Ask question

Ask question

All categories

3 years ago

12

Several applications are listed below. Determine the relative importance of the resilience and toughness of the materials chosen

for each application. Sort the items based on whether resilience is most important, toughness is most important, or both are equally important.
a. a non-critical spring that is used repeatedly
b. a high tension music wire
c. a one use safety device that absorbs impact energy
d. a burst panel designed to rupture at
e. a certain pressure
d. a structural member in a building

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

they are important together, but if you want to use just one future you must think about which one is first needed. and then try to learn for economical so don't use more money

You might be interested in

Thermodynamics deals with the macroscopic properties of materials. Scientists can make quantitative predictions about these macr

Arlecino [84]

Answer:

Explanation:

From the information given:

a.

Using the equipartition theorem, the average energy of a molecule dor each degree of freedom is:

$U = \dfrac{1}{2}k_BT$

$U = \dfrac{1}{2}nRT$

For s degree of freedom

$U = \dfrac{1}{2}snRT$

However, the molar specific heat $C_v = \dfrac{1}{n} \dfrac{dU}{dT}$

Therefore, in terms of R and s;

$C_v = \dfrac{1}{n} \dfrac{d}{dT} \begin{pmatrix} \dfrac{1}{2} snRT \end {pmatrix}$

$C_v = \dfrac{Rs}{2}$

b.

Given that:

Cv=70.6Jmol⋅K and R=8.314Jmol⋅K

Then; using the formula $C_v = \dfrac{Rs}{2}$

$70.6 \ J/mol.K = \dfrac{(8.314 \ J/mol.K)\times s}{2}$

$70.6 \ J/mol.K \times 2= (8.314 \ J/mol.K)\times s$

$s= \dfrac{70.6 \ J/mol.K \times 2}{ (8.314 \ J/mol.K) }$

s = 16.983

s $\simeq$ 17

7 0

3 years ago

g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is

boyakko [2]

Answer:

a) $\dot W = 0.417\,kW$ , b) $COP_{R} = 2.198$ , c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

$\dot W = \dot Q_{H} - \dot Q_{L}$

$\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)$

$\dot W = 0.417\,kW$

b) The Coefficient of Performance of the refrigerator is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}$

$COP_{R} = 2.198$

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

$COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}$

$COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}$

$COP_{R,ideal} = 6.218$

The refrigeration cycle is irreversible, as $COP_{R} < COP_{R,ideal}$ .

3 0

3 years ago

A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for paramet

Phoenix [80]

Answer:

(a) - A12 = A21 = 2.747

(b) - A12 = 2.148; A21 = 2.781

(c)- A12 = 2.781; A21 = 2.148

Explanation:

(a) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = A21 = 2.747

(b) - x1(a) = 0.2 | x2(a) = 0.8 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = 2.148; A21 = 2.781

(c) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.8 | x2(b) = 0.2

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = 2.781; A21 = 2.148

7 0

4 years ago

If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas

jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ( $\eta$ ), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$ , then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.

8 0

3 years ago

Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,

AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure ' also doubles from ,

Thus,Highest osmotic pressure that membrane may experience is, '=2*

=2*29.319 atm

' =58.638 atm

3 0

3 years ago