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drek231 [11]
3 years ago
14

8.17 Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 10 MPa, 4808C, and the condenser pre

ssure is 6 kPa. The turbine and pump have isentropic efficiencies of 80 and 70%, respectively. Determine for the cycle (a) the heat transfer to the working fluid passing through the steam generator, in kJ per kg of steam flowing. (b) the thermal efficiency. (c) the heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per kg of steam flowing.

Engineering
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

attached below

Explanation:

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The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphe
Sonbull [250]

Answer:

a. The time required for the tank to empty halfway is presented as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

b. The time it takes for the tank to empty the remaining half is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time 't', is presented as follows;

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}

\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} =  \dfrac{v_2}{2 \cdot g}

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh

The time for the tank to drop halfway is given as follows;

\int\limits^{t_1}_0 {} \,  dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_1  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) =  { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)

t_1   =   { \dfrac{\sqrt{2}  \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)The time required for the tank to empty halfway, t₁, is given as follows;

t_1   =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

\int\limits^{t_2}_0 {} \,  dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh

t_2  =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} }

The total time, t, to empty the tank is given as follows;

t = t_1 + t_2 =   \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2  = { \dfrac{ D_0^2  }{D} \cdot\sqrt{\dfrac{H}{g} } =  \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}

t =  \sqrt{2}  \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }

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View the picture below and then correctly answer the questions using the following words: Temperate Zone, Tropical Zone, Polar Z
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Can you provide the picture? Thanks !

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Explain the function of both of the organelles pictured below.
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Fatigue failure occurs under the condition of (a) High elastic stress (b) High corrosivity (c) High stress fluctuations (d) High
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Answer:

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Explanation:

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