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3 years ago

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You gently place several steel needles on the free surface of the water in a large tank. The needles come in two lengths: some a

re 5 cm long, and some are 10 cm long. Needles of each length are available with diameters of 1 mm, 2.5 mm, and 5 mm.
Make a prediction as to which needles, if any will float.

Assume contact angle theta = 0. (For water, rho=1000kg/m3, sigma=0.0728 N/m, for steel, SG= 0.83).

1 answer:

Stolb23 [73]3 years ago

4 0

Answer:

attached below

Explanation:

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A cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's ve

Musya8 [376]

Answer:

The surface temperature is 921.95°C .

Explanation:

Given:

a=25 cm ,P=350 hp⇒P=260750 W

Power transmitted $0.95\times 260750$ W and remaining will lost in the form of heat.This heat transmitted to air by the convection.

h=230 $\frac{W}{m^2-K},\eta =0.95$

Actually heat will be transmit by the convection.

In convection Q=hA $\Delta T$

So $P=\Delta T\times Q$

$0.05\times 260750=230\times0.25^2\(T-15)$

T=921.95°C

So the surface temperature is 921.95°C .

6 0

3 years ago

A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang

AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

$\omega =\dfrac{d\theta}{dt}$

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

$V=\sqrt{\dfrac{GM}{R}}$

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

$V=\sqrt{\dfrac{GM}{R}}$

$V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}$

V=7059.44 m/s

iii)

We know that centripetal fore given as

$F=\dfrac{mV^2}{R}$

Here given that m= 200 kg

R= 8000 km

so now by putting the values

$F=\dfrac{mV^2}{R}$

$F=\dfrac{200\times 7059.44^2}{8000\times 10^3}$

F=1245.8 N

3 0

3 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

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4 years ago

Indicates the design of the building and<br> adjoining areas

Lelu [443]

Answer:

Architectural Plan

Explanation:

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3 years ago

g A heat exchanger is designed to is to heat 2,500 kg/h of water from 15 to 80 °C by engine oil. The configuration of the heat e

sergey [27]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file gave a detailed solution of the problem.

8 0

3 years ago