Theoretical maximum CoP of a vapour absorption system where, TG= Temperature of generator, TE=Temperature of evaporator, TO= Environmental temperature is:
-
Solution
Theoretical maximum CoP of vapour absorption system,
(CoP)max=\(\frac{T_{E}}{T_{G}}\left ( \frac{T_{G}-T_{O}}{T_{O}-T_{E}} \right )\)
6 kj of heat transfer has to takes place in 10 min from one end to other end of a metal cylinder of 10 cm2 area of cross-section,length 1m and thermal conductivity as 100 w/mk. The temperature difference between the two ends of cylinder bar will be______.
-
Solution
Given: Heat transfer (Q) = 6 kg = 6 × 103J
thermal conductivity (k) = 100 w/mk
Area of cross section (A) = 10 cm2
Length (Δx) =1m
Time duration = 10 min = 10 × 60 = 600 seconds
Rate of heat transfer q(rate)=\(\frac{6\times 10^{3}}{600}= 10J/S\)
Now,using formula,
Thermal conductivity (k) =\(\frac{^{q}rate^{\Delta x}}{\Delta T\times A}\)
100 =\(\frac{10\times 1}{\Delta T\times 10\times 10^{-4}}\Rightarrow \Delta T=\frac{10}{1000\times 10^{-4}}= 100°C\)
The air with enthalpy of 200kJ/kg is compressed by an air compressor to a pressure and y temperature at which its enthalpy becomes 400 kJ/kg. The loss of heat is 80kJ/kg from the compressor as the air passes through it. Neglecting K.E and P.E, the power required for an air mass flow of 0.5 kJ/s is _______.
-
Solution
Writing steady flow energy equation,
h1+ q= h2 + w
where, h1 and h2= enthalpies initially and finally respectively.
q = heat flow
w = work
Given: h1= 200 kJ/kg,h2= 400 kJ/kg,Q = –80 kJ/kg (loss).
200 – 80 = 400 +w
w = 120 – 400 = – 280 kJ/kg.
Power required for an air mass flow= –280 × 0.5 = –140 KW≡140 KW
The deflection of the spring after suspending mass of 25kg from a spring is 24mm. If the motion is aperiodic at a speed of 2 mm/s and maximum disturbing force and frequency of vibration of force acting on a body are 200 N and 124 z respectively. Then the amplitude of ultimate motion will be______.
-
Solution
Given: Mass of body (m) = 25kg
deflection of spring (dx) = 24mm
Disturbing force (F) = 200N
frequency (f) = 124z
using formula, f(x) =\(\frac{1}{2\pi }\sqrt{\frac{g}{\delta _{x}}}=\frac{1}{2\pi }\sqrt{\frac{10}{24\times 10^{-3}}}=3.25Hz\)
If the motion is aperiodic, then,w = wx
\(\sqrt{\frac{g}{\delta _{x}}}=\sqrt{\frac{10}{24\times 10^{-3}}}\)
= 20.41 rad/s
Now,damping coefficient (e)= critical damping coefficient (Cc)
= 2 mwx
= 2× 25 × 20.41
= 1020.5 N/m/s
Amplitude of ultimate motion = (A)
=\(\overset{F}{\sqrt{(x-mw^{2})^{2}+(CW)^{2}}}\)
w = 2πf = 2π× 12 = 75.36 rad/s
f (x) =1⁄2π\(\sqrt{\frac{x}{m}}\Rightarrow 3.25=\frac{1}{2\pi }\sqrt{\frac{x}{25}}\)
x = 10,414.2 N/m
then,
A=\(\overset{200}{\sqrt{\left [ 10414.2-25\times (75.36)^{2} \right ]^{2}+[1020.5\times 75.36]^{2}}}\)
=\(\frac{200}{4.81\times 10^{3}}=\frac{0.2}{4.81}= 0.0416 mm.\)
A particle of mass 4kg is resting on a surface. The force is acting on the particle and the equation of force as a function of time is given below: f = t2 i + (10 t – 12)j + 2.4 t3kN where‘t’ is time in seconds. Then the velocity of particle after 18 seconds will be ______.
-
Solution
Given : Mass = 4kg,
f = t2i + (10t – 12) j + 2.4t3
on integrating the above force equation with respect to time
(t) Impulse of force = ∫f dt=\(\int_{0}^{18}\left \{ t^{2}i+(10t-12)j+2.4t^{3} \right \}dt\)
=\(\left [ \frac{t^{3}}{3}i+\left ( 5t^{2}-12t \right )j+0.6t^{4} \right ]_{0}^{18}dt\)
= 1944i + 1402 j +62985.6 kNS
Now,using impulse-momentum equation,
1944i + 1402j + 62985.6 = mn = 4v
v =1⁄4[1944i + 1402j +62985.6]
v =486i + 350.5j + 15746.4
N=\(\sqrt{(486)^{2}+(350.5)^{2}+(15746.4)^{2}}\)
=\(\sqrt{2483.09\times 10^{5}}\)
= 15 × 103 m/s
The Z-transform of \(\left ( \frac{1}{4} \right )^{n}\) (u[n] – u[n– 5])
-
Solution
X(Z) =\(\sum_{n=0}^{4}\left ( \frac{1}{4} z^{-1}\right )^{4}\)
=\(=\frac{1-\left ( \frac{1}{4}Z^{-1} \right )^{5}}{1-\left ( \frac{1}{4}Z^{-1} \right )}\)
=\(=\frac{Z^{5}-(0.25)^{5}}{Z^{5}-(0.25)^{5}},all\, Z\)
The area bounded by the curve y=\(2\sqrt{x}\),x= 1 and x= 4 is __________ .
-
Solution
y2 = 4x
y= – x
x2 = 4x
x= 0, 4\(\int_{1}^{4}\int_{-x}^{2\sqrt{x}}dydx=\int_{1}^{4}[y]_{-x}^{2\sqrt{x}}dx\)
=\(\int_{1}^{4}[2\sqrt{x}+x]dx=\int_{1}^{4}(2x^{1/2}+x)dx\)
=\(\left [ 2\frac{x^{3/2}}{3/2}+\frac{x^{2}}{2} \right ]_{1}^{4}\)
= 16.83
-
Solution
Let f(z) = sinπ z2
\(\frac{1}{(z-2)(z-1)}=\frac{1}{z-2}-\frac{1}{z-1}\)
\(\int \frac{f(z)dz}{(z-2)}-\int \frac{f(z)dz}{(z-1)}\)
z= 1, 2 both lies inside the circle.
∴2π f(2) – 2π f(1)
2πsin 4π– 2πsin π
0 – 0 =0
The real root of the equation xex– 2= 0 correct to three decimal places by Newton’s Raphson method is…………….
-
Solution
Let f (x) = xex – 2
f'(x) = xex + ex
xn+1=xn-\(\frac{f(x_{n})}{f'(x_{n})}\)
xn+1=xn-\(\frac{(x_{n}e^{x_{n}}-2)}{x_{n}e^{x_{n}}+e^{x}}=\frac{x_{n}^{2}e^{x_{n}+2}}{x_{n}e^{x_{n}}+e^{x_{n}}}\)
f (0) < 0 f (1) > 1
So, root lies between 0 and 1.
An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second,third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. The probability that the gun hits the plane is ……………. .
-
Solution
P1= 0.4, P2= 0.3, P3= 0.2, P4= 0.1
P (the gun hits atleast once) = P (the gun hits the plane)
= 1– P (the plane is hit in none of the shots)
= 1 – (1 – P1) (1– P2) (1– P3) (1– P4)
= 1 – (1 – 0.4) (1 – 0.3) (1 – 0.2) (1 – 0.1)
= 1 – (0.6) (0.7) (0.8) (0.9)= 0.6976