The engine of a 1500-kg automobile has a power rat- ing of 75 kW. Determine the time required to accelerate this car from rest t
1 answer:
Answer:
7.72 seconds
The answer is not realistic as the time is very less to reach to a speed of 100 km/hr
also, we have not taken other factors into consideration like wind drag etc.
Explanation:
Data provided in the question:
Mass of the engine = 1500 kg
Power rating = 75 kW = 75,000 W
Initial speed, v₁ = 0
Final speed = 100 km/hr = = 27.78 m/s
Now,
Power = Work done ÷ Time
also,
Work done = Final energy - Initial energy
=
=
= 578703.70 J
thus,
75,000 = 578703.70 ÷ time
or
time = 7.716 s ≈ 7.72 seconds
The answer is not realistic as the time is very less to reach to a speed of 100 km/hr
also, we have not taken other factors into consideration like wind drag etc.
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Answer:
110 m or 11,000 cm
Explanation:
- let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
- let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
- let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively
M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s
C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c
<u>Using effectiveness-NUT method</u>
- <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>
C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c
C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c
From the result above cold fluid heat capacity rate is smaller
Dimensionless heat capacity rate, C = minimum capacity/maximum capacity
C= C<em>min</em>/C<em>max</em>
C = 5.016/8.62 = 0.582
.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>
Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)
Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW
.<em>3 Third, we determine the actual heat transfer rate, Q</em>
Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)
Q = (5.016 kW/°c)(80 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW
.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε
ε<em> </em>= Q/Qmax
ε <em>= </em>303.66 kW/702.24 kW
ε = 0.432
.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>
NTU =
NTU =
NTU = 0.661
<em>.6 sixth, we determine Heat Exchanger surface area, As</em>
From the question, the overall heat transfer coefficient U = 640 W/m²
As =
As =
As = 5.18 m²
<em>.7 Finally, we determine the length of the heat exchanger, L</em>
L =
L =
L= 109.91 m
L ≅ 110 m = 11,000 cm