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3 years ago

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Air enters the compressor of a regenerative gas-turbine engine at 300 K and 100 kPa, where it is compressed to 800 kPa and 580 K

. The regenerator has an effectiveness of 72%, and the air enters the turbine at 1200 K. For a turbine efficiency of 86%, draw the cycle on T-s diagram and find: a) Heat transfer in the regenerator. b) Thermal efficiency. (114.2kJ/kg,, 36%)

1 answer:

Bad White [126]3 years ago

3 0

Answer:

a) The amount of heat transfer in the regenerator, q = 114.12 kJ/kg

b) Thermal efficiency = 35.9%

Explanation:

The calculations are neatly handwritten and attached as files to this solution for easiness of expression and clarity. The cycle is also drawn on the T - S diagram and included in the attached files. Check the files below for the complete calculation.

I hope this helps!

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3 years ago

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A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.

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Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e = $\frac{V_v}{V_s}$ = $\frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)$

Specific gravity $G_{s}$ = $\frac{P_s}{P_w} =$ $\frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)$

% Saturation S = $\frac{V_w}{Vv}$ × $\frac{100}{1}$ = $\frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right}$ × $\frac{100}{1}--------(3)$

water content w = $\frac{M_w}{M_s}$ = $\frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)$

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w = $\frac{M_w}{Ms}$

Recall; mass = Density × volume

w = $\frac{V_wP_w}{V_sP_s} ------(5)$

From eqn. (2) $G_{s}$ = $\frac{P_s}{P_w}$

∴ $\frac{1}{G_s} = \frac{P_w}{P_s} ------(6)$

putting eqn. (6) into (5)

w = $\frac{V_w}{V_sG_s} -----(7)$

Again, from eqn (1)

$V_s = \frac{V_v}{e}$

substituting into eqn. (7)

$w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v} = S$

∴ $w = \frac{Se}{G_s} -----(8)$

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, $G_{s} = 2.7, e= 0.9$

∴ $w= \frac{0.9*1}{2.7} = 0.333$

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

$w= \frac{0.01*0.9 }{2.7} = 0.0033$

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil ( $P_w=\frac{1_g}{cm^{3} }$ )

Given, $V_{s} = 100 cm^{3 }$ , S = 50% = 0.5

%S = $\frac{V_w}{V_v}$ × $\frac{100}{1}$ = $\frac{V_w}{eV_s}$ × $\frac{100}{1}$

0.50 = $\frac{V_w}{0.9* 100} = 45cm^{3}$

mass of water = $P_wV_w= 1 * 45 = 45_{g}$

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Over 30 day period, a lake surface area is 1260 acres. The inflow is 36 cfs, thee outflow is 30 cfs. Seepage loss is 1.5 in. The

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Answer:

-0.1 inches

Explanation:

The net inflow is ...

36 cfs -30 cfs = 6 cfs

The number of seconds in 30 days is ...

(3600 s/h)(24 h/da)(30 da) = 2,592,000 . . . . seconds/(30 days)

Then the volume of inflow is ...

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The number of square feet in 1260 acres is ...

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So, the increase in depth due to the inflow is ...

(15,552,000 ft^3)/(54,885,600 ft^2) ≈ 0.283353 ft ≈ 3.4002 in

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3 years ago