The fouling factor of the tube is 0.0097 (min × ft2 × °F)/BTU.
To calculate the fouling factor, we first need to determine the overall heat transfer coefficient (U). We can use the following equation:
Q = U × A × LMTD
where Q is the heat transferred, A is the inner surface area of the tube, LMTD is the logarithmic mean temperature difference, and U is the overall heat transfer coefficient.
We know that the inner surface area of the tube is 50 ft2, and we can assume that the length of the tube (L) is 1 ft for simplicity. The LMTD can be calculated using the following equation:
LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
where ΔT1 is the temperature difference between the hot and cold fluids at the inlet, and ΔT2 is the temperature difference between the hot and cold fluids at the outlet. In this case, ΔT1 = 122 - 70 = 52°F and ΔT2 = 122 - 70 = 52°F.
Plugging in the values, we get:
Q = U × 50 × 1 × (52 / ln(52/52)) = U × 50
We also know that the flow rate of the cold fluid (water) is 100 gal/min, which is equivalent to 12.5 ft3/min. Using the specific heat of water (1 BTU/lb°F), we can calculate the heat transferred as:
Q = m × cp × ΔT = 12.5 × 8.34 × (122 - 70) = 5205 BTU/min
Equating the two expressions for Q, we get:
U × 50 = 5205
Solving for U, we get:
U = 104.1 BTU/(min × ft2 × °F)
Now we can calculate the fouling factor (Rf) using the following equation:
Rf = 1 / U - 1 / Ui
where Ui is the clean inner surface heat transfer coefficient, which can be estimated based on the properties of the fluids and the tube geometry. For a typical shell-and-tube heat exchanger, Ui is usually in the range of 200-400 BTU/(min × ft² × °F).
Assuming Ui = 300 BTU/(min × ft² × °F), we get:
Rf = 1 / 104.1 - 1 / 300 = 0.0097 (min × ft² × °F)/BTU
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This manipulator is used to establish a field width for the value that follows it: field_width set field setw iomanip None of these
The manipulator used to establish a field width for the value that follows it is setw from the iomanip library in C++. It allows you to specify the number of characters that should be used for the output of the value.
For example, if you want to output a value with a field width of 5, you would use set w(5). This will ensure that the value is formatted with 5 characters, adding spaces as necessary to achieve the desired width. So the correct option in this case would be "set field setw iomanip". The term "set field set w iomanip" does not make sense in this context.
The manipulator used to establish a field width for the value that follows it is "setw". This term is part of the iomanip library in C++. So, the correct answer is "setw". The other terms mentioned in the question, such as "field_width", "set field", and "setw ioman ip", are not relevant to the context of the question.
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The manipulator in question is called "setw" and it is a part of the "iomanip" library in C++. option c is correct.
This manipulator is used to set the field width for the value that follows it. The value can be a string, integer, or any other data type.
The purpose of setw is to make the output more readable and organized by aligning the data in columns. For example, if you are printing a table with multiple columns, you can use setw to set the width of each column so that the data is aligned and easy to read.
The syntax for using setw is as follows:
cout << setw(10) << "Hello";
In this example, setw(10) sets the field width to 10 characters for the string "Hello". This means that "Hello" will be printed with a width of 10 characters, even if it is shorter than that. If the string is longer than 10 characters, it will be truncated to fit within the allotted space.
Overall, setw is a useful manipulator for formatting output in C++. It helps to make the output more readable and organized, which is important when dealing with large amounts of data.
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Note: The complete question would de as bellow,
This manipulator is used to establish a field width for the value that follows it:
a)field_width
b)set field
c)setw
d)iomanip
e)None of these
Assume:
A = 1101 0011 1111 0110
B = 0110 1101 1101 1110
Write the series of operations necessary to pack A into B (and store the result in C), where the 8 lowest order bits of B are stored in the 8 highest order bits of C, and the 8 highest order bits of A are stored in the 8 lowest order bits of C.
To pack A into B and store the result in C, the following operations can be performed.
What is the explanation for the above response? Perform a logical shift right on A by 8 bits, which will result in 0000 0000 1101 0011.Perform a logical shift left on B by 8 bits, which will result in 1011 0110 1101 1110.Perform a logical OR operation between the results of step 1 and step 2, which will result in 1011 0110 1111 1110.Perform a logical shift left on A by 8 bits, which will result in 0000 0000 0000 0000.Perform a logical shift right on B by 8 bits, which will result in 0000 0000 0110 1101.Perform a logical OR operation between the results of step 4 and step 5, which will result in 0000 0000 0110 1101.Perform a logical OR operation between the results of step 3 and step 6, which will result in C = 1011 0110 1111 1110 0000 0000 0110 1101.Learn more about operations at:
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