If 180 g of potassium Iodide is dissolved in 100 CM three of water at 30°C and blank solution is formed

A cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's ve

Musya8 [376]

Answer:

The surface temperature is 921.95°C .

Explanation:

Given:

a=25 cm ,P=350 hp⇒P=260750 W

Power transmitted $0.95\times 260750$ W and remaining will lost in the form of heat.This heat transmitted to air by the convection.

h=230 $\frac{W}{m^2-K},\eta =0.95$

Actually heat will be transmit by the convection.

In convection Q=hA $\Delta T$

So $P=\Delta T\times Q$

$0.05\times 260750=230\times0.25^2\(T-15)$

T=921.95°C

So the surface temperature is 921.95°C .

6 0

3 years ago

Single point cutting tool removes material from a rotating work piece to generate a cylinder is called • Facing Tuming • Both 1

madam [21]

Answer:Turning

Explanation: Turning is the process in which the work piece is subjected to machining so that excess part is removed with the help of rotation by turning machine or lathe machine.The cutter tool is used for cutting the excess of the work piece and it is mostly single-pointed so that give accurate removal of the excess of work piece.At times , according to the requirement multi-pointed tool is also used Therefore, the correct option is turning.

6 0

3 years ago

) A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging proc

bonufazy [111]

Answer:

it will be a scattered plot

Explanation:

5 0

2 years ago

Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)

vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0

3 years ago

Which option should the engineers focus on as they develop the train in the following scenario?

pav-90 [236]

Answer:

Engineers can design a train with a regenerative braking system

Explanation:

Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.

5 0

3 years ago