Answer:
In C++:
#include <iostream>
#include <cmath>
using namespace std;
int main(){
double total = 0;
for(int i = -170; i<=-130;i++){
if(i%2 == 0 && i%7==0){
double angle = i*3.14159/180;
total+=sin(angle); } }
cout<<"Total: "<<total;
return 0; }
Explanation:
This initializes the total to 0
double total = 0;
This iterates from -170 to - 130
for(int i = -170; i<=-130;i++){
This checks if the current number is even and is divided by 7
if(i%2 == 0 && i%7==0){
This converts the number from degrees to radians
double angle = i*3.14159/180;
This adds the sine of the number
total+=sin(angle); } }
This prints the calculated total
cout<<"Total: "<<total;
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at X kPa, and exits at 800 kPa and 60oC. The refrigerant leaves the condenser as saturated liquid at 800 kPa. The number of letters in your first name multiplied by 10 plus 50
Answer:
Hello your question has some missing information below are the missing information
The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump
answer : 2.49
Explanation:
For vapor-compression refrigeration cycle
P1 = P4 ; P1 = 140 kPa
P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa
From pressure table of R 134a refrigerant
h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg
h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )
= 296.8kJ/kg
h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg
also h4 = 95.47 kJ/kg
To determine the coefficient of performance
Cop = ( h1 - h4 ) / ( h2 - h1 )
∴ Cop = 2.49
A chemical process stream enters a shell-and-tube exchanger at a temperature of 200.0°Fand does two passes on the shell side, exiting the exchange at 170.0°F. The Heat exchanger Spring 2021 has 200 stainless steel tubes that are 2-in.ODand 10.0ft long. Indicate whether the temperature of the process stream will increase, decrease, or remain the sameunder the following scenarios. You must justify your answerto receive full credit.a)The flow rate of the cooling fluid is increased.b)There are 200 tubes that are 1.0-in. OD and 20.0ft long.c)The number of shell passes is doubled.d)The tube material is changed to copper.
Answer:
a) Decrease
b) Decrease
c) Decrease
d) Decrease
Explanation:
Ti= 200°F ,
Te = 170°F
Area of heat exchanger = [tex]\pi *(2 )* 10[/tex] = 20π
A) when The flow rate of the cooling fluid is increased
Temperature of process stream will decrease and this is because the tube side heat transfer coefficient will increase and this will increase the rate of heat transfer thereby decreasing the temperature of the process stream.
B) when There are 200 tubes that are 1.0-in. OD and 20.0ft long
The temperature of the process stream will decrease and this is because the heat transfer coefficient will increase likewise the heat transfer rate
C) When The number of shell passes is doubled
This will cause an increase in the overall length of the shell, an increase in velocity of constant volumetric flowrate, hence the Temperature of the process steam will decrease as well
D) When The tube material is changed to copper.
Due to the high thermal conductivity of copper when compared to steel , switching to copper will cause a decrease in the temperature of the process steam
what technology has been used for building super structures
Answer: Advanced technologixal machines
Explanation: such as big cranes, multiple workers helping creat said structure, and big bull dozers
a) The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate in a fluidized bed dryer is 0.1 kg water removed/m2-s, determine the time required for the product to begin the falling-rate drying period. The product has a cube shape with 5-cm sides; the initial product density is 950 kg/m3.
Answer:
≈ 53 seconds
Explanation:
calculate Time required for the product to begin the falling-rate drying period
Initial moisture content = 0.77 kg water /kg of product
= 3.35 kg water /kg solids
Critical moisture content = 0.3 kg water / kg product
= 0.43 kg water / kg solids
∴ amount of water to be removed = 3.35 - 0.43 = 2.95kg water /kg solids
next: calculate surface are a of product during drying
= (0.05 * 0.05 ) * 6
= 0.015 m^3
Drying rate = 0.1 kg water m^2.s^-1 * 0.015 m^3 = 1.5 * 10^-3 kg water s^-1
applying product density
initial product mass = 0.11875 * 0.23 = 0.0273kg solid
hence total amount of water to removed = 2.92 * 0.0273 = 0.07972 kg
therefore : Time required for the product to begin the falling-rate drying period
= 0.07972 / 1.5 * 10^-3
= 53 seconds
An interest rate of norminal 12% per year , compounded weekly is
Answer: It is a nominal rate per year
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in oC, and the quality of the refrigerant at the exit of the expansion valve. Step 1 Determine the temperature of the refrigerant at the exit, in oC.
Answer:
[tex]T_{2}[/tex] = -9.24 °C
x = 0.1057
Explanation:
The tables used in this answer and explanation come from Fundamentals of Engineering Thermodynamics 9th Edition.
Using Table A-14: Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table and the given [tex]P_{2}[/tex], [tex]T_{2}[/tex] can be determined by finding the temperature that corresponds with [tex]P_{2}[/tex] on the table. In this case, [tex]T_{2}[/tex] = -9.24 °C.
The quality of the refrigerant can be determined by using data from the same table and [tex]h_{2} =274.26[/tex] kJ/kg.
Necessary data (P=3bar):
[tex]h_{f}=137.42[/tex] kJ/kg
[tex]h_{g}=1431.47[/tex] kJ/kg
The formula to calculate quality is [tex]h_{2} =h_{f}+x(h_{g}-h_{f})[/tex].
Rearranging for x:
[tex]x=\frac{h_{2}-h_{f} }{h_{g}-h_{f} }= \frac{274.26-137.42}{1431.47-137.42}=0.1057[/tex]
brainly and points if you want
Answer:
thank you
Explanation:
have a nice day
Answer:
thankd
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other examples of joption
Answer:
Profit from stock price gains with limited risk and lower cost than buying the stock outright.
Profit from stock price drops with limited risk and lower cost than shorting the stock.
Profit from sideways markets by selling options and generating income.
Get paid to buy stock.
Explanation:
I don’t know if you meant option but I hope this helps.
