can someone please define these three vocabulary words for my stem class i will give brainliest if i can figure out how

Air enters a turbine with a stagnation pressure of 900 kPa and a stagnation temperature of 658K, and it is expanded to a stagnat

bezimeni [28]

Answer:

12.332 KW

The positive sign indicates work done by the system ( Turbine )

Explanation:

Stagnation pressure( P1 ) = 900 kPa

Stagnation temperature ( T1 ) = 658K

Expanded stagnation pressure ( P2 ) = 100 kPa

Expansion process is Isentropic, also assume steady state condition

mass flow rate ( m ) = 0.04 kg/s

<u>Calculate the Turbine power </u>

Assuming a steady state condition

( p1 / p2 )^(r-1/r) = ( T1 / T2 )

= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )

= ( 9 )^0.285 = 658 / T2

∴ T2 = 351.22 K

Finally Turbine Power / power developed can be calculated as

Wt = mCp ( T1 - T2 )

= 0.04 * 1.005 ( 658 - 351.22 )

= 12.332 KW

The positive sign indicates work done by the system ( Turbine )

6 0

2 years ago

A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2

lora16 [44]

(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

Z = √[R²+(2πfL)²]........................ Equation 1

Where:

R = resistance of the coil
f = Frequency
L = Inductance of the coil
Z = Overall impedance of the circuit

From the question,

Given:

R = 7 ohms
L = 31.8 mH = 0.0318 H
f = 50 Hz
π = 3.14

Substitute these values into equation 1

Z = √[7²+(2×3.14×50×0.0318)²]
Z = √(49+99.7)
Z = √(148.7)
Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

I = V/Z.................. Equation 2

Where:

V = Voltage
I = current in the circuit.

Given:

V = 230 V.

Substitute into equation 2

I = 230/12.19
I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

∅ = Phase angle.

Substitute into equation 3

∅ = tan⁻¹(2×3.14×50×0.0318/7)
∅ = tan⁻¹(9.9852/7)
∅ = tan⁻¹(1.426)
∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

pf = cos∅............ Equation 4

Where:

pf = power factor.

Substitute the value of ∅ into equation 4

pf = cos(54.97°)
pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

P = V×I×pf................ Equation 5

Where:

P = The power consumed

Substitute the values into equation 5

P = 230(18.87)(0.574)
P = 2491.22 W

Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

5 0

2 years ago

What flow type occurs when the heat exchanger has multiple passes and baffles?

Vanyuwa [196]

I think that the answer would be B or C

4 0

3 years ago

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati

snow_tiger [21]

Answer:

$D=0.41m$

Explanation:

From the question we are told that:

Discharge rate $V_r=0.35 m3/s$

Distance $d=4km$

Elevation of the pumping station $h_p= 140 m$

Elevation of the Exit point $h_e= 150 m$

Generally the Steady Flow Energy Equation SFEE is mathematically given by

$h_p=h_e+h$

With

$P_1-P_2$

And

$V_1=V-2$

Therefore

$h=140-150$

$h=10$

Generally h is give as

$h=\frac{0.5LV^2}{2gD}$

$h=\frac{8Q^2fL}{\pi^2 gD^5}$

Therefore

$10=\frac{8Q^2fL}{\pi^2 gD^5}$

$D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}$

$D=0.41m$

8 0

3 years ago

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

2 years ago