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
motikmotik
3 years ago
7

A double-pipe heat exchanger is used to cool a hot fluid (cp = 3800 J/kg·K) entering the heat exchanger at 200°C with a flow rat

e of 0.4 kg/s. In the cold side, cooling fluid (cp = 4200 J/kg·K) enters the heat exchanger at 10°C with a mass flow rate of 0.5 kg/s. The double-pipe heat exchanger has a thin-walled inner tube, with convection heat transfer coefficients inside and outside of the inner tube estimated to be 1400 W/m2 ·K and 1100 W/m2 ·K, respectively. The heat exchanger has a heat transfer surface area of 2.5 m2 , and the estimated fouling factor caused by the accumulation of deposit on the surfaces is 0.0002 m2 ·K/W. (a) Determine the effectiveness values for the parallel- and counter-flow configurations. (b) Determine outlet temperatures of the hot fluid for the parallel- and counter-flow configurations. (c) Determine outlet temperatures of the cold fluid for the parallel- and counter-flow configurations.

Engineering
1 answer:
MAXImum [283]3 years ago
3 0

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem

You might be interested in
A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.
Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

S_{ut}=K \left(\cfrac ne \right)^n

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}

And we can replace values

K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87

Thus we get that the strength coefficient is K = 591.87 MPa

6 0
3 years ago
Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh
Nutka1998 [239]

Answer:

wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh

Explanation:

jwhwhwhwhwhwwhhahwhahahwh

6 0
3 years ago
Explain briefly the power analogy between the fluid “water” and the Word of God.
melomori [17]

Throughout the Word of God, water is a symbol, a "type," of the spirit of God, and is ... to represent holy spirit, that is, the divine nature and power of God. ... The average person can go about 60 days without food before he ... record illustrating the analogous relationship between water and the holy spirit.

3 0
3 years ago
An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in
Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, T_{a} = 140^{\circ}C = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, T_{i} = 60^{\circ}C = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

At room temperature

Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)

Thus

T_{e} = 109.71^{\circ}C = 273 + 109.71 = 382.71 K

where

T_{e} = Equilibrium temperature

Now,

To calculate the changer in entropy:

\Delta s = \Delta s_{a} + \Delta s_{i}

Now,

For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K

For Iron:

\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

6 0
3 years ago
The federal highway administration reports nearly
levacccp [35]

Explanation:

The federal highway administration reports nearly 800 work zone fatalities per year.

8 0
3 years ago
Other questions:
  • Who does the narrator blame for the loss of her job as editor-in-chief? <br> see if i care readworks
    8·2 answers
  • Write SQL queries to answer the following questions: What are the names of the course(s) that student Altvater took during the s
    13·1 answer
  • The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb&gt;ft. Determine t
    9·1 answer
  • A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str
    10·1 answer
  • Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers
    12·1 answer
  • List five areas that increased energy prices impact.
    9·1 answer
  • When you do a vehicle check, what do you NOT need to keep an eye on?
    9·1 answer
  • Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th
    10·1 answer
  • A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0
    14·1 answer
  • Witch measuring tool would be used to determine the diameter of a crankshaft journal
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!