1. The heat dissipation at any section of parabolic fin is given by
a) (t2 – t1) (b) (δ)
b) k (t2 – t1) (b) (δ)
c) k (t2 – t1) (δ)
d) k (t2 – t1) (b)
Explanation: Q = qx (A X) = k (t2 – t1) (b) (δ).
2. An air cooled cylindrical wall is to be fitted with triangular fins of 3 cm thickness at base and 12 cm in height. The fins are made from stainless steel with density 8000 kg/m3 and thermal conductivity 17.5 W/m K. The wall temperature is 600 degree Celsius and the fin is exposed to an environment with t a = 30 degree Celsius and h = 20 W/m2 K. What is the temperature distribution along the fin?
a) t = 10 + 250 I 0 [6.056 (x) 1/2].
b) t = 20 + 250 I 0 [6.056 (x) 1/2].
c) t = 30 + 250 I 0 [6.056 (x) 1/2].
d) t = 40 + 250 I 0 [6.056 (x) 1/2].
Explanation: α/α 0 = t – t 0/t 0 – t a = I 0 [2 B (x) ½]/ I 0 [2 B (l) ½]. Here B = (2 h l/k δ) ½ = 3.028.
3. Consider the above problem, make calculations for the rate of heat flow per unit mass of fin material used
a) 126.53 W/kg
b) 154.76 W/kg
c) 134.87 W/kg
d) 165.46 W/kg
Explanation: Q = b (2 h k δ) ½ α 0 I 1 [2 B (L) ½/ I 0 [2 B (L) ½ = 1822 W. Mass of fin per meter width = 14.4 kg. Therefore rate of heat flow per unit mass = 1822/14.4 = 126.53 W/kg.
4. The time constant of a thermocouple is the time taken to
a) Minimum time taken to record a temperature reading
b) Attain 50% of initial temperature difference
c) Attain the final value to be measured
d) Attain 63.2% of the value of the initial temperature difference
Explanation: The time constant of a thermocouple represents the time required to attain 63.2% value.
5. A thermocouple junction of spherical form is to be used to measure the temperature of the gas stream. The junction is at 20 degree Celsius and is placed in a gas stream which is at 200 degree Celsius. Make calculations for junction diameter needed for the thermocouple to have thermal time constant of one second. Assume the thermos-physical properties as given below
k = 20 W/ m K
h = 350 W/m2 K
c = 0.4 k J/kg K
p = 8000 kg/m3
a) 0.556 mm
b) 0.656 mm
c) 0.756 mm
d) 0.856 mm
Explanation: T = p V c/h A = p r c/3h. So, r = 3 h T/p c = 0.000328 m = 0.328 m.
6. A low value of time constant can be achieved for a thermocouple by
(i) Increasing the wire diameter
(ii) Increasing the value heat transfer coefficient
(iii) Use light metals of low density and low specific heat
a) ii and iii
b) i and iii
c) i and ii
d) i, ii and iii
Explanation: Diameter of wire should be less.
7. Which of the following has units of time constant? (Where, P is density, A is area, c is specific heat and V is volume)?
a) p V/h A
b) p c/h A
c) p V c/h A
d) V c/h A
Explanation: It has the unit of time and is time constant of the system.
8. Thermal radiation suffers no attenuation in a vacuum”
a) true
b) false
Explanation: It is gradual loss of intensity of any kind of flux.
9. How does the body temperature falls or rises with time?
a) Logarithmic
b) Parabolic
c) Linear
d) Exponentially
Explanation: The rate depends on the parameter h A/p V c.
10. The lumped parameter solution for transient conduction can be conveniently stated as
a) t – t a/t I – t a = 2 exponential (- B I F 0)
b) t – t a/t I – t a = exponential (- B I F 0)
c) t – t a/t I – t a = 3 exponential (- B I F 0)
d) t – t a/t I – t a = 6 exponential (- B I F 0)
Explanation: This is the general solution for lumped system parameter.