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3 years ago

1. You use

Engineering

1 answer:

lorasvet [3.4K]3 years ago

4 0

4-ways tell me if I’m wrong

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How does load transfer of space needle

UkoKoshka [18]

Answer:

The Space Needle is a cut away with minimal residual deflection due to load transfer.

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2 years ago

The less surface area of gasoline exposed to the air, the faster a given amount will burn. True of False

romanna [79]

Answer:

true

Explanation:

it depends on how fast the car goes because the gas burns faster or slower depending on the speed you go or drive. if your drive 50 MPH, the gas will burn slowly. if your drive 45 MPH, the gas will burn faster if you go slow too much. if you press the glass button and hold it for 5 minutes, the more you hold the gas for too long, the more the gas will burn inside the gas engine. go look at your car or your parents car and see how it goes and that will help.

4 0

2 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

3 years ago

2. The block is released from rest at the position shown, figure 1. The coefficient of

denis23 [38]

Answer:

Velocity = 4.73 m/s.

Explanation:

Work done by friction is;

W_f = frictional force × displacement

So; W_f = Ff * Δs = (μF_n)*Δs

where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24

Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;

mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).

Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;

ΔKE2 + (0.16)(mg cos 24)(2).

Plugging in the relevant values, we have;

1.22mg = ΔKE1 + 0.603mg

ΔKE1 = 1.22mg - 0.603mg

ΔKE1 = 0.617mg

Also,

0.813mg = ΔKE2 + 0.292mg

ΔKE2 = 0.813mg - 0.292mg

ΔKE2 = 0.521mg

Now total increase in Kinetic Energy is ΔKE1 + ΔKE2

Thus,

Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg

Putting 9.81 for g to give;

Total increase in kinetic energy = 11.164m

Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m

m cancels out to give; ½v² = 11.164

v² = 2 × 11.164

v² = 22.328

v = √22.328

v = 4.73 m/s.

5 0

3 years ago

Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −

sergij07 [2.7K]

Answer:

Given data:

Equation of the state $p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }$

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid $m^{3} / k g$

R = gas constant , $j/k g k$

a, b = Constants

Solution:

Specific heat difference, $\begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}$

According to cyclic reaction

$\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}$

Hence specific heat difference is

$c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}$

Equation of state, $p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}$

Differentiating the equation of state with respect to temperature at constant volume,

$\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}$

$\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}$

Differentiating the equation of the state with respect to volume at constant temperature.

$\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)$

Substituting both eq (3) and eq (4) in eq (2)

We get,

${cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

Specific heat difference equation,

$c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

7 0

3 years ago