Consider atmospheric air at 25 C and a velocity of 25 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 125 C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 105 , and 106 .
Answer:
Explanation:
Temperature of atmospheric air To = 25°C = 298 K
Free stream velocity of air Vo = 25 m/s
Length and width of plate = 1m
Temperature of plate Tp = 125°C = 398 K
We know for air, Prandtl number Pr = 1
And for air, thermal conductivity K = 24.1×10?³ W/mK
Here, charectorestic dimension D = 1m
Given value of Reynolds number Re = 105
For laminar boundary layer flow over flat plate
= 3.402
Therefore, hx = 0.08199 W/m²K
So, heat transfer rate q = hx×A×(Tp – To)
= 0.08199×1×(398 – 298)
The fins attached to a heat exchanger-surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?
Answer:
The rate of heat transfer has increased.
Explanation:
Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.
/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a
mean velocity of 7 m/s, and is cooled by the water outside. If the average heat
transfer coefficient is 85 W/m2
°C and the tube temperature is nearly equal to the
water temperature of 5°C, determine the exit temperature of air and the rate of heat
transfer.
Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above
Answer:
B) gate-source junction is reverse-biased
Explanation:
FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow" in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".
In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".
After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the set-up
Answer:
Move the buret clamp to a ring stand with a larger base.
Explanation:
A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.
The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.
The clamp is used to hold the burette in place.
If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.
The larger base provides a better center of gravity and stabilises the setup
Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperature of 80oC. Calculate the average heat transfer coefficient and the heat transfer rate per meter length of pipe.\
Answer:
Average heat transfer coefficient = 31 kw/m^2 k
Heat transfer rate per meter length of pipe = 116.808 KW
Explanation:
water temperature = 20⁰c,
free-stream velocity = 1.5 m/s
circular pipe diameter = 2.0 cm = 0.02 m
surface temperature = 80⁰c
A) calculate average heat transfer coefficient
we apply the formula below :
m = αAv
A (area) = [tex]\pi /4 (d)^2[/tex]
m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5
= 10^3 * 0.7857( 0.0004) * 1.5
= 0.4714 kg/s
Average heat transfer coefficient
h = [tex]\frac{m(cp)}{A}[/tex] , A = [tex]\pi DL[/tex]
L = 1 m , m = 0.4714 kgs , cp = 4.18
back to equation
h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex] = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k
B) Heat transfer rate per meter length of pipe
Q = ha( ΔT ), a = [tex]\pi DL[/tex]
= 31 * 0.0628 * ( 80 - 20 )
= 31 * 0.0628 * 60 = 116.808 KW
what scale model proves the initial concept?
Answer: A prototype
Explanation:
The scale model that proves the initial concept is called a domain model.
What is a scale model?A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.
A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.
A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.
Therefore, a domain model is the scale model that proves the initial concept.
To learn more about the scale model, refer to the below link:
https://brainly.com/question/14341149
#SPJ2
Write about traditional brick production in Pakistan
Answer:
Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.
Explanation:
In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about 1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.
There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.
Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 0.8m and 7m/s, respectively. Determine (a) the flow depth and the Froude number after the jump (b) the head loss (c) the dissipation ratio.
Answer:
A) Flow depth = 2.46 m, Froude number after jump = 0.464
B) head loss = 0.572 m
C) dissipation ratio = 0.173
Explanation:
Given data :
Velocity before jump ( v1 ) = 7 m/s
flow depth before jump ( y1 ) = 0.8 m
g = 9.81 m/s
Esi = 3.3 m ( calculated )
attached below is a detailed solution of the problem
Q1: You have to select an idea developing an application like web/mobile or industrial, it should be based on innovative idea, not just a simple CRUD application. After selecting the idea do the following: 1) How your project will be helpful and what problem this project addresses. (10-Marks) 2) Write down the requirements. (10Marks) 3) List the functional and non-functional requirements of your project. (10marks) 4) Which process model you will follow for this project and why? (10marks) 5) Draw the Level 0, and level 1 DFD of your application. (20marks)
Answer:
Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.
The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.
Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.
if I am right then make me brainliest
You have accumulated several parking tickets while at school, but you are graduating later in the year and plan to return to your home in another jurisdiction. A friend tells you that the authorities in your home jurisdiction will never find out about the tickets when you re-register your car and apply for a new license. What should you do?
Answer:
pay off the parking tickets
Explanation:
In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.
A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric field 4 cm from the wire is:
Answer:
New electric field = 18 N/C
Explanation:
Given:
Length (E1) = 2 cm
New length (E2) = 4 cm
Electric field = 36 N/C
Find:
New electric field
Computation:
New electric field = 36 [2 / 4]
New electric field = 36 [1/2]
New electric field = 18 N/C
