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3 years ago

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A heat pump with refrigerant-134a as the working uid is used to keep a space at 25C by absorbing heat from geothermal water that

enters the evaporator at 60C at a rate of 0.065 kg/s and leaves at 40C. Refrigerant enters the evaporator at 12C with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine (a) the mass ow rate of the refrigerant, (b) the rate of heat supply, (c) the COP, and (d) the minimum power input to the compressor for the same rate of heat supply.

1 answer:

Snezhnost [94]3 years ago

8 0

Answer:

A) 0.03382 kg/s

B) 7.0372 Kw

C) 4.3982

D) 0.7396 kw

Explanation:

Given data:

Evaporator at 60 C

Space temperature = 25 C

power consumed by compressor = 1.6 kw

T1( evaporator temperature ) = 12°C

attached below is the detailed solution

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Answer:

Check the explanation

Explanation:

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$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

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Read 2 more answers

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

<u>Answer</u>:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

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3 years ago