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IgorC [24]
3 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]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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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
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Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

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3 years ago
Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)
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The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

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A homogeneous 800kg bar AB is supported at either end by a cable asshown in the figure
aleksandr82 [10.1K]

The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of ​​the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N                       [n = Bronze]

Pₓ =  3924 N                             [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

To know more about normal stress, visit:

brainly.com/question/28012990

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