Answer:
13.95
Explanation:
Given :
Vector A polar coordinates = ( 7, 70° )
Vector B polar coordinates = ( 4, 130° )
To find A . B we will
A ( r , ∅ ) = ( 7, 70 )
A = rcos∅ + rsin∅
therefore ; A = 2.394i + 6.57j
B ( r , ∅ ) = ( 4, 130° )
B = rcos∅ + rsin∅
therefore ; B = -2.57i + 3.06j
Hence ; A .B
( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95
Answer:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
The pressure reduces to 2.588 bars.
Explanation:
According to Bernoulli's theorem for ideal flow we have

Since the losses are neglected thus applying this theorm between upper and lower porion we have

Now by continuity equation we have

Applying the values in the Bernoulli's equation we get
