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Brrunno [24]
3 years ago
9

Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa

ce of 5.4 mm (0.210 in.) is allowed between standard rails of length 11.9 m (39 ft), what is the highest possible temperature that can be tolerated without the introduction of thermal stresses
Engineering
1 answer:
DENIUS [597]3 years ago
8 0

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l )  = ∝ * ΔT

                                    ( 5.4 * 10^-3 / 11.9 )  = 12.1 * 10^-6 * ( T2 - 4 )

∴  ( T2 - 4 ) =  ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

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Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

6 0
3 years ago
A ladder logic program and the associated physical input/output components are given below. Lighting changes from full darkness
Katena32 [7]

Answer:

O2 is true.

Explanation:

8 0
4 years ago
Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The cha
Elza [17]

Answer:

V = 1.5062 m/s

Explanation:

look to the photos

5 0
3 years ago
Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload
Vinil7 [7]

Answer:

1) V_o = 10 liters

2) V_o = 12.26 liters

Explanation:

For isothermal process n =1

V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}

V_o = 10 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.03

actual \ volume = c1\times 10 = 10.3 liters

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.15

actual \volume = c1\times 10 = 11.5 liters

8 0
3 years ago
Calculate the density of a hydraulic oil in units of kg/m^3 knowing that the density is 1.74 slugs/ft^3. Then, calculate the spe
Amanda [17]

Answer:

Density of oil will be 897.292 kgm^3

And specific gravity of oil will be 0.897

Explanation:

We have given density of oil is 1.74 slugs/ft^3

We have to convert this slugs/ft^3 into kg/m^3

We know that 1 slugs = 14.5939 kg

So 1.74 slug = 1.74×14.5939 = 25.3933 kg

And 1 cubic feet = 0.0283 cubic meter

So 1.74slug/ft^3=\frac{1.74\times 14.5939kg}{0.0283m^3}=897.292kg/m^3

Now we have to calculate specific gravity it is the ratio of density of oil and density of water

We know that density of water = 1000 kg/m^3

So specific gravity of water =\frac{897.292}{1000}=0.897

7 0
4 years ago
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