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yarga [219]
4 years ago
15

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b

uilding are 675 and 695 mm Hg, respectively, determine the height of the building in m. Take the densities of air and mercury to be 1.18 kg/m3 and 13,600 kg/m3.
Engineering
1 answer:
Levart [38]4 years ago
8 0

Answer:

230.51 m

Explanation:

Pb = 695 mmHg

Pt = 675 mmHg

Pb - Pt = 20 mmHg

Calculate dP:

dP = p * g * H = (13600)*(9.81)*(20/1000) = 2668.320 Pa

Calculate Height of building as dP is same for any medium of liquid

dP = p*g*H = 2668.320

H = 2668.32 / (1.18 * 9.81) = 230.51 m

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Tech A says you can find the typical angle of a V-block engine by dividing the number of cylinders by 720
Lady_Fox [76]

Answer:

Tech A is correct

Explanation:

Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.

Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.

8 0
3 years ago
A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature
mafiozo [28]

Answer:

a)31%

b)34MW

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=<u>34.077MW</u>

6 0
4 years ago
A 220-V electric heater has two heating coils that can be switched such that either coil can be used independently or the two ca
Nana76 [90]

Answer:

The resistances of both coils are 131.7 Ω and 29.64 Ω.

Explanation:

Since, there are two coils, they can be used independently or in series or parallel. The power is given as:

Power = P = VI

but, from Ohm's Law:

V = IR

I = V/R

therefore,

P = V²/R

R = V²/P

Hence, the resistance (R) and (P) are inversely proportional. Therefore, the maximum value of resistance will give minimum power, that is, 300 W. And the maximum resistance will be in series arrangement, as in series the total resistance gets higher than, any individual resistance.

Therefore,

Rmax = V²/Pmin = R1 + R2

R1 + R2 = (220 V)²/300 W

R1 + R2 = 161.333 Ω    ______ en (1)

Similarly, the minimum resistance will give maximum power. And the minimum resistance will occur in parallel combination. Because equivalent resistance of parallel combination is less than any individual resistance.

Therefore,

(R1 R2)/(R1 + R2) = (220 V)²/2000 W

using eqn (1), we get:

(R1 R2) / 161.333 Ω = 24.2 Ω

R1 R2 = 3904.266 Ω²

R1 = 3904.266 Ω²/R2  _____ eqn (2)

Using this value of R1 in eqn (1), we get:

3904.266/R2 +R2 = 161.333

(R2)² - 161.333 R2 +3904.266 = 0

Solving this quadratic eqn we get two values of R2 as:

R2 = 131.7 Ω     OR     R2 = 29.64 Ω

when ,we substitute these values in eqn (1) to find R1, we get get the same two values as R2, alternatively. This means that the two coils have these resistance, and the order does not matter.

<u>Therefore, the resistance of both coils are found to be 131.7 Ω and 29.64 Ω</u>

7 0
3 years ago
A rocket is to be designed to produce 5 MN of thrust at sea level. The pressure in the combustion chamber is 7 MPa and the tempe
tiny-mole [99]

Answer:

a) specific impulse = 203 s

(b) mass flow rate = 2511 kg/s

(c) throat diameter = 0.774 m

(d) exit diameter = 1.95 m

(e) thrust =  5.3*10^6 N/m^2

(f) thrust at sea level = 15.6*10^6 N

(g) thrust with hydrogen = 5*10^6 N

(h) thrust with stagnation =  5*10^6 N

Explanation:

See the attached file for explanation.

7 0
4 years ago
Question 9 Two cells are connected in parallel. All cells are rated 1.5 V at 10 A. Et = ? (V)​
Eddi Din [679]
Parallel hookups increase amp hour capacity but voltage remains the same... 1.5 volts

Series the two and voltage is 3.0 volts.
7 0
4 years ago
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