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Citrus2011 [14]
2 years ago
9

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part

. The gas-turbine cycle of a combined gas–steam power plant has a pressure ratio of 8. Air enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 17 kg/s. Assume isentropic efficiencies of 100 percent for the pump, 85 percent for the compressor, and 90 percent for the gas and steam turbines.
Determine the rate of total heat input.
Engineering
1 answer:
ra1l [238]2 years ago
4 0

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

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prove that the heat transfer at the constant pressure is given by the enthalpy change during the process​
mihalych1998 [28]

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

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5 0
3 years ago
In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain
swat32

Answer:n=0.973

Explanation:

Given

When True strain\left ( \epsilon _T_1\right )=0.171

at \sigma _1=263.8 MPa

When True stress\left ( \sigma _2\right )=346.2 MPa

true strain \left ( \epsilon _T_2\right )=0.226

We know

\sigma =k\epsilon ^n

where \sigma=True stress

\epsilon=true strain

n=strain hardening exponent

k=constant

Substituting value

263.8=k\left ( 0.171\right )^n------1

346.2=k\left ( 0.226\right )^n-----2

Divide 1 & 2 to get

\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n

1.312=\left ( 1.3216\right )^n

Taking Log both side

ln\left ( 1.312\right )=nln\left ( 1.3216\right )

n=0.973

6 0
3 years ago
A long corridor has a single light bulb and two doors with light switch at each door.
Zigmanuir [339]

Answer:

The answer is below

Explanation:

Let A represent the first switch, B represent the second switch and C represent the bulb. Also, let 0 mean  turned off and 1 mean turned on. Since when both switches are  in the same position, the light is off. This can be represented by the following truth table:

A                    B                       C (output)

0                    0                        0

0                    1                          1

1                     0                         1

1                     1                          0

The logic circuit can be represented by:

C = A'B + AB'

The output (bulb) is on if the switches are at different positions; if the switches are at the same position, the output (bulb) is off. This is an XOR gate. The gate is represented in the diagram attached below.

6 0
3 years ago
If you are convicted of D.U.I. a second time in five years, your license may be revoked for up to __________ year/s.
LuckyWell [14K]
If you are convicted of a dui for a second time in five years, your license may be revoked for up to (2) years
5 0
3 years ago
The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve
stellarik [79]

Answer:

Explanation:

Given

velocity at A is v=4\ ft/s

For r=500\ ft

velocity is increasing at \dot{v}=0.004t\ ft/s^2

Tangential acceleration is given by

a_t=\frac{\mathrm{d} v}{\mathrm{d} t}

a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}

\int 0.004tdt=\int dv

\int dv=\int 0.004tdt

v=0.002t^2+c

at t=0\ v=4\ ft/s

4=0.002\cdot 0+c

c=4\ ft/s

thus v=0.002t^2+4

Velocity in terms of Displacement is given by

v=\frac{\mathrm{d} s}{\mathrm{d} t}

\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt

\Rightarrow s=\frac{0.002t^3}{3}+4t

When car has traveled \frac{3}{4} th of distance i.e.

s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}

s=750\pi

750\pi =\frac{0.002t^3}{3}+4t

\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0

on solving we get t=139.23\ s

Thus velocity at t=139.23\ s

v=42.76\ s

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}

a_n=3.658\ m/s^2

Tangential acceleration a_t at t=139.23\ s

a_t=0.556\ m/s^2

Net acceleration a_t=\sqrt{(a_n)^2+(a_t)^2}

a_n=\sqrt{(3.658)^2+(0.556)^2}

a_n=3.7\ m/s^2

   

8 0
3 years ago
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