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4 years ago

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What is the primary function of NCEES?

1 answer:

joja [24]4 years ago

7 0

National Council of Examiners for engineering and surveying a nonprofit organization

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3 years ago

Refrigerant R-12 is used in a Carnot refrigerator

Sphinxa [80]

Answer:

Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Explanation:

From figure

$Q_H$ is heat rejection process

$Q_L$ is heat transferred from the refrigerated space

$T_H$ is high temperature = 50 °C + 273 = 323 K

$T_L$ is low temperature = -20 °C + 273 = 253 K

$W_{net}$ is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator $(COP_{ref})$ is calculated as

$COP_{ref} = \frac{T_L}{T_H - T_L}$

$COP_{ref} = \frac{253 K}{323 K - 253 K}$

$COP_{ref} = 3.61$

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

$Q_H = 122.5 kJ/kg$

From coefficient of performance definition

$COP_{ref} = \frac{Q_L}{Q_H - Q_L}$

$Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L$

$Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)}$

$Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)}$

$Q_L = 95.93 kJ/kg$

Energy balance gives

$W_{net} = Q_H - Q_L$

$W_{net} = 122.5 kJ/kg - 95.93 kJ/kg$

$W_{net} = 26.57 kJ/kg$

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

$x = \frac{s_1 - s_f}{s_g - s_f}$

From figure

$s_1 = s_4 = 1.165 kJ/(K kg)$

Replacing with table values

$x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}$

$x = 0.37$

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

$h_1 = h_f + x \times (h_g - h_f)$

$h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg$

$h_1 = 241.58 kJ/kg$

By energy balance, $W_{t}$ turbine's work is

$W_{t} = |h_1 - h_4|$

$W_{t} = |241.58 kJ/kg - 249.7 kJ/kg|$

$W_{t} = 8.12 kJ/kg$

Finally, $W_{c}$ compressor's work is

$W_{c} = W_{net} + W_{t}$

$W_{c} = 26.57 kJ/kg + 8.12 kJ/kg$

$W_{c} = 34.69 kJ/kg$

6 0

3 years ago

A hollow, spherical shell with mass 2.00kg rolls without slipping down a slope angled at 38.0?.

mezya [45]

Answer:

$\mu = 0.31$

Explanation:

Given data:

mass = 2.00 kg

slope angle = 38.0

From figure

balancing force

$mgsin\theta - f = ma$ .....1

Balancing torque

$F_R = \frac{2}{3} mR^2 \alpha$ ......2

for pure rolling

$\alpha = \frac{a}{R}$

$F = \frac{2}{3} ma$

from 1 and 2nd equation

$mgsin\theta - \frac{2}{3}ma = ma$

$mgsin\theta = \frac{5}{3} ma$

$a = \frac{3}{5} g sin\theta$

$= \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2$

$F =\mu N$

$= \frac{2}{3} ma$

$= \frac{2}{3} 2\times 3.62 = 4.83 N$

N =normal force $= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N$

$\mu \times 15.44 = 4.83$

solving for coefficent of friction we get

$\mu = 0.31$

4 0

3 years ago