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Dmitry [639]
3 years ago
5

Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t

o the exit, where the pressure is 40 psia. If the isentropic nozzle efficiency is 95%, determine for the nozzle,
(a) the exit velocity of the steam in ft./sec, and

(b) the amount of entropy produced in BTU/ lbm R.
Engineering
1 answer:
almond37 [142]3 years ago
3 0

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

n = \frac{h_1 - h_2*}{h_1 - h_2}

n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}

solving for h2, we have

h_2 = 1197.77 Btu/Ibm

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

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cpct gvxjjxjhdfjokjdzfjiyddzzsjhxf

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2 years ago
A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini
Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

so  

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

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Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0
3 years ago
The alignment readings for the front of a vehicle are shown above. Camber and toe are within specification, caster is not. Techn
dlinn [17]

Answer:

B. B only

Given Information:

1. Camber and toe are within specification

2. Caster is not within specification

Technician A says that with the current settings, the left front tire tread may wear on the inside edge.

Technician B says that with the current settings, the vehicle may pull to the left

Explanation:

Lets discuss the effects of Camber, toe and caster misalignment

Effects of Camber and Toe misalignment:

Camber is the inward or outward tilt of the fron tires and is used to distribute load across the tread. Any misalignment causes uneven loading on the tires which results in tire wear on one edge.

The most common cause of tire wear on the inside edge is due to the camber misalignment which results in premature tire wear.

Another reason is of tire wear is vehicle’s toe. A slight misalignment of the toe reduces the life of the tire.

Since it is given that camber settings and toe settings are within specification therefore, tire tread wear on the inside edge cannot happen if camber and toe are within specification.

Technician A cannot be right.

Effects of Caster misalignment:

Whenever there is a misalignment of the castor then the vehicle will not be able to go in straight line rather it will pull to either left or right side. Caster misalignment also causes heavy or light steering depending upon the positive or negative misalignment of caster.

Since it is given that caster settings are not within specification therefore, the vehicle may pull to the left due to the caster misalignment.

Technician B must be right.

4 0
3 years ago
A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal
Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

  V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X_{L}, which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX_{L}

X_{L} = \frac{V}{I} ................ 2

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From equation 2,

X_{L}= \frac{240}{15}

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To determine the inductance of the inductor,

X_{L} = 2\pifL

L = \frac{X_{L} }{2 \pi f}

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The inductance of the inductor is 0.051H.

4 0
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insens350 [35]

Answer:

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cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m ,  Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr  expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

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2 years ago
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