Answer:
Explanation:
The detailed steps and appropriate calculation with analysis is as shown in the attachment.
Answer Explanation:
the efficiency of the the engine is given by=1-
where T₂= lower temperature
T₁= Higher temperature
we have given efficiency =70%
lower temperature T₂=27°C=273+27=300K
higher temperature T₁=627°C=273+627=900K
efficiency=1-
=1-
=1-0.3333
=0.6666
=66%
66% is less than 70% so so inventor claim is wrong
Answer:Counter,
0.799,
1.921
Explanation:
Given data




Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger
Equating Heat exchange
![m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]](https://tex.z-dn.net/?f=m_hc_%7Bph%7D%5Cleft%20%5B%20T_%7Bh_i%7D-T_%7Bh_o%7D%5Cright%20%5D%3Dm_cc_%7Bpc%7D%5Cleft%20%5B%20T_%7Bc_o%7D-T_%7Bc_i%7D%5Cright%20%5D)
=
we can see that heat capacity of hot fluid is minimum
Also from energy balance

=


NTU=1.921





Answer:
Fuel efficiency for highway = 114.08 miles/gallon
Fuel efficiency for city = 98.79 miles/gallon
Explanation:
1 gallon = 3.7854 litres
1 mile = 1.6093 km
Let's first convert the efficiency to km/gallon:
48.5 km/litre = (48.5 * 3.7854) km/gallon
48.5 km/litre = 183.5919 km/gallon (highway)
42.0 km/litre = (42.0 * 3.7854) km/gallon
42.0 km/litre = 158.9868 km/gallon (city)
Next, we convert these to miles/gallon:
183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon
183.5919 km/gallon = 114.08 miles/gallon (highway)
158.9868 km/gallon = (158.9868 /1.6093) miles/gallon
158.9868 km/gallon = 98.79 miles/gallon (city)
Answer:
11.2mm or 0.45in
Explanation:
The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.
Please go through the attached file for a step by step solution to this question.