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Scilla [17]
3 years ago
5

The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm

and 6 mm pitch.
force required at the rim of a 300mm diameter hand wheel, if there is a collar
bearing of 50 mm mean diameter provided in the arrangement to exert axial
load. Assume the coefficient of friction for the collar as 0.2.

Engineering
1 answer:
kykrilka [37]3 years ago
4 0

Answer:

See calculation below

Explanation:

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

<u><em>1. Force required at the rim of handwheel </em></u>

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d = \frac{do + dc}{2} = (75 + 69) / 2 = 72 mm

and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

∴ Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T =  W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

<u><em>2. Maximum compressive stress in the screw</em></u>

                         30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

                      π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

<u><em>3. efficiency of the straightener</em></u>

<u><em></em></u>

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

∴Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

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A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and gauge pressure of 300 kPa. The gas is h
Sergio [31]

Answer:

the final temperature is 77.1 °C

Explanation:

Given the data in the question;

Initial temperature; T₁ = 27°C = ( 27 + 273)K = 300 K

Initial absolute pressure P₁ = 300 kPa = ( 300 + 101.325 )kPa = 401.325 kPa

Final absolute pressure P₂ = 367 kPa = ( 367 + 101.325 )kPa = 468.325 kPa

Now, to calculate the final temperature, we use the ideal gas equation;

P₁V/T₁ = P₂V/T₂

but it is mentioned that the rigid tank is closed,

so the volume is the same both before and after.

Change in volume = 0

hence;

P₁/T₁ = P₂/T₂

we substitute

401.325 kPa / 300 K = 468.325 kPa / T₂

T₂ × 401.325 kPa  = 300 K × 468.325 kPa

T₂ = [ 300 K × 468.325 kPa ] / 401.325 kPa

T₂ = 140497.5 K / 401.325

T₂ =  350.08 K

T₂ = ( 350.08 - 273 ) °C

T₂ = 77.1 °C

Therefore, the final temperature is 77.1 °C

3 0
3 years ago
True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'
defon

Answer:

true

Explanation:

8 0
2 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
3 years ago
Construction support involves mostly what kind of work?
Paraphin [41]

Answer:

Outdoors

Explanation:

Construction workers perform outdoors.

6 0
2 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
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