1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mario62 [17]
4 years ago
14

An open glass tube is inserted into a pan of fresh water at 20 °C. What tube diameter is needed to make the height of capillary

rise equal to four times the tube diameter? State all assumptions.
Engineering
1 answer:
kiruha [24]4 years ago
5 0

Answer:

The tube diameter is 2.71 mm.

Explanation:

Given:

Open glass tube is inserted into a pan of fresh water at 20°C.

Height of capillary raise is four times tube diameter.

h = 4d

Assumption:

Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.

Take surface tension of water at 20°C as 72.53\times 10^{-3} N/m.

Take density of water as 100 kg/m3.

Calculation:

Step1

Expression for height of capillary rise is gives as follows:

h=\frac{4\sigma\cos\theta}{dg\rho}

Step2

Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:

4d=\frac{4\times72.53\times10^{-3}\cos0^{\circ}}{d\times9.81\times1000}

d^{2}=7.39\times10^{-6}

d=2.719\times10^{-3} m.

Or

d=(2.719\times10^{-3}m)(\frac{1000mm}{1m})

d=2.719 mm

Thus, the tube diameter is 2.719 mm.

 

You might be interested in
Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost m
Ilia_Sergeevich [38]

Answer:

ordinary bulb total cost is $39.54

fluorescent bulb total cost is $13.05

amount save = 39.54 - 13.05 = $26.49

resistance = 626.1 ohm

Explanation:

in the 1st part

bulb on time = 3 year = 4380 hours

life of bulb = 750 h

so number of bulb required = \frac{4380}{750}

number of bulb required = 6

cost of 6 bulb is = 6 × 0.75 = $4.5

so

cost of operation is = 100 × 4380 × \frac{0.08}{1000}

cost of operation = $35.04

so total cost will be = $4.5 + $35.04  = $39.54

and

when compare with florescent bulb

time = 3 year = 4380 h

life of bulb = 10000 h

so number of bulb required = \frac{4380}{10000}

number of bulb required = 0.43 = 1

cost of 6 bulb is = 1 × 5 = $5

so

cost of operation is = 23 × 4380 × \frac{0.08}{1000}

cost of operation = $8.05

so total cost will be = $5 + $8.05  = $13.05

in part 2nd

total amount save while compare bulb is

amount save = 39.54 - 13.05 = $26.49

and in part 3rd

resistance of bulb is

resistance = \frac{v^2}{P}

resistance = \frac{120^2}{23}

resistance = 626.1 ohm

6 0
3 years ago
The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0
3 years ago
When a circuit has more resistance, what happens to current flow ?
zloy xaker [14]
The relationship between resistance and the area of the cross section of a wire is inversely proportional . When resistance is increased in a circuit , for example by adding more electrical components , the current decreases as a result.
6 0
3 years ago
A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th
trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

-  The mass flow rate of steam through the boiler

-  The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

                                        E_6 + E_2 = E_3

               flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

                                    y*h_6 + (1-y)*h_3 = h_3

                                  y*2810 + (1-y)*192.8 = 721

Compute y:                          y = 0.2018

- Heat produced by the boiler q_b:

                             q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

                    q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b:               q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

                                       q_c = (1-y)*(h_8 - h_1)

                                 q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c:               q_c = 1834.26 KJ/ kg

- Net power output w_net:

                                     w_net = q_b - q_c

                                w_net = 3303.58 - 1834.26

                                    w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

                                     flow(m) = P / w_net

compute flow(m)          flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

                                     u = 1 - (q_c / q_b)

                                     u = 1 - (1834.26/3303.58)

                                     u = 44.48 %

5 0
3 years ago
The best grade of hardwood lumber that is generally available is _____​
Vesnalui [34]

Answer:

FAS

Explanation:

first and second grade

5 0
2 years ago
Other questions:
  • The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting,
    8·1 answer
  • PLEASE HELP 100 POINTS!!! WILL MARK BRAINLIEST
    9·2 answers
  • Briefly describe the purpose of specifying boundary conditions.
    7·1 answer
  • 8–21 Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600 K. Calculate
    6·1 answer
  • What is the function of a regulator?
    8·1 answer
  • Mention verious medium level and higher level human resources related to engineering​
    6·1 answer
  • What was the most important thing you learned this school year in your engineering class and why did you choose this thing
    15·1 answer
  • Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo c
    11·1 answer
  • What are the nine Historical periods?
    9·1 answer
  • What is photosynthesis​
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!