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2 years ago

12

Which of the following statements about PAC storage and housekeeping is correct? (Welding)

1 answer:

elena-s [515]2 years ago

7 0

The answer would be C.the compressed air line is normally left connected

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3 years ago

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zhenek [66]

Answer

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Step-by-step explanation:

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3 years ago

What is your name in face book

nekit [7.7K]

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2 years ago

Read 2 more answers

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

3 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

2 years ago

Read 2 more answers