Consider a layer of insulation installed around a steam pipe. The radius of the pipe is R1 and the thickness of the insulation is R3-R1.The stream is maintained at a temperature TW and the air surrounding the insulation is at a temperature To and flows cross wise over the pipe. The air is flowed over the steam pipe at high enough velocity so a thermal boundary layer develops over the surface described by a heat transfer coefficient h.
Required:
Beginning with the heat equation, calculate the total quantity of heat being conducted per unit time (heat flow) through the insulation, Q (units: energy/time).
Answer:
hmmmmmmmmmmmmmmmmmmmmmmmmmmm
Explanation:
........
Vehicles begin arriving at an amusement park 1 hour before the park opens, at a rale of four vehicles per minute. The gale to the parking lot opens 30 minutes before the park opens. If the total delay to vehicles entering the parking lot is 3600 vehicle-minutes, (a) how long after the first vehicle arrival will the queue dissipate and (b) what is the average service rate at the parking lot gate? Assume a D/D/l queueing system at the parking lot gate.
Answer:
a) ≈ 30 mins
b) 8 vpm
Explanation:
a) Determine how long after the first vehicle arrival will the queue dissipate
The time after the arrival of the first vehicle for the queue to dissipate
= 29.9 mins ≈ 30 mins
b) Determine the average service rate at the parking lot gate
U = A / t
where : A = 240 vehicles , t = 30
U = 240 / 30 = 8 Vpm
attached below is a detailed solution of the given problems above
In order to cool a cylindrical steel rod of diameter D, length L and uniform temperature T1, it is placed in a well-mixed water bath at initial temperature T0 and volume V0. The heat transfer coefficient between the steel rod surface and the water is h1. Assume the specific heat, density and conductivity of the steel rod and water are known and constant. Write down heat transfer differential V1, P1, T1 equation and boundary and initial conditions for temperature distribution in both steel rod and water bath
Answer:
attached below
Explanation:
Cylindrical steel rod : diameter ( D ) , length L, uniform temperature T
Initial temp of water : To, heat transfer coefficient between steel rod surface and water : h1
The energy balance equation can be written as :
Rate of convectional heat loss = Rate of decrease in internal energy with respect to time
cp = heat capacity of body, v = volume , р = density of body
attached below is the heat transfer differential equation and boundary and initial conditions
Tech A says that a mechanical pressure regulator exhausts excess fluid back to the transmission pan. Tech B says that if the transmission pan is removed, the magnet must be replaced. Who is correct?
Answer:
A
Explanation:
For binary flash distillation, we discussed in class that there are 8 variables (F, ZA, V, ya, L, XA, P and T) and 4 equations derived from VLE and mass balances. Thus, we typically require 4 of these variables to be given so that we can obtain a unique solution to the problem. Let's say, your manager tells you that he has a feed mixture with 2 components (given F, za) and he requires you to come up with a flash column that can produce a certain desired amount of Vapor product (thus V, ya are specified). Identity of both components is known and all VLE data has been provided to you. Has the manager given you enough data? If yes, give a step-by-step description of how would you go about designing the flash column (basically find P and T)? If no, why?
Answer:
yes
Explanation:
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Applying fatigue failure criteria in 3D, determine the final relation (equation) for the following cases: - Considering only the internal pressure fluctuating from Pmax to Pmin Numerical Application: Pmax=1.0P and Pmin=0.2P σu= 690 MPa σe= 345 MPa Kf=1 ; Define P ? - Considering completely reversed internal pressure (P) and completely reversed bending moment (M) Write just the final equation (no numerical application)
Answer:
Hello your question is incomplete attached below is the complete question
answer :
I) P = t/R * 492.85
II) The final equation : PR / t + 4M/πR3 = б e
Explanation:
attached below is a detailed solution to the given problem
i) P = t/R * 492.85
ii) Final equation : PR / t + 4M/πR3 = б e
A circular sheet metal duct carries refrigerated air to a cold storage room for apples. The duct itself is 300 mm in outer diameter. The duct wall thickness is 0.6 mm. To reduce the heat gain from the surrounding air, we need to wrap the duct with insulation. The flowing air maintains the inner surface of the duct at 0 C. The outer surface temperature of insulation would be maintained at 25 C by the room air. Thermal conductivity of the sheet metal is 100 W/m-K and that of insulation is 0.04 W/m-K. Assuming the heat transfer to be as steady state, what thickness of insulation should be put on the duct to keep the rate of heat gain by the refrigerated air per meter
Answer:
The thickness of the insulator is approximately 34.918 mm
Explanation:
From the question, we have;
The outer diameter of the duct, D = 300 mm
The wall thickness of the duct, t₁ = 0.6 mm
The temperature of the inner surface of the duct, [tex]T_C[/tex] = 0°C
The temperature of the outer surface, [tex]T_H[/tex] = 25°C
The thermal conductivity of sheet metal, k₁ = 100 W/m-K
The thermal conductivity of insulation, k₂ = 0.04 W/m-K
Assumed rate of heat transfer through the walls of the refrigerator per second, Q = 30 W = 30 J/s
Therefore, we have;
[tex]Q = \dfrac{T_H - T_C}{R_{total}}[/tex]
[tex]\therefore R_{total} = \dfrac{T_H - T_C}{Q} = \dfrac{25 ^{\circ} - 0^{\circ}}{30 \, W} =\dfrac{5}{6} \ ^{\circ}C/W[/tex]
The outside radius, r₂ = 300 mm/2 = 150 mm
The inner diameter of the pipe, d = D - 2·t₁
∴ d = 300 mm - 2 × 0.6 mm = 298.8 mm
The inside radius, r₁ = d/2 = 298.8mm/2 = 149.4 mm
The heat resistance of the pipe, R₁, is given as follows;
[tex]R_1 = R_{pipe} = \dfrac{ln\left (\dfrac{r_2}{r_1} \right) }{2\cdot \pi \cdot k_1\cdot L}[/tex]
Where;
r₁, r₂, and k₁ are as defined above;
L = The length of the pipe = 1 m
Therefore, we have;
[tex]R_1 = R_{pipe} = \dfrac{ln\left (\dfrac{150}{149.4} \right) }{2\cdot \pi \cdot 100\cdot 1} \approx 6.378964 \times 10^{-6}[/tex]
[tex]R_{total}[/tex] = [tex]R_{insltor}[/tex] + [tex]R_{pipe}[/tex]
∴ [tex]R_{insltor}[/tex] = [tex]R_{total}[/tex] - [tex]R_{pipe}[/tex]
[tex]R_{insltor}[/tex] = 5/6 - 6.378964 × 10⁻⁴ ≈ 0.862695
The heat resistance of the insulator, R₂ = [tex]R_{insltor}[/tex] ≈ 0.862695 °C/W
The heat resistance of the insulator, R₂, is given as follows;
[tex]R_1 = R_{insltor} = \dfrac{ln\left (\dfrac{r_3}{r_2} \right) }{2\cdot \pi \cdot k_2\cdot L}[/tex]
Therefore;
[tex]R_2 = R_{insltor} = 0.862695 = \dfrac{ln\left (\dfrac{r_3}{150} \right) }{2\cdot \pi \times 0.04\times 1}[/tex]
[tex]0.832695 \times 2\times \pi \times 0.04\times 1 = {ln\left (\dfrac{r_3}{150} \right) }{}[/tex]
[tex]0.209279 = {ln\left (\dfrac{r_3}{150} \right) }{}[/tex]
[tex]e^{0.209279} = \dfrac{r_3}{150} \right) }{}[/tex]
r₃ = 150 × [tex]e^{0.209279}[/tex] = 184.918
The outer radius of the insulator, r₃ ≈ 184.918 mm
The thickness of the insulator, t₂ = r₃ - r₂
∴ The thickness of the insulator, t₂ ≈ 184.918 mm - 150 mm = 34.918 mm.
We have a parallel-plate capacitor with plates of metal each having a width W and a length L. The plates are separated by the distance d. Assume that L and W are both much larger than d. The maximum voltage that can be applied is limited to V max =K d, in which K is called the breakdown strength of the dielectric. Derive an expression for the maximum energy that can be stored in the capacitor in terms of K and the volume of the dielectric. If we want to store the maximum energy per unit volume, does it matter what values are chosen for L, W, and d? What parameters are important?
Answer:
max energy = [tex]W_{max} = \frac{1}{2}[/tex]*εo*εr*k^2
Explanation:
Given data:
weight of plates = W
length of plates = L
distance of separation = d
max voltage ( Vmax ) = Kd
Area ( A ) = WL
The values chosen for L, W, and d matters, although the maximum energy stored in the capacitor is independent of L, W, and d. but at a constant volume and a larger value for W and L which is > d, the value of the dielectric (εrK^2 ) should be a larger value '
The important parameters are : εrK^2 , k , d and Area
attached below is the remaining part of the solution
Please I need help with this
Photosynthesis energy is stored in the cells of green plants through
process called ______?
a crawler tractor is operated by a 180-hp diesel engine. calculate the probable gallons of fuel consumed per hr for each of the given conditions: a) when operating at an average of 60 percent of its capacity for 50 min per hr. b) when operating at 100 percent of its capacity for 15 min per hr, 60 percent capacity for 30 min per hr, and 20 percent of its capacity for 15 min per hr.
Answer:
3.59
4.32
Explanation:
We find the time factors and engine factor to solve for this.
Engine factor = (100%*15/60) + (60%*30/60) + (20%*15/60)
= 1x0.25 + 0.6x0.5 + 0.2x0.25
= 0.25 + 0.30 + 0.05
= 0.6
A. We find time factor
= 50/60 = 0.83 minutes
We then get consumption
= Time factor x engine x hp x .04
= 0.83 x 0.6 x 180 x 0.04
= 3.585
3.59 gallons in 1 hr
B. Time factor = 60/60 = 1
Consumption =
1x0.6x180x0.04
= 4.32 gallons in 1 hour
PLZ HELP I GIVE BRAINLIEST!!
Can someone help me plz!!!
Answer:
15 000 000 Ohms
Explanation:
1 Mega Ohm = 1 000 000 Ohms
So,
15 Mega ohms =15 000 000 Ohms
g ) Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater heater. Superheated vapor enters the turbine at 16 MPa, 560 oC, and the condenser pressure is 8 kPa. The cycle has a closed feedwater heater using extracted steam at 1 MPa. Condensate drains from the feedwater heater as saturated liquid at 1 MPa and is trapped into the condenser. The feedwater leaves the heater at 16 MPa and a temperature equal to the saturation temperature at 1 MPa. The mass flow rate of steam entering the first-stage turbine is 120 kg/s. Determine (a) the net power developed, in kW.
Answer:
146494 kw
Explanation:
Given data:
Turbine inlet pressure ( p1 ) = 16 MPa
Turbine inlet temperature ( T1 ) = 560°C
condenser pressure : P3 = 8Kpa
Extracted steam at pressure ( P2 ) = 1 MPa
mass flow rate of steam ( m ) = 120 kg/s
a) The net power developed
= 146494 kw
note: values of h1, h2, y, Wp are all calculated values not included in the solution to make the solution less cumbersome.
stole energy is also called
Answer:
Hey mate....
Explanation:
This is ur answer.....
Energy theft, also called energy diversion, occurs when individuals tamper with electric meters or electric power lines. Energy theft ranges from tapping into a neighbor's energy source to illegally adjusting a meter. Meter tampering occurs in homes, grocery stores, restaurants and other commercial establishments.
Hope it helps you,
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4.In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is generated. The electric power you get out from the 80% efficiency turbine is known to be 50 MW, what is the rate of irreversible loss in the piping system (in MW unit)
Answer:
The rate of irreversible loss will be "55.22 MW".
Explanation:
The given values are:
Elevation,
h = 120 m
Flow of water,
Q = 100 m³/s
Efficiency,
= 80%
i.e,
= 0.8
Efficiency turbine,
= 50 MW
Now,
Without any loss,
The power generated by turbine will be:
⇒ [tex]P=\delta gQh[/tex]
On substituting the values, we get
⇒ [tex]=1000\times 9.8\times 100\times 120[/tex]
⇒ [tex]=117.72 \ MW[/tex]
Power generated in actual will be:
= [tex]\frac{50}{0.8}[/tex]
= [tex]62.5 \ MW[/tex]
Hence,
Throughout the piping system,
The rate of irreversible loss is:
= [tex]Power \ generated \ by \ turbine-Power \ generated \ in \ actual[/tex]
= [tex]117.72-62.5[/tex]
= [tex]55.22 \ MW[/tex]
in 1960 a magnetic azimuth of 90 degrees and a magnetic declination of 1degrees east.given that magnetic declination was changing at a rate of 5 degrees west.what is the magnetic azimuth in 1983?
Answer:
4c4u4civ4j4
Explanation:
x24uu37z3x3uzu2
2.3
What does NBT stand for?
Answer:
Named Binary Tag ( NBT)
Explanation:
The format is designed to store data in a tree structure made up of various tags
The meaning of the given word NBT is called; National Benchmark Test.
What is the meaning of NBT?The meaning of the given word NBT is called National Benchmark Test.
NBT ( national benchmark test ) is defined as a national test which must be written and passed by High school students that intend to go to the university ( i.e. passport to the university ).
Read more about NBT at; https://brainly.com/question/26887507
Question text
A man leaves home for a cycle ride and comes back home after a half-an-hour ride covering a distance of one km. What is the average velocity of the ride?
Answer: Hello, The average velocity of the ride is 0 km/h. Your Welcome! Mark me as Brainliest!
Explanation: