Ask question

Ask question

All categories

3 years ago

10

Water is the working fluid in an ideal Rankine cycle. Superheatedvapor enters the turbine at 10MPa, 480°C, and the condenser pre

ssure is6 kPa. Isentropic efficiencies of the turbine and pump are 80% and 75%,respectively. Determine for the cyclea.the actual heat transfer to the working fluid passing through thesteam generator, in kJ per kg of steam flowing.b.the thermal efficiency.c.the actual heat transfer from the working fluid passing through thecondenser to the cooling water, in kJ per kg of steam flowing

1 answer:

choli [55]3 years ago

6 0

Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

$h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K$

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

$s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}$

$s_2 =0.51624 + x(7.82)$

$s_2 =0.51624 + 7.82x$

From the values obtained;

$s_1 =s_2= 6.52846 \ kJ/kg.K$

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222 = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

$h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 ) \\ \\ h_2 =2010.4084 \ kJ/kg$

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

$n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}$

$0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}$

$0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}$

$0.8 * {1311.6116}= {3322.02-h_{2a}$

$1049.28928= {3322.02-h_{2a}$

$h_{2a}= {3322.02- 1049.28928$

$h_{2a}= 2272.73072$ kJ/kg

The work pump is calculated by applying the formula:

$w_p = v_{f \ 6 kpa} (p_4-p_3)$

$w_p = 0.0010062 * (10000-6)$

$w_p = 0.0010062 *9994$

$w_p = 10.0559628 \ kJ/kg$

However;

$w_p = h_4 -h_3$

From the process;

$h_3 = h_{f(6 kpa)} = 150.15 \ kJ/kg$

$10.0559628 = h_4 - 150.15$

$10.0559628+ 150.15 = h_4$

$160.2059628= h_4$

$h_4= 160.2059628 \ kJ/kg$

The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

$n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}$

You might be interested in

What is the definition of comma

masya89 [10]

Com·ma
/ˈkämə/
NOUN
1. a punctuation mark (,) indicating a pause between parts of a sentence. It is also used to separate items in a list and to mark the place of thousands in a large numeral.

2. a minute interval or difference of pitch.

5 0

3 years ago

Locate the centroid y¯ of the composite area. Express your answer to three significant figures and include the appropriate units

german

Answer:

Please see the attached Picture for the complete answer.

Explanation:

4 0

3 years ago

When does the vc-turbo engine use lower compression ratios?.

Veronika [31]

Answer:

Explanation:

The VC-T engine (for "variable compression, turbocharged") can adjust its compression ratio between 8:1 and 14:1 on the fly, offering high-compression efficiency under light loads and the low compression needed for turbocharged power under hard acceleration.

7 0

2 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

3 years ago

Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i

leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

$R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}$

$R_{convection}=\frac{1}{h(2\pi r_{2}L)}$

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0

3 years ago