There is a picture linked for the question I have if anyone can help me with my whole quiz I’ll try to give 60 points

A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

Furkat [3]

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

5 0

3 years ago

The U.S. Coast Guard requires certain equipment be carried onboard a vessel based on the vessel's length. How must an operator m

nevsk [136]

Answer:

The answer is option B(:

Explanation:

3 0

3 years ago

The ignition temperature of 87-octane gasoline vapor is about 430 ∘C and, assuming that the working gas is diatomic and enters t

Flura [38]

Answer:

I have solved the problem below. I hope it will let you clear the concept.

For any inquiries ask me in the comments.

Explanation:

6 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$