Answer:
a) 1253 kJ
b) 714 kJ
c) 946 C
Explanation:
The thermal efficiency is given by this equation
η = L/Q1
Where
η: thermal efficiency
L: useful work
Q1: heat taken from the heat source
Rearranging:
Q1 = L/η
Replacing
Q1 = 539 / 0.43 = 1253 kJ
The first law of thermodynamics states that:
Q = L + ΔU
For a machine working in cycles ΔU is zero between homologous parts of the cycle.
Also we must remember that we count heat entering the system as positiv and heat leaving as negative.
We split the heat on the part that enters and the part that leaves.
Q1 + Q2 = L + 0
Q2 = L - Q1
Q2 = 539 - 1253 = -714 kJ
TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:
η = 1 - T2/T1
T2/T1 = 1 - η
T2 = (1 - η) * T1
The temperatures must be given in absolute scale (1453 C = 1180 K)
T2 = (1 - 0.43) * 1180 = 673 K
673 K = 946 C
True
Suspension is the system of tires, tire air, springs, shock absorbers and linkages that connects a vehicle to its wheels and allows relative motion between the two.[1] Suspension systems must support both road holding/handling and ride quality
Answer: 
Explanation:
Given
Discharge is 
Diameter of pipe 
Distance between two ends of pipe 
friction factor 
Average velocity is given by
Pressure difference is given by
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
The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.
Define work. Explain the rate of doing work.
Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.
Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.
Solution Explained:
Given,
Weight = 1000N and distance = 5m
A/Q, the work here is done in lifting then
Work = (weight) × (distance moved)
= 1000 X 5
= 5000Nm or 5000J = 5kJ
Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.
To learn more about work, use the link given
brainly.com/question/25573309
#SPJ9
<u />
