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IgorC [24]
2 years ago
12

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
Engineering
1 answer:
Semenov [28]2 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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Blababa [14]
Automobiles and Airplanes is what I can think of
3 0
3 years ago
The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
3 years ago
A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
hichkok12 [17]

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

                                =900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

6 0
3 years ago
‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

E=12\times 9.8\times 25\\\\E=2940\ J

So, the potential energy of the mass is 2940 J.

3 0
3 years ago
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