Ask question

Ask question

All categories

4 years ago

7

A home’s temperature is maintained at 25◦C by an AC system with a COP of 3.2 which cools outside air (38◦C with a 30°C Wet Bulb)

to 10 C and injects the cooled air into the home. The 25 C air exits the home. A) draw the process on the attached psychrometric chart. B) If the home gains 50,000 KJ/hr what is the required power in kW? C) How much water is condensed if any? D) Without changing the COP or the heat gain of the home what could be done to lower the power requirement of the AC system and what is the minimum power required?

1 answer:

motikmotik4 years ago

3 0

Answer:

Detailed solution is given below:

You might be interested in

Which professional draws maps and plans for projects involving structures other than buildings, such as bridges?

kolbaska11 [484]

Answer:

I believe it is a civil drafter

Explanation:

O*NET site

3 0

4 years ago

Read 2 more answers

To begin manufacturing we would start with a piece of 48" x 96" x 1⁄2” plate, LIST 3 MACHINES that might be used to produce this

7nadin3 [17]

Answer:

search google for this!

Explanation:

i found it online

5 0

3 years ago

A short-circuit experiment is conducted on the high-voltage side of a 500 kVA, 2500 V/250 V, single-phase transformer in its nom

Anarel [89]

Given Information:

Primary secondary voltage ratio = 2500/250 V

Short circuit voltage = Vsc = 100 V

Short circuit current = Isc = 110 A

Short circuit power = Psc = 3200 W

Required Information:

Series impedance = Zeq = ?

Answer:

Series impedance = 0.00264 + j0.00869 Ω

Step-by-step explanation:

Short Circuit Test:

A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.

The series impedance in polar form is given by

Zeq = Vsc/Isc < θ

Where θ is given by

θ = cos⁻¹(Psc/Vsc*Isc)

θ = cos⁻¹(3200/100*110)

θ = 73.08°

Therefore, series impedance in polar form is

Zeq = 100/110 < 73.08°

Zeq = 0.909 < 73.08° Ω

or in rectangular form

Zeq = 0.264 + j0.869 Ω

Where Req is the real part of Zeq and Xeq is the imaginary part of Zeq

Req = 0.264 Ω

Xeq = j0.869 Ω

To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.

Turn ratio = a = Vp/Vs = 2500/250 = 10

Zeq2 = Zeq/a²

Zeq2 = (0.264 + j0.869)/10²

Zeq2 = (0.264 + j0.869)/100

Zeq2 = 0.00264 + j0.00869 Ω

Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.

4 0

3 years ago

If a seat could sit on a chair would a chair be able to sit on a seat?

mars1129 [50]

Answer:

no

Explanation:

a seat goes on a chair and never goes off. also the chairs legs are to wide to sit on a seat.

5 0

3 years ago

Read 2 more answers

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago