for t>0 is calculated as follows;

V = L(di/dt)

Where L = 50μH

di/dt = 18(e^-10t + (-10)te^-10t)

di/dt = 18e^-10t(1 - 10t)

Substitute 50μH for L and 18e^-10t(1 - 10t) for di/dt in V = L(di/dt)

V = 50μH * 18e^-10t(1 - 10t)

V = 50 * 10^-6(18e^-10t(1 - 10t))

V = 0.9e^-10t(1-10t)

Hence, the voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Find the power (in microwatts) at the terminals of the inductor when t=200 ms.

Given that t = 200ms = 200 * 10^-3s = 0.2s

Power, p is calculated using the following formula;

p = Li(di/dt)

p = 50 * 10^-6(18te^-10t)18e^-10t(1-10t)

p = 50 * 10^-6 * (18 * 0.2 * e^-(10*0.2)) * (18 * e^(-10 * 0.2) * (1-10*0.2)

p = -5.93E5W

p = -59.3μW

c. Is the inductor absorbing or delivering power at 200 ms?

Because of the negative sign, the inductor is delivering power.

d. Find the energy (in microjoules) stored in the inductor at 200 ms.

Energy is calculated as ½Li²

= ½ * 50 * 10^-6 * (18te^-10t)²

= ½ * 50 * 10^-6 * (18 * 0.2 * e ^ (-10 * 0.2))²

= 0.0000059342669999498J

= 5934.3nJ

e. Find the maximum energy (in microjoules) stored in the inductor and the time (in milliseconds) when it occurs.

Calculating the derivation in (a)

di/dt = 0

18e^-10t(1-10t) = 0

1 - 10t = 0

-10t = -1

t = 1/10

t = 100ms

To calculate the energy, first we need to calculate the current

I(t=100) = 18 * 0.1 * e^(-10(0.1)

I = 0.662182994108596

I = 6621.82mA

The energy is calculated as follows;

w = ½ * 50 * 10^-6 * (6.621)²

w = 0.001095941025

w = 1095941.025nJ

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3 years ago
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