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3 years ago

Takt time is the rate at which a factory must produce to satisfy the customer's demand. a)- True b)- False

Engineering

1 answer:

laila [671]3 years ago

6 0

Answer: a)True

Explanation: Takt time is defined as the average time difference between the production of the two consecutive unit of goods by the manufacturer and this rate is matched with the demand of the customer. This is the time which is calculated to find the acceptable time for which the goods unit must be produced by the factory to meet the needs of the customer. Therefore , the statement is true that takt time is the rate at which a factory must produce to satisfy the customer's demand.

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3 years ago

Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic

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3 years ago

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:

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

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

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

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

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3 years ago

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SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

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moment of inertia I = 144 inc^4

y = 3.96 in

$\sigma_b = \frac{M}{I} \times y$

putting all value to get bending stress

$\sigma_b = \frac{50.112}{144} \times 3.96$

$\sigma_b = 1.37 psi$

Bending stress at point 3.96 is $\sigma_b$ = 1.37 psi

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3 years ago

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blagie [28]

Answer:

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Explanation:

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2 years ago