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Nezavi [6.7K]
2 years ago
8

The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-

stress distribution along a radial line of the shaft where the shear stress is maximum. Set T1
Engineering
1 answer:
jarptica [38.1K]2 years ago
5 0

Answer:

Max shear = 8.15 x 10^7 N/m2

Explanation:

In order to find the maximum stress for a solid shaft having radius r, we will be applying the Torsion formula which can be written as;

Allowable Shear Stress = Torque x Radius / pi/2 x radius^4

Putting the values we have;

T = 2000 N/m

Radius = Diameter/2 = 0.05 / 2 = 0.025 m

Putting values in formula;

Max shear = 2000 x 0.025 / 3.14/2 x (0.025)^4

Max shear = 8.15 x 10^7 N/m2

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A steam power plant operates on an ideal Rankine cycle with two stages of reheat and has a net power output of 120 MW. Steam ent
uysha [10]

Answer:

a) 40.6%

b)72.19kg/s

Explanation:

The Rankine cycle with two reheat stages has 9 stages in total.

The maximum pressure will be at the first inlet stage of the HP turbine which is stage 3. The minimum pressure will be the exit stage of the condenser because the condenser operates under vacuum pressure which is stage 1.

The following assumptions can be made:

1 - Each component in the cycle is analyzed as an open system operating at steady-state.

  2 - All of the processes are internally reversible.

  3 - The turbine and pump operate adiabatically and are internally reversible, so they are also isentropic.

   4 - Condensate exits the condenser as saturated liquid.

  5 - The effluent from the HP turbine is a saturated vapor.

  6 - No shaft work crosses the system boundary of the boiler or condenser.

   7 - Changes in kinetic and potential energies are negligible

a) The thermal efficiency of the cycle is defined as the work of the cycle divided by the total heat input to the system. The stages that have heat input is stages 2-3, 4-5, 6-7.

For stage 2:

s₁=s₂ assuming isentropic

s_1=0.4762 @ P_1=15MPa

enthalpy will be a compressed liquid so after interpolation

h_2=97.93+(0.4762-0.2932)((180.77-97.93)/(0.5666-0.2932))=153.38kJ/kg

For stage 3:

Superheated steam @ T=500⁰C and P=15MPa

h_3=3310.8kJ/kg

Stage 4:

superheated vapor

P=5MPa

s₃=s₄=6.3480 kJ/kg, we must use interpolation to find h₄

h_4=2925.7-(6.348-6.2111)((3069.3-2925.7)/(6.4516-6.2111))=3007.44kJ/kg

Stage 5:

Superheated steam @ T=500⁰C and P₄=P₅=5 MPa

h_5=3434.7kJ/kg

Stage 6:

Superheated steam at P₆= 1MPa

s₅=s₆

s_6=6.9781

We find h₆ using interpolation from the steam tables:

h_6=2943.1-(6.9781-6.9265)((3051.6-2943.1)/(7.1246-6.9265))=2970.67kJ/kg

Stage 7:

P₇=P₆=1MPa

T=500⁰C superheated steam

h_7=3479.1kJ/kg

The heat into the cycle is:

=(h_3-h_2)+(h_5-h_4)+(h_7-h_6)

=(3310.8-153.38)+(3434.7-3007.44)+(3479.1-2970.67)=4108.74kJ/kg

We can determine the work out by the condenser from stage 9 to stage 1:

Stage 1:

saturated liquid P=5kPa

h_1=137.75kJ/kg

Stage 9:

We assume that its a saturated liquid with quality of 1 at 5kPa and

s₇=s₉ and after interpolation

h_9=2568.53kj/kgK

Qout = [/tex]2568.53-137.75=2430.79kJ/kg[/tex]

The thermal efficiency can be written in terms of qin and qout:

n=1-(q_o/q_i)=1-2430.79/4093.11=0.4061

Efficiency of 40.61%

b)

The mass flow rate can be calculated from the Wnet:

W_n=W_t-W_p

Work of the turbines minus the work of the pumps:

W_n=m((h_3-h_4)+(h_5-h_6)+(h_7-h_9)-(h_1-h_2)

120000=m(1662.33)

m=72.19

mass flow rate of steam is 72.19 kg/s

7 0
3 years ago
A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for paramet
Phoenix [80]

Answer:

(a) - A12 = A21 = 2.747

(b) - A12 = 2.148; A21 = 2.781

(c)-  A12 = 2.781; A21 = 2.148

Explanation:

(a) - x1(a) = 0.1 |  x2(a) = 0.9 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

  • x1(a)*γ1(a) = x1(b)γ1(b)

        x2(a)*γ2(a) = x2(b)γ2(b)

  • A12 = A21 = 2.747

(b) -  x1(a) = 0.2 |  x2(a) = 0.8 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

  • x1(a)*γ1(a) = x1(b)γ1(b)

        x2(a)*γ2(a) = x2(b)γ2(b)

  • A12 = 2.148; A21 = 2.781

(c) -  x1(a) = 0.1 |  x2(a) = 0.9 | x1(b) = 0.8 | x2(b) = 0.2

LLE equations:

  • x1(a)*γ1(a) = x1(b)γ1(b)

        x2(a)*γ2(a) = x2(b)γ2(b)

  • A12 = 2.781; A21 = 2.148
7 0
3 years ago
Which of the following characteristics would not give animals an advantage in the ocean?
Taya2010 [7]
I believes it’s long body hair
3 0
2 years ago
Describe a gear train that would transform a counterclockwise input rotation to a counterclockwise output rotation where the dri
masya89 [10]

Answer:

For a gear train that would train that transform a counterclockwise input into a counterclockwise output such that the gear that is driven rotates three times when the driver rotates once, we have;

1) The number of gears in the gear train = 3 gears with an arrangement such that there is a gear in between the input and the output gear that rotates clockwise for the output gear to rotate counter clockwise

2) The speed ratio of the driven gear to the driver gear = 3

Therefore, we have;

Speed \ Ratio =\dfrac{Speed \ of \ Driven \ Gear}{Speed \ of \ Driver \ Gear} = \dfrac{The \ Number \ of \ Teeth \ of \ Driver \ Gear}{The \ Number \ of \ Teeth \ of \ Driven \ Gear}

Therefore, for a speed ratio of 3, the number of teeth of the driver gear, driving the output gear, must be 3 times, the number of teeth of the driven gear

Explanation:

3 0
3 years ago
In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t
Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0
3 years ago
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