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Svetradugi [14.3K]
4 years ago
6

A load P is applied horizontally while the other end is fixed to a structure. A load P is applied horizontally while the other e

nd is fixed to a structure. How many bolts are required to make the assembly safe? Use 8 mm bolts (M8-1.25) that yield at 0.5 MPa. (You must provide an answer before moving to the next part.)
Engineering
1 answer:
sveta [45]4 years ago
8 0

Answer:

The number of bolt is 4.

Explanation:

Given that,

\tau = 0.5\ MPa

distance = 8 mm

Suppose, Load P = 205 N

We need to calculate the shear area of bolt

Using formula of shear stress

\tau=\dfrac{P}{A}

Here is double shear.

So, \tau will be double.

A=\dfrac{P}{2\tau}

Put the value into the formula

A=\dfrac{205}{2\times0.5\times10^{6}}

A=205\times10^{-6}\ m^2

We need to calculate the area of one bolt

Using formula of area

A'=\dfrac{\pi}{4}d^2

Put the value into the formula

A'=\dfrac{\pi}{4}\times(8\times10^{-3})^2

A'=50.26\times10^{-6}\ m^2

We need to calculate the number of bolt

Using formula of number of bolt

n=\dfrac{A}{A'}

Put the value into the formula

n=\dfrac{205\times10^{-6}}{50.26\times10^{-6}}

n=4.0

Hence, The number of bolt is 4.

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b. 5.6

c. 48

Explanation:

∆U= 120,000 KJ/h

Since

1 day =  24 hrs

14 days =24 x  14 hrs

14 days = 336 hrs

∆U = 120000 x 336 KJ

=40320000KJ

K = 1000

M = 1000000

∆U = 403200 MJ

Work done

W= 2000 KW.h    ( 1 h = 3600 s)

W= 7200 MJ

According to the first law of thermodynamics

∆U = Q+W

Q= 40320 - 7200 MJ

1)Qa=33120 MJ

Coefficient of performance

COP = ∆U/W

COP= 40320 / 7200

2)COP = 5.6

3)COP of ideal heat pump

Th/(Th - Tl)

Th = 15°C

Tl = 9°C

Convert Celsius to Kelvin

273 + 15 = 288

288/(15-9)

288/6

48

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A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associat
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The question is incomplete! Complete question along with answers and explanation is provided below.

Question:

A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.

a) Draw the circuit

b) Calculate the power drawn by the lamp

c) Calculate the apparent power

d) Calculate the power factor

e) Calculate the reactive power

f) Calculate the size of the capacitor necessary to provide unity power factor correction

Explanation:

a) draw the circuit

Refer to the attached image.

As you can see in the attached drawing, it is a series circuit containing  two resistors and one inductor.

In a series circuit, current remains same throughout the circuit

The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.

The current flowing in the circuit can be found by ohm's law

 I = V/Z

where V is the voltage and Z is the total impedance of the circuit

 Z = R + XL

where  XL is the inductive reactance

XL = j2 π f L

XL = j2*π*60*0.9

XL = j339.29Ω

Total resistance is

R =200 + 80 = 280 Ω

Total impedance is

Z = 280 + j339.29 Ω

b) Calculate the power drawn by the lamp

First calculate the current

I = V/Z

I = 120/(280 + j339.29)

I = 0.272<-50.46° A  (complex notation)

P = I²R

P = (0.272)²200

P ≈ 15 W

Power drawn by the circuit

P=V*I*cos(50.46°)

P=20.77 W

c) Calculate the apparent power

A = VI*

A = 120*0.272<50.46°

A = 32.64<50.46° VA

d) Calculate the power factor

PF = cos(50.46)

PF = 0.63

e) Calculate the reactive power

Q = VIsin(50.46)

Q = 120*0.272<-50.46*sin(50.46)

Q = 25.13<-50.46  VAR

f) Calculate the size of the capacitor necessary to provide unity power factor correction

The required reactive compensation power is

Qc = P (tan(old) - tan(new))

Qc = 20.77 (tan(50.46) - tan(0))

Qc = 25.16 VAR

C = Qc/2πfV²

C = 25.16/2*π*60*120²

C = 4.63 uF

Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.

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