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Marrrta [24]
3 years ago
6

The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False

Engineering
1 answer:
jeyben [28]3 years ago
4 0

Answer:

a). TRUE

Explanation:

Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.

Mathematically, thermal efficiency is

        η =  net work done / heat input

While heat rate is  the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.

Mathematically, heat rate is

       Heat rate = heat input / net work done

Thus from above we can see that heat rate is the reciprocal of thermal efficiency.

Thus, Heat rate is reciprocal of thermal efficiency.

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atroni [7]
Houses the CYLINDERS, Water Jacket & Crankcase
8 0
3 years ago
If the price of the car is less than or equal to your available cash, display "no". If the price of the car is more than your av
Ede4ka [16]

Answer:

function decision(car_price, available_cash) {

   if(car_price <= available_cash) {

   console.log("no");

   }

   else  {

   console.log("yes");

   }

   }

decision(car_price, available_cash); or decision(available_cash, car_price);

Explanation:

using functions in Javascript:

functions; this refers to dividing codes into reusable parts.

e.g function function_name() {

console.log("How are you?");

}

you can call or invoke this function by using its name followed by parenthesis, like this: function_name(). each time the function is called it will   print out "How are you?".

Parameters: these are variables that act as placeholders for the values that are to be input into a function when it is called

Arguments: The actual values that input or passed into a function when it is called.

e.g

function function_name(parameter1, parameter2) {

console.log(parameter1, parameter2);

}

then we call function_name: function_name("please", "leave"):we have passed two arguments, "please"  and "leave". Inside the function parameter1 equals "please" while parameter2 equals "leave".

Hence, from the question given the two parameters "car_price" and "available_cash" respectively, we write the function with name function_name:

function decision(car_price, available_cash) {

   if(car_price <= available_cash) {

   console.log("no");

   }

   else  {

   console.log("yes");

   }

   }

decision(car_price, available_cash); or decision(available_cash, car_price);

7 0
3 years ago
t is desired to produce aligned carbon fiber-epoxy matric composite having a longitudinal tensile strength of 800 MPa. Given (a)
Nesterboy [21]

Answer:

The value of critical length = 3.46 mm

The value of volume of fraction of fibers = 0.43

Explanation:

Given data

\sigma_{T} =  800 M pa

D = 0.017 mm

L = 2.3 mm

\sigma_{f} = 5500 M pa

\sigma_{m} = 18 M pa

\sigma_c = 13.5 M pa

(a) Critical fiber length is given by

L_{c} = \sigma_{f} (\frac{D}{2 \sigma_{c} } )

Put all the values in above equation we get

L_{c} =5500 (\frac{0.017}{(2) (13.5)} )

L_{c} = 3.46 mm

This is the value of critical length.

(b).Since this  critical length is greater than fiber length Than the volume fraction of fibers is given by

V_{f} = \frac{\sigma_T - \sigma_m}{\frac{L\sigma_c}{D} - \sigma_m }

Put all the values in above formula we get

V_{f} = \frac{800-18}{\frac{(2.3)(13.5)}{0.017} - 18 }

V_{f} = 0.43

This is the value of volume of fraction of fibers.

3 0
3 years ago
Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is
VARVARA [1.3K]

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}

Solve the following for the real stress and pressure for the stable.\sigma_{r1}=K(\varepsilon_{r1})^{n}

K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}

Solve the following for the true state stress and stress2.

\sigma_{r2}=K(\varepsilon_{r2})^n

     =\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa

4 0
3 years ago
The accompanying specific gravity values describe various wood types used in construction. 0.320.350.360.360.370.380.400.400.40
marysya [2.9K]

Answer:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

Explanation:

Given

0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,

0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,

0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,

0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.

0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,

0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.

0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.

0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.

0.78.

The 0.3's is will be plotted as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}

The 0.4's is as follows:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}

The 0.5's is as follows:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}

The 0.6's is as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}

Lastly, the 0.7's is as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

The combined steam and leaf plot is:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

8 0
3 years ago
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