Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T

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Answer 1
Answer:

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 +  (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C


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Explain why the torsion test can be considered to provide a better assessment of plasticity than the tensile test.

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Answer and Explanation:

The main objective that the torsion test serves is the determination of the material behavior or the behavior of the test sample when subjected to torsional stresses or forces due to the application of moments that results in shear stress along the axis.

Plasticity is the property of elastic material and tension or shear stresses leads to plasticity in a material where these links are the weakest, that gives torsion test a major advantage over the tension test.

Torsion tests are performed on materials to deduct  properties like the shear modulus of elasticity, the torsional strength, and the MOR, i.e., Modulus of Rupture.

This test can be used to obtain larger strain values of strain without any complexity as that in tension test.

This test provides a curve of shear-stress-shear strain which is more significant in determining the plasticity as compared to the curve of stress-strain in tension test.

Maximum torque for a given value of maximum stress will be 2 times higher in torsion as that of tension.

In torsion, for plastic flow, the threshold value of shear stress is achieved before the threshold value of normal stress for fracture whereas in tension the critical value of normal stress is achieved sooner than the critical shear for plastic flow.

The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal

Answers

Answer:

Isobaric process

Explanation:

The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.

At constant pressure, the work done is given by :

W=p* \Delta V

Where

W is the work done by the system

p is the constant pressure

\Delta V is the change in volume

So, the correct option is (c) " isobaric process ".

Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a number that divides into another without a remainder.

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Answer:

// Program is written in C++

// Comments are used to explain some lines

// Only the required function is written. The main method is excluded.

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

// The next line declares the final result of summation of divisors. The variable declared is also

//initialised to 0

int result = 0;

// find all numbers which divide 'num'

for (int i=2; i<=(num/2); i++)

{

// if 'i' is divisor of 'num'

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

}

cout<<result+1;

}

In this exercise, using the knowledge of computational language in C++, we have that this code will be written as:

The code is in the attached image.

We can write the C++  as:

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

int result = 0;

for (int i=2; i<=(num/2); i++)

{

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

}

cout<<result+1;

}

See more about C++ at brainly.com/question/19705654

Indicate whether the following statements are true or false for an isothermal process: (A) Q=T(∆S). (B) ∆U=0.(C) The entropy change of the system is always zero. (D) The total entropy change of the system and the surroundings is always zero. (E) The entropy change of the surroundings is negative. (F) Q=W.

Answers

Answer:

A=False

B=False

C=False

D=False

E=False

F=False

Explanation:

A. In an isothermal process, only the reversibly heat transfer is 0, Q_(rev)=T (\Delta S)

B. Consider the phase change of boiling water. Here, the temperature remains constant but the internal energy of the system increases.

C. This is not true even in reversible process, as can be inferred from the equation in part A.

D. This is only true in reversible processes, but not in all isothermal processes.

E. Consider the phase change of freezing water. Here, the surroundings are increasing their entropy, as they are taking in heat from the system.

F. This is not true if (\Delta U)\neq 0, like in answer B. One case where this is true is in the reversible isothermal expansion (or compression) of an ideal gas.

Currently, system administrators create Ken 7 users in each computer where users need access. In the Active Directory, where will system administrators create Ken 7 users? 2. How will the procedures for making changes to the user accounts, such as password changes, be different in the Active Directory? 3. What action should administrators take for the existing workgroup user accounts after converting to the Active Directory? 4. How will the administrators resolve the differences between the user accounts defined on the different computers? In other words, if user accounts have different settings on different computers, how will the Active Directory address that issue? 5. How will the procedure for defining access controls change after converting to the Active Directory?

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1. First, you would need to open Active Directory Users and Computers. You click on the folder in which you want to add an account, and point to new, and then user. You would fill in the new user's information, such as name and initials.

2. In Active Directory, you input the user logon name, click on the UPN suffix in the drop-down list. It will prompt you to input password and confirm it.

3. Administrators would need to create new user accounts for all users, then join these to the AD domain manually.

4. Administrators will have to manually change the permissions and privileges of the users in order to meet the new established requirements.

5. After converting to the Active Directory, access control will be administered at the object level by setting different levels of access.

(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinusodal force given by p(t) = P sin ωt. All positions are measured from equilibrium. Use m_1=1.5 kg, m_2=2 kg, k=7 N/m, b=3.2 (N∙s)/m, P=15 N, =12 rad/sec. Hint: first create the state space model for the system. Then use SS2TF to make the two transfer functions and then the two Bode plots (include with submission). Use the plots to find the steady-state equations.

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