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Nonamiya [84]
2 years ago
7

Technician A says that weld-on dent removal attachments may be used on a steel panel.

Engineering
1 answer:
spin [16.1K]2 years ago
7 0

Answer:

sjasfjajfhajshfdff

Explanation:

You might be interested in
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
Bad White [126]

Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume V_2 = 0.16 V_1

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature T_1 = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

V_1-V_2 = \dfrac{\pi}{4}D^2L

V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)

V_1-0.16V_1= 36.55714291

0.84 V_1 =36.55714291

V_1 =\dfrac{36.55714291}{0.84 }

V_1 =43.52040823 \ in^3 \\ \\  V_1 = 43.52 \ in^3

V_1 = 0.02518 \ ft^3

the mass in air ( lb) can be determined by using the formula:

m = \dfrac{P_1V_1}{RT}

where;

R = 53.3533 ft.lbf/lb.R°

m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R  \times 519 .67 ^0 R}

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

v_{r1} =158.58

u_1 = 88.62 Btu/lb

At state of volume 2; the relative volume can be determined as:

v_{r2} = v_{r1}  \times \dfrac{V_2}{V_1}

v_{r2} = 158.58 \times 0.16

v_{r2} = 25.3728

The specific energy u_2 at v_{r2} = 25.3728 is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

v_{r3} = 0.1828

u_3 = 1098 \ Btu/lb

To determine the relative volume at state 4; we have:

v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}

v_{r4} =0.1828 \times \dfrac{1}{0.16}

v_{r4} =1.1425

The specific energy u_4 at v_{r4} =1.1425 is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

W_{net} = Heat  \ supplied - Heat  \ rejected

W_{net} = m(u_3-u_2)-m(u_4 - u_1)

W_{net} = m(u_3-u_2- u_4 + u_1)

W_{net} = m(1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (410.08)

\mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the  four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

W = 4 \times N'  \times W_{net

where ;

N' = \dfrac{2400}{2}

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0
2 years ago
Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find
ahrayia [7]

Answer:

A.) Find the answer in the explanation

B.) Ua = 7.33 m/s , Vb = 7.73 m/s

C.) Impulse = 17.6 Ns

D.) 49%

Explanation:

Let Ua = initial velocity of the rod A

Ub = initial velocity of the rod B

Va = final velocity of the rod A

Vb = final velocity of the rod B

Ma = mass of rod A

Mb = mass of rod B

Given that

Ma = 2kg

Mb = 1kg

Ub = 3 m/s

Va = 0

e = restitution coefficient = 0.65

The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.

Please find the attached files for the solution

6 0
3 years ago
The temperature of a system rises by 10 °C during a heating process. Express the rise in temperature of K, R, and °F.
Lorico [155]

Explanation:

Given T = 10 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (10 + 273.15) K = 283.15 K

<u>T = 283.15 K </u>

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32  

So,

T (°F) = (10 × 9/5) + 32 = 50 °F

<u>T = 50 °F</u>

The conversion of T( °C) to T(R) is shown below:

T (R) = (T (°C) × 9/5) + 491.67

So,

T (R) = (10 × 9/5) + 491.67 = 509.67 R

<u>T = 509.67 R</u>

6 0
3 years ago
I have to find the critical points of this function of two variables <img src="https://tex.z-dn.net/?f=%5C%5Cf%28x%2Cy%29%3Dx%5E
liraira [26]

Answer:

no i dont think there is

Explanation:

because theres not

4 0
3 years ago
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

5 0
3 years ago
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