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3 years ago

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An ideal vapor-compression refrigeration cycle using refrigerant-134a as the working fluid is used to cool a brine solution to −

5°C. This solution is pumped to various buildings for the purpose of air-conditioning. The refrigerant evaporates at −10°C with a total mass flow rate of 7 kg/s, and condenses at 600 kPa. Determine the COP of the cycle and the total cooling lo

1 answer:

Semenov [28]3 years ago

8 0

Answer:

1141

7.2

Explanation:

The evaporator Temperature T1 = -10°C

h1 = 244.55kj/kg

S1 = 0.93782kj/kg

The enthalpy at second state from given condenser pressure from super heated refrigerant table

P2 = 600

S2 = S1

h2 = 267.19kj/kg

Enthalpy at 3 and 4 from condenser pressure

h3 = h4 = 81.5kj/kg

To get cooling load

QL = m(h1-h4)

M= 7

= 7(244.55-81.5)

= 7(163.05)

= 1141.35kw = 1141KW

COP

= h1-h4/h2 -h1

= 244.55-81.5/267.19-244.55

= 163.05/22.64

= 7.2

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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28

Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0

4 years ago

Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload

Vinil7 [7]

Answer:

1) $V_o = 10 liters$

2) $V_o = 12.26 liters$

Explanation:

For isothermal process n =1

$V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}$

$V_o = 10 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.03

$actual \ volume = c1\times 10 = 10.3 liters$

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

$V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}$

$V_o = 12.26 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.15

$actual \volume = c1\times 10 = 11.5 liters$

8 0

3 years ago

a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What

Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, $dq/dt = 100 W$

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, $T_0 = 20^0 C = 273 + 20 = 293 K$

Temperature of the inside of the box, $T_{in} = ?$

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

$dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C$

5 0

4 years ago

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

7 0

4 years ago

A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

4 0

4 years ago