Which option demonstrates what will happen once a prototype is developed

Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C.

algol [13]

This question is incomplete, the complete question is;

Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.

Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.

Answer:

- the quality of the steam exiting the second stage of the turbine is 0.9329

- the thermal efficiency is 36.05%

Explanation:

get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .

Specific enthalpy h1= 3192.3 kJ/kg

Specific entropy s1 = 5.9566 kJ/kg.K

Process 1 to 2s is isentropic expansion process in the turbine

S1 = S2s

get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K

h2s = 2822.2 kJ/kg

get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)

0.78 = (3192.3 - h2) / (3192.3 - 2822.2)

h2 = 2903.6 kJ/kg

get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C

h3 = 3422.2 kJ/kg

s3 = 6.8803 kJ/kg.K

Process 3 to 4s is isentropic expansion process in the turbine

S3 = S4s

get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K

h4s = 2118.8 kJ/kg

get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)

0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )

h4 = 2405.5 kJ/kg

get the properties at pressure, p5 = 6 kPa

h5 = hf

= 151.53 kJ/kg

v5 = Vf

= 0.0010064 m³/kg

get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at

p6 = p1 = 28 MPa

np = v5( p6 - p5) / (h6 - h5)

0.82 = ((0.0010064)( 28000 - 6)) / (h6 - 151.53)

h6 = 185.89 kJ/kg

Now to find the quality of the steam at the exit of the second stage of the turbine

At stat4, p4 = 6kPa

h4f = 151.53 kJ/kg

h4fg = 2415.9 kJ/kg

h4 = h4f + x4h4fg

2405.5 = 151.53 + (x4 (2415.9))

x4 = 0.9329

the quality of the steam exiting the second stage of the turbine is 0.9329

Also to find the efficiency of the power plant, we use the following equation;

n = Wnet / Qin

= (Wt1 + Wt2 - Wp) / (Q61 + Q23)

= [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]

[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]

= 0.3605

n = 36.05%

therefore the thermal efficiency is 36.05%

3 0

3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0

3 years ago

What are employers required to do to keep employees safe from caught-in and -between hazards from hand-held power tools?

nika2105 [10]

Answer:

Employees who use hand and power tools and who are exposed to the hazards of falling, flying, abrasive and splashing objects, or exposed to harmful dusts, fumes, mists, vapors, or gases must be provided with the appropriate equipment needed, including Personal Protective Equipment, to protect them from the hazard.

Explanation:

8 0

3 years ago

______number can be used to describe the relative growth of the hydraulic boundary layer and the thermal boundary layer. a) Reyn

weqwewe [10]

Answer: d) Prandtl number

Explanation: Prandtl number is basically defined as the ratio between the fluid's viscosity to the thermal conductivity.It doesn't have any sort of dimension. The fluids which are discovered with the small Prandtl numbers are considered as good fluids as they have a smooth rate of flow and as the number increases the fluid are not considered as reliable. Thus,option (d) is the correct option.

3 0

4 years ago

To purchase a new car, you borrow $20,000. The bank offers a 6-year loan at an interest rate of 3.25% compounded annually. If yo

Natalka [10]

Answer:

SPCA factor

Single payment compound amount factor.

Total amount pay A = $24,230.95 (Approx)

Interest paid = $4,230.95 (Approx)

Explanation:

Given:

P = $20,000

n = 6 year

r = 3.25%

Find:

Total amount pay A

Computation:

A=p(1+r)ⁿ

A=20,000[1+3.5%]⁶

A=20,000[(1.0325)⁶]

Total amount pay A = $24,230.95 (Approx)

Interest paid = $24,230.95 - 20,000

Interest paid = $4,230.95 (Approx)

6 0

3 years ago