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Mashcka [7]
3 years ago
5

An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m

aintain its temperature constant at 20. if the temperature of the outside is 35 the power required to operate this air conditioning system is
(a) 0.58 kW
(b) 3.20 kW
(c) 1.56 kW
(d) 2.26 kW
(e) 1.64 kW
Engineering
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

COP = \frac{T_{L}}{T_{H}-T_{L}}

COP = \frac{293.15\,K}{308.15\,K-293.15\,K}

COP = 19.543

Lastly, the power required to operate the air conditioning system is:

\dot W = \frac{\dot Q_{L}}{COP}

\dot W = \frac{32\,kW}{19.543}

\dot W = 1.637\,kW

Hence, the answer is E.

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The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg wa
BigorU [14]

Answer:

HB = 3.22

Explanation:

The formula to calculate the Brinell Hardness is given as follows:

HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }

where,

HB = Brinell Hardness = ?

P = Applied Load in kg = 500 kg

D = Diameter of Indenter in mm = 10 mm

d = Diameter of the indentation in mm = 1.55 mm

Therefore, using these values, we get:

HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }

<u>HB = 3.22 </u>

4 0
3 years ago
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