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Anna11 [10]
2 years ago
5

The value of Rth is *

Engineering
1 answer:
Vladimir79 [104]2 years ago
4 0
Johnjjjjjjhhhhhhhhjjjjjjjjjjj
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Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values
Nataliya [291]

Answer:

Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.

Explanation:

Solution

Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating  pressure shortens the hose life.

In systems where pressure peaks are severe, select or pick a hose with higher maximum operating  pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.

Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.

7 0
3 years ago
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws
ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\

E=52000Hp.h

E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\

E=38724920Wh

E=52000Hph(\frac{1977378.4  ft lb}{1Hph}

E=1.028x10^11 ftlb

3 0
3 years ago
______________ help protect the lower legs and feet from heat hazards like molten metal and welding sparks.
astraxan [27]

Answer:

leggings

Explanation:

they allow the metal or sparks to not hurt you can the leggings can be easily and quickly removed.

8 0
3 years ago
Consider an infinitely thin flat plate of chord c at an angle of attack α in a supersonic flow. The pressure on the upper and lo
amm1812

Answer:

X_cp = c/2

Explanation:

We are given;

Chord = c

Angle of attack = α

p u (s) = c 1

​p1(s)=c2,

and c2 > c1

First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.

I've attached the solution

4 0
3 years ago
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
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