When is a handrail required for stairs?

Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure

iris [78.8K]

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3

P5/P6 = P7/P8 = √9 =3

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg

Calculate the net work done

Wnet = (Wturb)out - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

4 0

3 years ago

Differences between acidic and basic Bessemer process

dybincka [34]

Answer:

In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.

5 0

2 years ago

What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?

Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

$A=\dfrac{\pi}{4}\times 0.25^2\ m^2$

$A=0.049\ m^2$

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0

3 years ago

A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu

krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

$\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}$

$10*2 = \sqrt{1 + 8f^2 - 1$

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

$\frac{V_1}{\sqrt{g*y_1}} = 7.416$

$V_1 = 7.416 *\sqrt{32.2*1}$

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

$V_2 = \frac{3666.66}{80*10}$

= 4.208 ft/sec

Froude number, F2 =

$\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}$

F2 = 0.235

d) $El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}$

$El = \frac{(10-1)^3}{4*1*10}$

$= \frac{9^3}{40}$

= 18.225ft

e) for critical depth, we use :

$y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3$

= 3.80 ft

7 0

3 years ago

Read 2 more answers

Based on the graphs of stress-strain from the V-MSE site, how would you characterize the general differences between polymers an

Pepsi [2]

Answer:

Option A

Explanation:

Alloys are metal compounds with two or more metals or non metals to create new compounds that exhibit superior structural properties. Alloys have high level of hardness that resists deformation thereby making it less ductile compared to polymers. This is due to the varying difference in the chemical and physical characteristics of the constituent metals in the alloy.

6 0

3 years ago