9. A piece of Cherry wood is 5/4 x 4" X 4'<br> What is the length in inches?

The fins attached to a surface are determined to have an effectiveness of 0.9. Do the rate of heat transfer from the surface dec

Nesterboy [21]

Answer: The rate of heat transfer decreases.

Explanation:

Fin effectiveness is defined as the ratio of heat transfer rate from a finned surface to the heat transfer rate from the same surface if there were no fins. Its value is expected to be greater than 1.

Having effectiveness value of 0.9 which is less than 1 indicates that the heat transfer rate will decrease since a fin effectiveness smaller than 1 shows that the fin acts as insulation (thermal insulation).

8 0

3 years ago

Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati

Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0

3 years ago

Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running

Nikolay [14]

Answer:

Explanation:

The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.

Round Robin:

we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

A B C D E A B C* D E A B D E A B D* E A B E A B* E A E A E* A A

C* = 8

D*=17

B*=23

E*=28

AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES

B) PRIORITY SCHEDULING:

1-6 7-14 15-24 25-26 27-30

B E A C D

AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.

C)FCFS

1-10 11-16 17-18 19-22 23-30

A B C D E

AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES

D)SJF

1-2 3-6 7-12 13-20 21-30

C D B E A

AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.

3 0

3 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago

5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h

frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

V2max = √2 * 9.81 * 3

V2max = √58.86 = 7.67 m/s

maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

Qmax = 0.00071 * 7.67 = 0.00545 m³/s

Qmax = 5.45 L/s

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Actual flow rate will be less because of heads such as friction head and pressure head.

7 0

3 years ago

9. A piece of Cherry wood is 5/4 x 4" X 4' What is the length in inches?