The inverted U-tube is used to measure the pressure difference between two points A and B in an inclined pipeline through which water flows. The differenceof level h=0.4m ; X=0.2; and Y=0.3m. Calculate the pressure difference between point B and A if the top of the manometer is filled with:
i) Air
ii) paraffin of relative density of 0.75

Answers

Answer 1

Answer:

i) 0.610 m or 610 mm

ii) 0.4 m or 400 mm

Explanation:

The pressure difference between the pipes is

a) Air

Pa + πha +Ha = Pb + πhb +Hb

Pa - Pb = π(hb-ha) + Hb-Ha

Relative density of air = 1.2754 kg /m3

Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm

b) paraffin of relative density of 0.75

Pa - Pb = π(hb-ha) + Hb-Ha

Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm


Related Questions

Stefan Lano needs displays that will show the musical instrument inventory in his chain of music stores that caters to musicians playing in world-class symphony orchestras in Basel, Switzerland; Buenos Aires, Argentina; and Philadelphia and New York in the United States. Design a form-fill interface for the inventory control of musical instruments in all four stores that could be used on a computer display screen. Assume that English will be the interface language.

Answers

Answer:

attached below is an example of  the form

Explanation:

This type of form-fill can be described/developed based on the requirements or information needed by the organization in order to perfectly fulfill an order  and also to retain customers by making the form very easy and interactive attached below is an example of such form fill  interface

A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 19,815 lb, a wingspan of 53.3 ft, and an average chord length of 6 ft. The Oswald efficiency factor is 0.81 and the zero-lift drag coefficient is equal to 0.02. The jet has two turbofan engines, each producing a maximum thrust of 3,650 lb at sea level.

Required:
a. Create a plot of the drag polar for this aircraft for CL from 0 to 5. Plot CL on the vertical axis, CD on the horizontal axis, and do not include negative CL values.
b. What is the total drag coefficient at the flight condition described above?
c. What is the required thrust for level flight at this altitude in lb?
d. If the pilot runs the engines at maximum thrust, what is the instantaneous rate of climb at this altitude and velocity?

Answers

Answer:

a) attached below

b) 0.0337

c) 2730.206 Ib

d) 2320.338 ft/min

Explanation:

a) Plot of the drag polar for this aircraft

first we will calculate :

Wing area (s) = Wing span (b) * Average chord length(c)

                       = 53.3 * 6 = 319.8 ft^2

Aspect ratio =  b^2 / s = 8.883

K = 1 / [tex]\pi[/tex]eAR = 1 /

Drag polar ( Cd ) = 0.02 + 0.044 C^2L

attached below is a plot of the drag polar

Attached below is the detailed solution of the remaining part of the question

Q.13 In order to produce maximum starting torque in a split-phase motor, how many degrees out of phase should the start- and run-winding currents be with each other?
Select one:
A. 180°
B. 0°
C. 120°
D. 90°

Answers

Answer:

D. 90 degrees.

Explanation:

Torque is a rotational force which moves an object in other direction. There should be 90 degrees out of phase to start, run winding currents with each other. Torque is produced by the rotational motion of an object. The angle of the object must be 90 degrees set in order to create torque.

Hot engine oil with heat capacity rate of 4440 w/k (product of mass flow rate and specific heat) and an inlet temperature of 150°c flows through a double pipe heat exchanger. the double pipe heat exchanger is constructed of a 1.5-m-long copper pipe (k = 250 w/m·k) with an inner tube of inside diameter 2 cm and outside tube diameter of 2.25 cm. the inner diameter of the outer tube of the double pipe heat exchanger is 6 cm. oil flowing at a rate of 2 kg/s through inner tube exits the heat exchanger at a temperature of 50°c. the cold fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.

Answers

Explanation:

fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.xgjicbb .follow me

8. Which of these plastics is a themoplastic - melts or softens with heat.
Acrylic
Bakelite
Polyester resin
Melamine
Epoxy Resin

Could someone please help me, this is very urgent, I will pay an extra bonus if done correct.

Answers

Answer:

Acryclic

Explanation:

Hope it helps you!

Technician A says that ultra-high-strength steels (UHSS) are generally defined as those steels with tensile strengths greater than 700 MPa.
Technician B says that advanced high-strength steels (AHSS) are generally defined as those steels with tensile strengths from 375 MPa up to 1700
Who is right?
O A only
OB only
O Both A and B
O Neither A norb
Table of Contents

Answers

b only i think i’m not fully sure.

There are gauges available for checking panel flushness. TRUE OR FALSE?

Answers

Answer: true

Explanation:

because flushness means your dead   so you are dead now

There are gauges available for checking panel flushness. to allow for front-to-back movement. to allow for side-to-side movement. for height adjustments. So yes it is true. Hope this helps!

A hydraulic jump is formed in a 4m wide outlet just downstream of a control gate, which is located at the upstream end of the outlet. The flow depth upstream of the gate is 20 m. Assume there are no losses in the flow through the gate. If the outlet discharge is 40 m3/s, determine (6 points)a. Flow depth downstream of the jump (y1)b. Energy losses in the jump (between y1 and y2).

Answers

Answer:

Width w = 4m

Glow depth = y1 = 20m

Outlet discharge = 40m

V1= velocity of flow = 40/20*4 = 1/2 = 0.5m/s

Froud number = v1/√gy1

= 0.5/√9.81x20 = 0.0356

1. Y2/20 = 1/2[-1+√1+8*(0.0356)²]

Y2 = 0.05

2. Energy loss in the jump = (20-0.05)²/4x20x0.05

= 1985m

Which part of a machine control unit interacts with the machine tools through electric signals?=]

A machine control unit is the electronic hardware that stores information and controls the machine tools. This unit contains a data processing unit that stores and manipulates data, and a ___________ that interacts with the machine tools through electrical signals.

Answers

Answer:

control loop unit

Explanation:

Edmentum/Plato

A baker deck decorates 42 cupcakes and 30 minute she decorated cupcakes at a constant rate how many cups cage the baker decorate pre-minute

Answers

Answer: 1.4 cupcakes per minute.

Explanation:

Here's the correct question:

A baker decorates 42 cupcakes in 30 minutes. How many cupcakes can the baker decorate per minute?

Since the baker decorates 42 cupcakes in 30 minutes, the number of cupcakes that the baker decorates for each minutes since it's at a constant rate will be the total number of cupcakes made divided by the number of minutes. This will be:

= 42cupcakes / 30minutes

= 1.4 cupcakes per minute.

Therefore, the number of cupcakes that the baker can decorate per minute is 1.4cupcakes/minute

7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 veh/h. The approach is controlled by a pretimed signal with a cycle length of 60 seconds and D/D/1 queuing holds. Local standards dictate that signals should be set such that all approach queues dissipate 10 seconds before the end of the effective green portion of the cycle. Assuming that approach capacity exceeds arrivals, determine the maximum length

Answers

Answer:

23.34 seconds

Explanation:

Flow rate = 1500

Arrival = 800 vehicle per hour

Cycle c = 60 seconds

Dissipation time = 10 seconds

Arrival time = 800/3600 = 0.2222

Rate of departure = 1500/3600 = 0.4167

Traffic density p = 0.2222/0.4167 = 0.5332

Real time = r

r + to + 10 = c

to = c-r-10 ----1

t0 = p*r/1-p ----2

Equate both 1 and 2

C-r-10 = p*r/1-p

60-r-10 = 0.5332r/1-0.5332

50-r = 0.5332r/0.4668

50-r = 1.1422r

50 = 1.1422r + r

50 = 2.1422r

r = 50/2.1422

r = 23.34 seconds

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