Whats the best used for cable -stayed bridge

An isentropic steam turbine processes 5.5 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100°C. Five percent of this f

bija089 [108]

Answer:

The answer is 1823.9

Explanation:

Solution

Given that:

m = 5.5 kg/s

= m₁ = m₂ = m₃

The work carried out by the energy balance is given as follows:

m₁h₁ = m₂h₂ +m₃h₃ + w

Now,

By applying the steam table we have that<

p₃ = 50 kPa

T₃ = 100°C

Which is

h₃ = 2682.4 kJ/KJ

s₃ = 7.6953 kJ/kgK

Since it is an isentropic process:

Then,

p₂ = 500 kPa

s₂=s₃ = 7.6953 kJ/kgK

which is

h₂ =3207.21 kJ/KgK

p₁ = 3HP0

s₁ = s₂=s₃ = 7.6953 kJ/kgK

h₁ =3854.85 kJ/kg

Thus,

Since 5 % of this flow diverted to p₂ = 500 kPa

Then

w =m (h₁-0.05 h₂ -0.95 )h₃

5.5(3854.85 - 0.05 * 3207.21 - 0.95 * 2682.4)

5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)

5.5 ( 123363249.32 -0.95 * 2682.4)

w=1823.9

3 0

3 years ago

The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh

lapo4ka [179]

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, $N_{A}$ = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Now we know that for maximum velocity,

$\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }$

$\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }$

$N_{B}$ = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Therefore moment of inertia of flywheel, I = m. $k^{2}$

=30 X $0.1^{2}$

= 0.3 kg- $m^{2}$

Now torque on the output shaft

T₂ = I x ω

= 0.3 X 1064.2 rpm

= $0.3\times \frac{2\pi \times 1064.1}{60}$

= 33.429 N-m

Torque on the Shaft B is 33.429 N-m

4 0

3 years ago

A PMOS device with VT P = −1.2 V has a drain current iD = 0.5 mA when vSG = 3 V and vSD = 5 V. Calculate the drain current when:

Ksju [112]

The answer is b ! Hope I helped

7 0

2 years ago

Assuming that the following three variables have already been declared, which variable will store a Boolean value after these st

Yuki888 [10]

Answer:

C

Explanation:

Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.

5 0

3 years ago

Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in

Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time = ∑ $(\frac{Clock cyles}{Clock rate})$

Clock cycles can be determined using following formula

Clock cycles = ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

Clock cycles = ( 50 x $10^{6}$ x 1) + ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

= $\frac{512(10^{6}) }{2(10^{9}) }$

Initial execution Time = 256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

( $\frac{Clockcycles}{2}$ )= ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

$CPI_{FP improved}$ x No. FP instructions = ( $\frac{Clockcycles}{2}$ ) -[ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = ( $\frac{512(10)^{6} }{2}$ ) - [ ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = - 206 x $10^{6}$

$CPI_{FP improved}$ = - 206 x $10^{6}$ / 50 x $10^{6}$

$CPI_{FP improved}$ = - 4.12 < 0

3 0

3 years ago