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Mars2501 [29]
3 years ago
14

A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surfac

e is 5 m. The atmospheric pressure is 95 kPa. Considering the atmospheric pressure, the hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.

Engineering
1 answer:
UkoKoshka [18]3 years ago
3 0

Answer:

864 KN

Explanation:

Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.

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

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A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of
yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

V_d=2.5\times 10^{-3} m^3

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0
3 years ago
Situation: Peter is designing a new hybrid car that functions on solar power. He is currently working on sketches of his design
givi [52]

Answer:

whats the question?

Explanation:

8 0
2 years ago
(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
g A primary sedimentation basin is designed for an average flow of 0.3 m3/s. The TSS concentration in the influent is 240 mg/L.
yarga [219]

Answer:

(a) 0.243 m3/day

(b) 96 mg/l

(c) 0.426 m3/min

Explanation:

The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240= 9.6 mg/L

Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day

To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3

Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day

Approximately, the volume of sludge is 0.243 m3/day.

(b)

Efficiency of 60 percent is equivalent to 0.6

Efficiency=(influent concentration- flow rate)/influent concentration

0.6=(240-flow rate)/240

Flow rate= 96 mg/l

(c)

Cycle time= 0.243/0.57=0.4263157894736 m3/min

Rounded off, cycle time is 0.426 m3/day

3 0
3 years ago
Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is
frozen [14]

Answer:

x = 0.944

Explanation:

Steam at outlet is an superheated steam, since T > T_{sat}. From steam tables, the specific enthalpy is:

h_{out}=2682.4\,\frac{kJ}{kg}

The throttle valve is modelled after the First Law of Thermodynamics:

h_{in} = h_{out}

Hence, specific enthalpy at inlet is:

h_{in}=2682.4\,\frac{kJ}{kg}

The quality in the steam line is:

x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }

x = 0.944

6 0
3 years ago
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