The relationship between weight and speed is what kind of relationship

Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi

VladimirAG [237]

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

$\eta_{th} = 1 - \frac{T_{L}}{T_{H}}$

Where:

$T_{L}$ - Temperature of the cold reservoir, in K.

$T_{H}$ - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

$\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}$

$\eta_{th} = 0.466 \,(46.6\,\%)$

Source 2

$\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}$

$\eta_{th} = 0.0798 \,(7.98\,\%)$

The power produced by each device is presented below:

Source 1

$\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot W = 85.433\,kW$

Source 2

$\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot W = 146.3\,kW$

The source 2 produces the largest amount of power.

8 0

3 years ago

determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004

Afina-wow [57]

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

6 0

3 years ago

An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box

True [87]

Answer:

Rate of Entropy =210.14 J/K-s

Explanation:

given data:

power delivered to input = 350 hp

power delivered to output = 250 hp

temperature of surface = 180°F

rate of entropy is given as

$Rate\ of\ entropy = \frac{Rate\ of \ heat\ released}{Temperature}$

T = 180°F = 82°C = 355 K

Rate of heat = (350 - 250) hp = 100 hp = 74600 W

Rate of Entropy $= \frac{74600}{355} = 210.14 J/K-s$

8 0

3 years ago

A bar having a length of 5 in. and cross-sectional area of 0.7 i n . 2 is subjected to an axial force of 8000 lb. If the bar str

barxatty [35]

Answer:

E=1.969 × 10¹¹ Pa

Explanation:

The formula to apply is;

E=F*L/A*ΔL

where

E=Young modulus of elasticity

F=Force in newtons

L=Original length in meters,m

A=area in square meters m²

ΔL= Change in length in meters,m

Given

F= 8000 lb = 8000*4.448 =35584 N

L= 5 in = 0.127 m

A= 0.7 in² =0.0004516 m²

ΔL = 0.002 in = 5.08e-5 m

Applying the formula

E=(35584 * 0.127)/(0.0004516*5.08e-5 )

E=1.969 × 10¹¹ Pa

8 0

4 years ago

Foundation dampproofing is most commonly installed: Select one: a. on both the inside and outside of the basement wall b. Baseme

puteri [66]

Answer:A

Explanation:

Damp roof is generally applied at basement level which restrict the movement of moisture through walls and floors. Therefore it could be inside or the outside basement walls.

4 0

3 years ago