'Energy' has the potential to:

A closed tank contains ethyl alcohol to a depth of 66 ft. Air at a pressure of 23 psi fills the gap at the top of the tank. Dete

irina1246 [14]

Answer:

639.4psi

Explanation:

Pressure at the closed valve = Air pressure+ (density*gravity*height)

=23psi+(49.27lb/ft^3*32.17ft/s^2*56ft^3)

=23psi+88760.89psft

=23psi+(88760.89/144)psi

=23+616.4

=639.4psi

3 0

3 years ago

All of these are true about steel EXCEPT that:

natta225 [31]

Answer:B) heat tends to strengthen high-strength steel.

Explanation: You are right it is B

3 0

2 years ago

Copper is generally considered easy to weld because of its high thermal conductivity: (a) true or (b) false?

inn [45]

Answer:

b)False

Explanation:

Those material have high thermal it is very difficult to weld because due to high thermal conductivity it transmit the heat in to the surrounding and can not reach a particular temperature required to melt the material.And when material does not melt then there is no possibility to weld the material.

So from above we can say that it is very difficult to weld the copper material due to high thermal conductivity.Generally welding of copper done by usiong gas welding technique.

So our option b is right.

8 0

3 years ago

A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti

Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = $\frac{0.03194}{0.315}$

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m

here h is heat required and m is mass flow rate

H = 2680.36 × 0.10139

H = 271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = $\frac{H}{0.90}$

net boiler heat = $\frac{271.75}{0.90}$

net boiler heat = 301.94 kW

5 0

3 years ago

A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e

mash [69]

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³ mm), it is necessary to calculate the change in diameter Δd for these four alloys.

Then we use the equation υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒ υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.

For the Steel alloy

∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm

0.005 mm < 7.5×10 ⁻³ mm

Therefore, the steel is a candidate.

For the Titanium alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm

0.0111 mm > 7.5×10 ⁻³ mm

Hence, the Titanium alloy is not a candidate.

7 0

3 years ago