What do u call a bad bird

To make 1000 containers of ice cream you need: 600 gallons of milk, 275 gallons of cream, and 120 gallons of flavor. Each ingred

Kamila [148]

Answer:

For the cream, 32 gallons should be reduced and 12 gallons should be decreased for flavor.

Explanation:

To prepare a total of 1000 gallons of ice cream you need 600 gallons of milk, 275 gallons of cream and 120 gallons of flavor, therefore we must calculate the percentages of each ingredient, as follows:

%milk=(600/1000)x100=60%

%cream=(275/1000)x100=27.5%

%flavor=(120/1000)x100=12%

If you reduce the amount of milk by 10% you have:

Milk quantity=600 gallons-(600x0.1)=540 gallons

To maintain the same percentages of each ingredient, you must make a rule of three to know the amount of cream and flavor that would need to be used with the 540 gallons of milk. The rule of three is as follows:

540 gallons of milk------------------60%

x gallons of cream--------------------27.5%

Clearing the x:

x gallons of cream=(540x27.5)/60=243 gallons

In the same way for flavor:

540 gallons of milk------------------60%

x gallons of flavor--------------------12%

Clearing the x:

x gallons of flavor=(540x12)/60=108 gallons

Verifying that they meet the percentages that were calculated before:

Total amount of ice cream=540+243+108=891 gallons

Calculate the percentages of each ingredient:

%milk=(540/891)x100=60.6%

%cream=(243/891)x100=27.3%

%flavor=(108/891)x100=12.1%

As can be seen, it is found that approximately the same percentages calculated above are met. Therefore, we can already calculate the amount by which the cream should be reduced and the flavor.

For the cream:

Gallons of cream=275-243=32 gallons

For the flavor:

Gallons of flavor=120-108=12 gallons

3 0

3 years ago

Question Completion Status:

Mashutka [201]

Answer: c. Centre of pressure

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

6 0

2 years ago

Superheated water vapor at a pressure of 20 MPa, a temperature of 500oC, and a flow rate of 10 kg/s is to be brought to a satura

katrin2010 [14]

Answer:

1.96 kg/s.

Explanation:

So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;

=> Superheated water vapor at a pressure = 20 MPa,

=> temperature = 500°C,

=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."

=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."

K1 = 3241.18, k2 = 93.28 and 2725.47.

Therefore, m1 + m2= m3.

10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.

=> 1.96 kg/s.

7 0

3 years ago

An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred b

irinina [24]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

4 0

3 years ago

Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure

iris [78.8K]

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3

P5/P6 = P7/P8 = √9 =3

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg

Calculate the net work done

Wnet = (Wturb)out - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

4 0

3 years ago