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3 years ago

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Water enters a leaky cylindrical tank (D = 1 ft) at a rate of 8 ft3/min. Water leaks out of the tank at a rate of 17% of the flo

w into the tank. At what rate will water rise in the tank (answer in ft/min)?

1 answer:

stellarik [79]3 years ago

3 0

Answer:

Water would rise in the tank at a rate of 8.45 ft/min

Explanation:

Diameter of leaky cylindrical tank (D) = 1 ft

Base area = πD^2/4 = 3.142×1^2/4 = 0.7855 ft^2

Volumetric flow rate at which water enters the tank = 8 ft^3/min

Volumetric flow rate at which water leaks out = 0.17 × 8 = 1.36 ft^3/min

Volumetric flow rate at which water rises = 8 - 1.36 = 6.64 ft^3/min

Rate at which water would rise in ft/min = volumetric flow rate at which water rises ÷ base area of the cylinder = 6.64 ft^3/min ÷ 0.7855 ft^2 = 8.45 ft/min

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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

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

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

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4 years ago

The article provides information by using a list. What does it list? A. Thanksgiving food B. places where clams can be found C.

Gelneren [198K]

Answer:

C

Explanation:

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Which option identifies what engineers will do to help in the following scenario?

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Answer:

a

Explanation:

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Read 2 more answers

Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12

Katarina [22]

Answer:

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

$- \dot W_{out} + \dot m \cdot (h_{in} - h_{out}) = 0$

The turbine power output is:

$\dot W_{out} = \dot m\cdot (h_{in}-h_{out})$

The volumetric flow is:

$\dot V = \frac{\pi}{4} \cdot \left( \frac{2}{12}\,ft \right)^{2}\cdot (620\,\frac{ft}{s} )$

$\dot V \approx 13.526\,\frac{ft^{3}}{s}$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,\frac{ft^{3}}{lbm}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu}$

$\dot m = \frac{13.526\,\frac{ft^{3}}{s} }{1.33490\,\frac{ft^{3}}{lbm} }$

$\dot m = 10.133\,\frac{lbm}{s}$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,\frac{BTU}{lbm}$

State 2 (Saturated Vapor)

$h = 1146.1\,\frac{BTU}{lbm}$

The turbine power output is:

$\dot W_{out} = (10.133\,\frac{lbm}{s} )\cdot (1479.1\,\frac{BTU}{lbm}-1146.1\,\frac{BTU}{lbm})$

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

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Answer:

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